Download Chapter 4 Compact Topological Spaces

Chapter 4
Compact Topological Spaces
4.1 Compact Spaces and Related Results
Definition 4.1.1. A subset K of a topological space (X, J ) is said to be a compact
set if A is a collection of open sets in X such that K ⊆ ∪ A then there exists
n
A∈A
n ∈ N and A1 , A2 , A3 , . . . , An ∈ A such that K ⊆ ∪ Ai . That is K is a compact
i=1
subset of a topological space (X, J ) if and only if A is any open cover for K implies
A has a finite subcollection say Af that will also cover K.
Note. If A is a collection of open sets in (X, J ) and K ⊆ X is such that K ⊆ ∪ A,
A∈A
then we say that A is an open cover for K.
?
Example 4.1.2. Let X be nonempty set and J be a topology on X. Let K be a
finite subset of X.
Case 1: K = φ.
Then verify that K is a compact set (exercise).
Case 2: K is a nonempty finite set.
In this case, there exists n ∈ N and x1 , x2 , . . . , xn ∈ X such that K = {x1 , x2 , . . . , xn }.
Now suppose A is a collection of open sets in X and {x1 , x2 , . . . , xn } = K ⊆ ∪ A.
A∈A
Then for each i ∈ {1, 2, . . . , n}, xi ∈ Ai for some Ai ∈ A . (Note that i 6= j need not
imply Ai 6= Aj .) Now Af = {A1 , A2 , . . . , An } is a finite subcollection of A such that
K = {x1 , x2 , . . . , xn } ⊆ A1 ∪ A2 ∪ · · · ∪ An = ∪ A. That is, we started with an open
A∈Af
77
cover A for K and we could get a finite subcollection Af of A that also covers K.
Hence by the definition, K is a compact subset of (X, J ).
Example 4.1.3. Let X be a nonempty set and Jf = {A ⊆ X : Ac = XKA is a
finite set or Ac = X}. That is, Jf is the cofinite topology on X. We have proved in
example 4.1.2 that every finite subset of any topological space is compact.
So, let us assume that K is an infinite subset of X. Now consider a collection
A of open sets such that K ⊆
∪ A. Since we have assumed K is an infinite
A∈A
set, K 6= φ. Take an element say x0 ∈ K. Now x0 ∈ K ⊆ ∪ A. This implies
A∈A
there exists A0 ∈ A such that x0 ∈ A0 . Now A0 is a nonempty open set in the
cofinite topological space (X, Jf ) implies XKA0 = Ac0 is a nonempty finite set (or
Ac0 = φ ⇒ A0 = X ⇒ Af = {A0 }). Also K ⊆ A0 ⇒ Af = {A0 }. Let K ∩ Ac0 =
{x1 , x2 , . . . , xn }. Since K ⊆ ∪ A each xi ∈ Ai , i = 1, 2 . . . , n. Now K ⊆ X =
A0 ∪ Ac0
⇒K⊆
A∈A
c
(A0 ∪ A0 ) ∩ K = (A0 ∩ K) ∪ (Ac0 ∩ K)
⊆ A0 ∪ A1 ∪ A2 ∪ · · · ∪ An . That
is, Af = {A0 , A1 , A2 , . . . , An } or Af = {A0 } is a finite subcollection of A that also
covers K. Hence K is a compact subset of (X, Jf ). That is, in a cofinite topological
space every subset is a compact set.
Remark 4.1.4. In a cofinite topological space (X, Jf ), if A is a nonempty open set
then A will almost cover any K ⊆ X. That is, maximum finitely many elements may
not be in A and hence every subset K becomes a compact set.
Example 4.1.5. Let X be any set and J be a topology on X. Note that J ⊆ P(X),
the collection of all subsets of X. If J is a finite set then every subset K of X is
compact in (X, J ).
Also note that in a cofinite topological space (X, Jf ) every finite subset of
X is closed and if F 6= X, F is not a finite set then F is not closed. Now let us
78
prove that R with usual topology Js is not a compact space. That is (R, Js ) is not
a compact space. Now we want to prove that the subset R of the topological space
∞
(R, Js ) is not compact. Note that R ⊆ ∪ (−n, n). That is, A = {(−n, n) :
n=1
n ∈ N} is an open cover for R. Suppose this open cover has a finite subcollection
k
say Af = {A1 , A2 , . . . , , Ak } such that R ⊆ ∪ Ai . If Ai ∈ A this implies there exists
i=1
ni ∈ N such that Ai = (−ni , ni ). So, R ⊆ (−n1 , n1 ) ∪ (−n2 , n2 ) ∪ · · · ∪ (−nk , nk ). Let
n0 = max{n1 , n2 , . . . , nk }. Then R ⊆ (−n0 , n0 ), a contradiction. Note that n0 +1 ∈ R
but n0 + 1 ∈
/ (−n0 , n0 ). We could arrive at this contradiction by assuming that A
has a finite subcollection that also covers R. Hence such an assumption is wrong.
That is, this particular collection A = {(−n, n) : n ∈ N} is an open cover for R. But
this cannot have any finite subcover. Therefore (R, Js ) is not a compact space. Note
that R with cofinite topology Jf is a compact space. That is, (R, Jf ) is a compact
topological space but (R, Js ) is not a compact topological space.
Also note that if A is any unbounded subset of R then A is not a compact
∞
subset of the topological space (R, Js ). Note that A ⊆ ∪ (−n, n) but A is not a
n=1
bounded set (that is A is unbounded set) implies there cannot exist any n0 ∈ N such
that A ⊆ (−n0 , n0 ). (Recall: A is a bounded subset of R if and only if there exists
n0 ∈ N such that |x| < n0 ∀x ∈ A. That is A is bounded if and only if A ⊆ (−n0 , n0 )
for some n0 ∈ N.)
Now let us prove that a closed subset of a compact topological space is
compact.
Theorem 4.1.6. If A is a closed subset of a compact topological space (X, J ) then
A is a compact set in (X, J ).
79
Proof. Let A be a collection of open sets in X such that A ⊆ ∪ B. Now A 0 =
B∈ A
c
A ∪ {A } is a collection of open sets such that X = A ∪ A ⊆ A ∪ ( ∪ B). That
c
c
B∈A
is, A is an open cover for the compact space (X, J ) and hence there exists n ∈ N
0
and A1 , A2 , . . . , An ∈ A 0 such that X ⊆ A1 ∪ A2 ∪ · · · ∪ An . If one of Ai = Ac , then
{A1 , A2 , . . . , Ai−1 , Ai+1 , . . . , An } is a finite subcover of A that also covers A. If none
n
of Ai = Ac then {A1 , A2 , . . . , An } ⊆ A such that A ⊆ ∪ Ai . Hence in any case every
i=1
open cover A of A has a finite subcollection that also covers A. This implies that A
is a compact subset of (X, J ).
Recall that a topological space (X, J ) is said to be a Hausdorff topological
space if x, y ∈ X, x 6= y then there exist open sets U , V in X such that x ∈ U, y ∈ V
and U ∩ V = φ.
In a cofinite topological space we have proved that every subset is compact.
In particular (X, Jf ) is a compact topological space for any set X. But if A 6= X is
an infinite subset of X then A is a compact set but it is not a closed set. Now let us
prove that such a thing cannot happen in a Hausdorff topological space.
Theorem 4.1.7. If K is a compact subset of a Hausdorff topological space then K is
a closed set.
Proof. Let us prove that K c = XKK is an open set. So take x ∈ K c . Our aim is to
prove that x is an interior point of K c . That is, we will have to find an open set Ux in
X such that x ∈ Ux ⊆ K c . Now x 6= y for each y ∈ K. Hence (X, J ) is a Hausdorff
space implies there exist open sets Uy , Vy such that
x ∈ Uy ,
y ∈ Vy ,
80
Uy ∩ Vy = φ.
(4.1)
Now {Vy : y ∈ K} is an open cover for the compact set K. Hence this implies there
n
n
i=1
i=1
n
exist y1 , y2 , . . . , yn ∈ K such that K ⊆ ∪ Vy i . Let Ux = ∩ Uy i (refer Eq. (4.1))
n
then Ux is an open set containing x and Ux ∩ K ⊆ Ux ∩ ( ∪ Vy i ) = ∪ (Ux ∩ Vy i ) ⊆
i=1
n
i=1
c
∪ (Uy i ∩ Vy i ) = φ. This implies Ux ∩ K = φ ⇒ Ux ⊆ K . Therefore, each x ∈ K c is
i=1
an interior point of K c . Hence K c is an open set and therefore K is a closed set. Note. Let (X, J ) be a topological space and Y ⊆ X. Then JY = {A ∩ Y : A ∈ J }
is a topology on Y.
?
So, now it is easy to prove:
Theorem 4.1.8. A subset Y of a topological space (X, J ) is compact if and only if
whenever A is a collection of open sets in (Y, JY ) such that Y = ∪ A then there
n
A∈A
exists n ∈ N and A1 , A2 , . . . , An ∈ A such that Y = ∪ Ai .
i=1
Proof. Let us assume the given hypothesis. That is, assume that whenever A is
a collection of open sets in the topological space (Y, JY ) then this open cover A
for Y has a finite subcover. Now we will have to prove that the given subset Y of
the topological space (X, J ) is compact. So start with a collection say B of open
sets in (X, J ) satisfying the condition that Y ⊆ ∪ B (recall the definition). Now
B∈B
B ∈ B ⊆ J ⇒ B ∩ Y ∈ JY . Hence Y ⊆ ∪ B implies Y = ∪ B ∩ Y . That
B∈B
B∈B
is, A = {B ∩ Y : B ∈ B} is a collection of open sets in (Y, JY ) which also covers
Y . Hence by the given hypothesis there exists n ∈ N and B1 , B2 , . . . , Bn ∈ B such
n
n
i=1
i=1
that Y = ∪ (Bi ∩ Y ). This implies that Y ⊆ ∪ Bi . Now we have proved: whenever
B is a collection of open sets in (X, J ) which covers Y then there exists n ∈ N and
n
B1 , B2 , . . . , Bn ∈ B such that Y ⊆ ∪ Bi . Therefore by our definition the given subset
i=1
Y of (X, J ) is a compact set. The proof of Y is a compact subset of (X, J ) implies
81
that the given hypothesis is satisfied follows in similar lines and hence the proof is
left as an exercise.
From what we have proved, we observe that a subset Y of a topological
space (X, J ) is compact if and only if, with respect to the induced topology JY ,
the topological space (Y, JY ) is compact.
Now let us prove that continuous image of a compact space is compact.
Theorem 4.1.9. Let (X, J ) be a compact topological space and (Y , J 0 ) be any other
topological space. Let f : (X, J ) → (Y , J 0 ) be a continuous function. Then the
image f(X) is a compact subset of Y.
Proof. To prove the subset f (X) of (Y, J 0 ) is a compact set, we start with a collection
say A of open sets in (Y, J 0 ) which satisfies
f (X) ⊆ ∪ A.
(4.2)
A∈A
Now A ∈ A implies A is open in (Y, J 0 ).
Hence f : (X, J ) → (Y, J 0 ) is
a continuous function implies that f −1 (A) is open in (X, J ). From Eq. (4.2)
−1
−1
X = f (f (X)) ⊆ f
∪ A = ∪ f −1 (A). This implies A0 = {f −1 (A) : A ∈ A }
A∈A
A∈A
is an open cover for the compact topological space (X, J ). Hence there exists n ∈ N
n
n
and A1 , A2 , . . . , An ∈ A such that X ⊆ ∪ f −1 (Ai ) (here X = ∪ f −1 (Ai )). This
i=1
i=1
n
n
n
−1
−1
implies that f (X) = f ∪ f (Ai ) = ∪ f (f (Ai )) ⊆ ∪ Ai . We have proved:
i=1
i=1
i=1
any arbitrary open cover A of f (X) has a finite subcover. Hence by the definition,
f (X) is a compact subset of (Y, J 0 ).
Using the above theorem and the result that every compact subset of a
Hausdorff space is closed we prove:
82
Theorem 4.1.10. Let (X, J ) be a compact topological space and (Y, J 0 ) be a
Hausdorff topological space. Let f : (X, J ) → (Y, J 0 ) be a bijective continuous
map. Then the inverse map f −1 : (Y, J 0 ) → (X, J ) is also a continuous map. That
is f is a homeomorphism.
Proof. Take an open set A in (X, J ). Now f is a bijective map implies that
(f −1 )−1 (Ac ) = f (Ac )
(4.3)
Note that Ac is a closed subset of the compact space implies Ac is a compact set
implies f (Ac ) is a compact subset of the Hausdorff space. This implies f (Ac ) is a
closed set in Y .
Hence from Eq. (4.3) (f −1 )−1 (Ac ) is a
−1
f (A) = (f −1 )−1 (A) = Y K (f −1 )
closed set implies
(Ac ) is an open set.
We have proved: A is an open set in (X, J ) implies (f −1 )−1 (A) is an open set
in Y . Hence f −1 : (Y, J 0 ) → (X, J ) is a continuous map.
Remark 4.1.11. To prove the above theorem it is also enough to prove that if B is
a closed subset of X then f (B) is a closed subset of Y .
Definition 4.1.12. A collection F of subsets of a given set X is said to have
finite intersection property (f.i.p) if for any n ∈ N and F1 , F2 , . . . , Fn ∈ F
n
then ∩ Fi 6= φ.
i=1
Theorem 4.1.13. A topological space (X, J ) is a compact space if and only if
whenever F is a collection of closed subsets of X which has f.i.p then ∩ F 6= φ.
F ∈F
Proof. Assume that (X, J ) is a compact topological space.
collection F of closed subsets of X which has the f.i.p.
Now start with a
Our aim is to prove
∩ F 6= φ. To achieve this, let us use the method of proof by contradiction.
F ∈F
83
c
Suppose ∩ F = φ. Then by the DeMorgan’s law
F ∈F
c
∩ F
F ∈F
= ∪ F c = X. This
F ∈F
implies {F : F ∈ F } is an open cover for the compact space. Hence there exists
n
c
n
n
n ∈ N and F1 , F2 , . . . , Fn ∈ F such that X = ∪ Fic ⇒ X c = ∪ Fic = ∩ Fi .
i=1
n
i=1
i=1
Therefore ∩ Fi = φ, a contradiction to the fact that F has the finite intersection
i=1
property. We arrived at this contradiction by assuming that ∩ F = φ. Hence this
F ∈F
is not a valid assumption. This implies ∩ F 6= φ. Let us leave the converse part as
F ∈F
an exercise.
Now we prove that real valued continuous function on a compact topological
space attains its maximum.
Theorem 4.1.14. Let (X, J ) be a compact topological space and Js be the usual
topology on R. Let f : (X, J ) → (R, Js ) be a continuous function. Then there exists
x0 ∈ X such that f (x) ≤ f (x0 ) for all x ∈ X. That is f attains its maximum at x0 .
Proof. Let us use the method of proof by contradiction. Then for a given a ∈ X, f
cannot attains its maximum at a. Hence there exists a0 ∈ X such that f (a) < f (a0 ).
This means that f (a) ∈ (−∞, f (a0 )). (Fix any one a0 ∈ X satisfying f (a) < f (a0 ).)
Hence f (X) ⊆ ∪ (−∞, f (a0 )).
a∈X
f(a')
f(a)
Figure 4.1
This implies that A = {(−∞, f (a0 )) : a ∈ X} is an open cover for the compact
subspace f (X) of R (continuous image of a compact space is compact). Hence there
n
exist a1 , a2 , . . . , an ∈ X such that f (X) ⊆ ∪ (−∞, f (a0i )). This implies f (X) ⊆
i=1
(−∞, f (a0 )) for some a0 ∈
{a01 , a02 , . . . , a0n }.
84
Hence for this a0 ∈ X by our assumption
there exists a a0 ∈ X such that f (a0 ) < f (a00 ). But f (X) ⊆ (−∞, f (a0 )) implies
f (a00 ) < f (a0 ). Hence we have got a contradiction. This means f (a) < f (a0 ) cannot
be true for all a ∈ X and hence there should exist at least one x0 ∈ X such that
f (x) ≤ f (x0 ) for all x ∈ X.
Remark 4.1.15. In a similar way we can prove that continuous image of a compact
set attains its minimum at a point y0 ∈ X.
Theorem 4.1.16. Tychonoff. Let X and Y be compact topological spaces. Then
the product space X × Y is compact.
Proof. For each x0 ∈ X, y → (x0 , y) is a surjective continuous function and Y is a
compact space implies x0 × Y is a compact subset of X × Y. Let A be a collection of
basic open sets such that X × Y =
∪
U ×V ∈A
U × V.
Y
d
For
X = [a, b]
Y = [c, d]
Wx Y
0
c
a
x0
(
(
X
b
Figure 4.2
This implies that x0 ×Y ⊆
∪
U ×V ∈A
U ×V implies there exist U1 ×V1 , U2 ×V2 , · · · , Un ×Vn
such that
x0 × Y ⊆ (U1 × V1 ) ∪ (U2 × V2 ) ∪ · · · ∪ (Un × Vn ).
85
(4.4)
Also if for some i, (Ui × Vi ) ∩ (x0 × Y ) = φ, then we do not require to include such an
Ui ×Vi in our finite subcover {Ui ×Vi }ni=1 . So assume that each (Ui ×Vi )∩(x0 ×Y ) 6= φ.
This in turn implies that xo ∈ Ui , ∀ i = 1, 2, . . . , n and hence x0 ∈ Wx0 =
n
∩ Ui . Now it is clear that Wx0 × Y ⊆ (U1 × V1 ) ∪ (U2 × V2 ) ∪ · · · ∪ (Un × Vn ).
i=1
Consider (x, y) ∈ Wx0 × Y. Then x ∈ Ui for all i and y ∈ Y. Hence from Eq. (4.4),
(x0 , y) ∈ Uj × Vj for some j. This implies (x, y) ∈ Uj × Vj for the same j. That is for
each x0 ∈ X, the tube Wx0 × Y is covered by finitely many members of A.
Now let us prove that X × Y is covered by finitely many such tubes Wx × Y.
Now {Wx : x ∈ X} is an open cover for X. Hence X is a compact space implies there
k
exist x1 , x2 , . . . , xk ∈ X such that X = ∪ Wxi . Now (x, y) ∈ X × Y ⇒ x ∈ Wxi for
i=1
k
some i, 1 ≤ i ≤ k and hence (x, y) ∈ Wxi × Y . This implies that X × Y ⊆ ∪ Wxi × Y
i=1
and hence X × Y is covered by finitely many members of A. This proves that X × Y
is a compact topological space.
4.2 Local Compactness
A Hausdorff topological space (X, J ) is said to be locally compact if and only
if for each x ∈ X and for each open set U containing x there exists an open set V
containing x such that V is compact and V ⊆ U . Now it is easy to prove that a
Hausdorff topological space (X, J ) is locally compact if and only if for each x ∈ X
there exists an open set V such that x ∈ V and V is a compact set in X.
Examples 4.2.1. (i) If X is a compact Hausdorff space then X is locally compact.
(ii) Rn with Euclidean topology is locally compact but not compact. Here n ∈ N
and J is the topology induced by the metric d((x1 , x2 , . . . , xn ), (y1 , y2 , . . . , yn )) =
n
1
P
|xk − yk |2 2 .
k=1
86
Also it is easy to prove that if, for 1 ≤ p ≤ ∞, dp ((x1 , x2 , . . . , xn ), (y1 , y2 , . . . , yn )) =
n
1
P
|xk − yk |p p and d∞ ((x1 , x2 , . . . , xn ), (y1 , y2 , . . . , yn )) = max{|xk − yk | : k =
k=1
1, 2, . . . , n} then dp is a metric on Rn . (Note. Proof of dp (x, y) ≤ dp (x, z) + d(z, y) for
all x, y, z ∈ Rn is not that easy.)
For x = (x1 , x2 , . . . , xn ) ∈ Rn , y = (y1 , y2 , . . . , yn ) ∈ Rn , kx + yk ≤ kxk + kyk
is known as Minkowski’s inequality, 1 ≤ p ≤ ∞.
If we use this inequality then dp (x, y) = kx − ykp = k(x − z) + (y − z)kp
≤ kx − zkp + kz − ykp = dp (x, z) + dp (z, y).
Also it is to be noted that the topology Jp on Rn induced by the metric dp is
same as J2 = J .
Definition 4.2.2. A topological space (X, J ) is said to be limit point compact if
every infinite subset of X has a limit point.
Theorem 4.2.3. Every compact topological space (X, J ) is limit point compact.
Proof. Let A be an infinite subset of X. Suppose A0 = φ. That is A does not have
any limit point. Note that A = A ∪ A0 = A implies A is a closed set, then x ∈ A
implies x ∈
/ A0 implies there exists an open set Ux such that x ∈ Ux , Ux ∩ AK{x} = φ.
(That is Ux ∩ A ∩ {x}c = φ ⇒ Ux ∩ A = {x}. Now {Ux : x ∈ A} is an open cover
for the closed subset A of the given compact topological space. Hence there exists a
natural number n and x1 , x2 , . . . , xn ∈ A such that A ⊆ Ux1 ∪· · ·∪Uxn . This gives that
A = (Ux1 ∪ · · · ∪ Uxn ) ∩ A = (Ux1 ∩ A) ∪(Ux2 ∩ A) ∪ · · · ∪ (Uxn ∩ A) = {x1 , x2 , . . . , xn }.
Hence we have arrived at a contradiction by assuming A0 = φ. Therefore A0 6= φ.
That is we have proved that every infinite subset A of the given compact topological
space has at least one limit point. This means that (X, J ) is a limit point compact
topological space.
87
What about the converse of the above theorem? Is every limit point compact
topological space compact? Limit point compact does not imply compact.
Example 4.2.4. Let X = {0, 1}, J = {φ, X} and Y = N = {1, 2, . . .}, the set of all
natural numbers and J 0 = P(N), that J 0 is discrete topology on N. Let X0 = X × Y
be the product space. Here {X × {n}} is an open cover for X × Y . But for any fixed
k
k ∈ N, X × Y = X × N * (X × {1}) ∪ · · · ∪ (X × {k}) (note: (1, k + 1) ∈
/ ∪ X × {j}).
j=1
This gives that X × Y is not a compact topological space.
Now let A be a nonempty subset of X × Y . Then there exists k ∈ N such
that (0, k) ∈ A or (1, k) ∈ A. Let us say (0, k) ∈ A. In this case we claim that
(1, k) ∈ A0 . Take a basic open set U containing (1, k) then U = X × {k}. Now
(0, k) ∈ U ∩ AK{(1, k)} =
6 φ. Hence we have proved that (1, k) is a limit point of A.
Note that if (1, k) ∈ A then we can prove that (0, k) is a limit point of A. So we have
proved that every nonempty subset A of X × Y has a limit point. In particular every
infinite subset of X × Y has a limit point. Therefore X × Y is a limit point compact.
4.3 One Point Compactification of a Topological Space (X, J )
It is given that (X, J ) is a non compact Hausdorff topological space. Our
aim is take an element say ∞ (just a notation) which is not in X. For each x ∈ X,
we have open sets containing x. For ∞ ∈ X ? = X ∪ {∞} we aim to define open sets
satisfying: If U is an open set containing ∞ then each such open set is so large that
the complement of U (with respect to X ∗ ) is rather a small set. That is we want
that X ? KU = C, where C is a compact set in (X, J ) and since ∞ ∈ U , ∞ ∈
/ C. So,
if we start with a collection A of open sets in our new topological space (note: we
have not yet defined such a topology on X ? ) which covers X ? , then ∞ ∈ A0 for some
88
A0 ∈ A. Fix one such A0 . Now X ? KA0 = C
a compact subset of (X, J ).
So, if our proposed topology say J ? on X ? is such that JX? = J then A is also an
open cover for C and C is a compact subspace of (X, J ) implies C is also a compact
subspace of (X ? , J ? ). Hence there exists n ∈ N and A1 , A2 , . . . , An ∈ A such that
C ⊆ A1 ∪ A2 ∪ · · · ∪ An . Therefore X ? = (X ? KA) ∪ A = C ∪ A0 ⊆ A1 ∪ · · · ∪ An ∪ A0 .
Hence every open cover A of (X ? , J ? ) has a finite subcover. So we see that if we
could define such a topology J ? on X ? such that JX? = J then (X ? , J ? ) is a compact
topological space.
We also want to retain the Hausdorff property. So keeping these points in
mind we define J ? as follows:
• if a subset A of X ? is such that ∞ ∈
/ A then A ⊆ X. In such a case A ∈ J ? if
and only if A ∈ J ,
• if ∞ ∈ A then, A ∈ J ∗ if and only if A = X ? KC, for some compact subset
C of X.
Now it is easy to prove that J ? is a topology on X ? .
Theorem 4.3.1. Let (X, J ) be a locally compact Hausdorff space. Then there exists
a topological space (X ? , J ∗ ) satisfying the following conditions:
(i) (X, J ) is a subspace of (X ? , J ? ),
(ii) X ? KX is a set containing exactly one element,
(iii) (X ? , J ? ) is a compact Hausdorff space.
Proof. Keeping the above requirements in mind we have defined J ∗ , we have taken
care that J ⊆ J ? and JX∗ = J . Also X ? KX = {∞}.
89
Now let us prove that (X ? , J ? ) is a compact Hausdorff space. So start with
an collection A of open sets in (X ? , J ? ) (that is A ⊆ J ? ) such that X ? = ∪ A.
A∈A
?
Now ∞ ∈ X implies ∞ ∈ A0 for some A0 ∈ A. It is quite possible that ∞ belongs
to more than one such A ∈ A. From such A just fix one A0 ∈ A. By the definition
of J ∗ , X ? KA0 = C is a compact subset of X. Note that A is a collection of open
sets in X ? and C is a compact subspace of X, and hence of X ∗ . Now C ⊆ ∪ A
A∈A
implies there exists n ∈ N and A1 , A2 , . . . , An ∈ A such that C ⊆ A1 ∪ A2 ∪ · · · ∪ An
implies X ? = (X ? KC) ∪ C ⊆ A0 ∪ A1 ∪ · · · ∪ An (X ? KA0 = C implies A0 = X ? KC)
(it is possible that A0 = Aj , for some j ∈ {1, 2, . . . , n}. So we have proved that
X ? ⊆ A0 ∪ A1 ∪ · · · ∪ An . This means that the started open cover A of X ? has finite
subcover {A0 , A1 , A2 , . . . , An }. Hence (X ? , J ? ) is a compact space.
Now let us prove that (X ? , J ? ) is a Hausdorff space. So start with x, y ∈ X ?
with x 6= y.
Case 1: x, y ∈ X (means x 6= ∞, y 6= ∞).
Now x, y ∈ X, (X, J ) is a Hausdorff topological space implies there exist U, V ∈ J
such that (i) x ∈ U , y ∈ V, (ii) U ∩ V = φ. But J ⊆ J ? . Hence we have U, V ∈ J ?
satisfying (i) and (ii) and this is what we wanted to prove.
Case 2: x ∈ X, y = ∞.
Here we require the fact that (X, J ) is locally compact space. Now x ∈ X, (X, J )
is locally compact Hausdorff space implies there exists an open set U containing x
such that C = U is a compact subset of (X, J ) (here U is the closure of U in X).
Hence by the definition of J ? , V = X ? KC is an open set containing ∞ and U is an
open set containing x. Further U ∩ V = φ and this is what we wanted to prove.
90
Remark 4.3.2. It is easy to prove that ∞ is a limit point of X. To prove this, start
with an open set U containing ∞. Then we have to prove that U ∩ X 6= φ. If X
is not a compact space, then U ∩ XK{∞} = U ∩ X 6= φ. Hence if (X, J ) is not a
compact Hausdorff space then ∞ is a limit point of X.
Examples 4.3.3. (i) Take X = (a, b] and consider X as a subspace of R here
a, b ∈ R, a < b). Now X is considered as a subspace of R, is a locally compact
Hausdorff space of R. Also X is not a compact space.
What is the one point compactification of X ? Here our X = (a, b] and a ∈
/ X.
So take ∞ = a. Note that while defining the one point compactification of X we
just took an object or (say an element) which we denoted by ∞ and ∞ ∈
/ X. So
what we need is ∞ ∈
/ X. Now what are the open sets containing our ∞ = a in X ? .
U ⊆ X ? is an open set containing a if and only if X ? KU = C is a compact subset of
X. Now it is easy to prove that (X ? , J ∗ ) is homeomorphic to [a, b] as f (x) = x for
all x ∈ X ∗ . Now let us prove that f is a homeomorphism. Here it is enough to prove
that f is a continuous map. So start with a nonempty open set U in [a, b] (here [a, b]
is considered as a subspace of R).
Case 1: a ∈
/ U.
Then U ⊆ (a, b]. In this case by our definition U is open in X ? . That is f −1 (U ) = U
is an open set in X ? .
Case 2: a ∈ U .
It is enough to consider a basic open set containing a. Hence U = [a, a + ) for some
0 < < b − a. Is f −1 (U ) = f −1 ([a, a + )) is an open set in (X ∗ , J ∗ ). Note that U is
an open set containing a if and only if X ? KU = [a, b]K[a, a + ) is a compact subset
of X(X = (a, b]). In our case X ? KU = [a + , b] which is a compact subset of (a, b].
91
Hence from our definition of J ∗ , f −1 (U ) is an open set in X ∗ . So from cases 1 and 2
we see that f is a continuous map. Now f : (X ? J ? ) → [a, b] such that
• f is bijective and continuous.
• (X ? , J ? ) is a compact space.
• [a, b] is Hausdorff space implies f is a homeomorphism.
Hence we have proved that there is a homeomorphism between the one-point
compactification of (X ? J ? ) and the compact Hausdorff space [a, b] and our X = (a, b]
is such that X is proper subspace of [a, b] whose closure equals [a, b]. In such a case
we say that [a, b] is a compactification of X.) So we define compactification of a
topological space (X, J ) as follows:
Definition 4.3.4. A compact Hausdorff topological space (Y , J ∗ ) is said to be a
compactification of a topological space (X, J ) if and only if
(i) (X, J ) is a proper subspace of (Y , J ? ),
(ii) X = Y .
N
x
x
f(N)=∞
s
f (x)
f (s)
Figure 4.3
92
f (x)
Note.
If Y KX is a single point then we say that (Y, J ∗ ) is the one point
compactification of (X, J ). Now it is easy to prove the following statements:
(i) the one point compactification of R is homeomorphic to the unit circle
S 1 = {(x1 , x2 ) ∈ R2 : x21 + x22 = 1},
(ii) the one point compactification of R2 is homeomorphic
to the sphere
S 2 = {(x1 , x2 , x3 ) ∈ R2 : x21 +x22 +x23 = 1}. So if we identify R2 with the complex plane
(there is a homeomorphism between R2 and C) then the one point compactification
C ∪ {∞} of C is known as the Riemann sphere or the extended complex plane.
?
4.4 Tychonoff Theorem for Product Spaces
Now let us prove that if (Xα , Jα ), α ∈ J is an arbitrary collection of compact
topological spaces then the product space Π Xα is also a compact topological space.
α∈J
This theorem is due to Tychonoff and different proofs are available in the literature.
To prove Tychonoff theorem we will use Zorn’s lemma. Let us recall the following:
Definition 4.4.1. Let X be a nonempty set and R ⊆ X × X, that is R is a relation
on X. If (x, y) ∈ R then we say that x is related to y and write x ≤ y. The pair
(X, R) is said to be a partially ordered set if and only if
(i) x ≤ x (≤ is a reflexive),
(ii) for x, y ∈ X, x ≤ y and y ≤ x ⇒ x = y. (That is ≤ is against symmetry in the
sense that x ≤ y and y ≤ x can happen only when x = y.) In this case we say that
≤ is antisymmetry,
(iii) for x, y, z ∈ X x ≤ y and y ≤ z ⇒ x ≤ z. (≤ is transitive.)
In this case we say that (X, ≤) is a partially ordered set (PO set).
93
Definition 4.4.2. Let (X, ≤) be a partially ordered set and A be a nonempty subset
of X. Then an element x ∈ X (note: x need not be in A) is called an upper bound
of A if and only if a ≤ x for all a ∈ A. An element y ∈ Y is called a lower bound of
A if and only if y ≤ a for all a ∈ A. If there exists an x0 ∈ X such that (i) x0 is an
upper bound of A, (ii) x ∈ X is an upper bound of A implies x0 ≤ x then such an
upper bound x0 is called the least upper bound (lub) of A and we can easily show
that l.u.b of A is unique, when it exists. An element x0 ∈ X is called the greatest
lower bound (glb) of A if it satisfies the following: (i) x0 is a lower bound of A, (ii)
if y0 ∈ X is a lower bound of A implies y0 ≤ x0 .
Definition 4.4.3. An element x0 ∈ X of a partially ordered set is called a maximal
element of X if x ∈ X is such that x0 ≤ x then x = x0 . An element y0 ∈ X is called
a minimal element of X if y ∈ X is such that y ≤ y0 then y = y0 .
Example 4.4.4. Let X = {1, 2, 3, 4, 5}, R = {(1, 2), (3, 4), (n, n) : n ∈ {1, 2, 3, 4, 5}}.
If (x, y) ∈ R then we say that x ≤ y. Here 2,4, 5 ∈ X and they are maximal elements
of X. Note that (2,3) ∈
/ R and hence 2 is not related to 3. That is 2 ≤ 3 is not true.
Similarly 2 is not related to 4 and 2 is not related to 5. So 2 is not smaller than other
elements of X and hence 2 is a maximal element of X. Since 3 ≤ 4 and 3 6= 4, 3 is not
maximal element of X. If y0 ∈ X is such that y0 is not larger than any other element
of X then we say that y0 is a minimal element of X. That is if there exists y ∈ X
such that y ≤ y0 then y = y0 .
A nonempty subset A of X is said to be a chain (also known as totally ordered
set) if for x, y ∈ A, x ≤ y or y ≤ x. That is any pair of elements x, y in A are
comparable.
Now we are in a position to state Zorn’s lemma:
94
Lemma 4.4.5. Zorn’s Lemma. Let (X, ≤) be a partially ordered set. Further
suppose every chain C ⊆ X has an upper bound in X. Then X will have at least one
maximal element.
We observe the following: A topological space (X, J ) is compact if and only
if whenever A is a collection of subsets of X which has finite intersection property
(f.i.p) then ∩ A 6= φ.
A∈A
Theorem 4.4.6. Tychonoff theorem. Let (Xα , Jα ), α ∈ J be a collection of
compact topological spaces. Then the product space ( Π Xα , J ) is also a compact
α∈J
space.
Proof. Start with a collection A of subsets of X = Π Xα which has f.i.p. Then we
α∈J
aim to prove that ∩ A 6= φ.
A∈A
Step 1:
Let F = {D : D is a collection of subsets of X containing A and D has f.i.p }.
For D1 , D2 ∈ F, define D1 ≤ D2 if D1 ⊆ D2 . Then (F, ≤) is a partially ordered
set. Now let C be a chain in F and A0 = ∪ D (here C ⊆ F and D ∈ F). It is
D∈C
easy to prove that A0 is an upper bound for C. For this, we will have to prove that
A0 ∈ F and D ≤ A0 for all D ∈ C. First let us prove that A0 has f.i.p. Let Aj ∈ A0
for j = 1, 2, . . . , n. Then Aj ∈ Dj , for some D1 , D2 , . . . , Dn ∈ C. As C is a chain for
j ∈ {1, 2, . . . , n} either Di ⊆ Dj or Dj ⊆ Di . Hence there exists k, 1 ≤ k ≤ n such
that Dj ⊆ Dk for all j ∈ {1, 2, . . . , n}. Then Aj ∈ Dk for all j and Dk has f.i.p implies
n
∩ Aj 6= φ. Also A ⊆ A0 . Hence A0 ⊆ F. By the definition of A0 , D ⊆ A0 for all
j=1
D ⊆ C. This proves that A0 ∈ F is an upper bound for C.
Now we have proved that every chain C in F has an upper bound in F.
Therefore by Zorn’s lemma F will have a maximal element say B ∈ F. This B ∈ F is
95
such that (i) A ⊆ B, B has f.i.p, (ii) whenever A0 is a collection of subsets of X such
that A ⊆ A0 , A0 has f.i.p then A0 ⊆ B.
Step 2: Now let us prove that B has the following properties:
(i) For n ∈ N, A1 , A2 , . . . , An ∈ B implies A1 ∩ A2 ∩ · · · ∩ An ∈ B.
(ii) If A is subset of X such that A ∩ B 6= φ, for all B ∈ B then A ∈ B.
To prove (i), let A0 = A1 ∩ A2 ∩ · · · ∩ An and B0 = B ∪ {A0 }. Then B0 ∈ F
and B ⊆ B0 . Since B is maximal, B = B0 . This proves that A0 ∈ B.
To prove (ii), take B0 = B ∪ {A}. Then B0 ∈ F and hence by step 1, A ∈ B.
Step 3: Let us prove that ∩ A 6= φ.
A∈B
For each α ∈ J, {Pα (A) : A ∈ B} is a collection of subsets of (Xα , Jα ). If
A1 , A2 , . . . , An ∈ B,
then B
has f.i.p
n
n
j=1
i=1
and ∩ Aj 6= φ. Let x ∈ ∩ Aj . Now
Pα (x) ∈ Pα (Aj ) for all j = 1, 2, . . . , n. Hence {Pα (A) : A ∈ B} is a collection of
subsets of the compact topological space (Xα , Jα ). Further this collection has f.i.p.
This gives that ∩ Pα (A) 6= φ. Let xα ∈ ∩ Pα (A) and x = (xα )α∈J . (That is,
A∈B
A∈B
we define f : J → ∪ Xα as f (α) = xα ∈ Xα and we identify f with x.) Now
α∈J
we aim to prove that x ∈ A, for each A ∈ B. So fix A ∈ B and let Pβ−1 (Vβ ) be
a subbasic open set containing x. Now x = (xα ) ∈ Pβ−1 (Vβ ) implies xβ ∈ Vβ . We
have xβ ∈ Pβ (A) and hence Vβ is an open set in (Xα , Jα ) containing xβ implies
Vβ ∩ Pβ (A) 6= φ implies there exists y ∈ A such that Pβ (y) ∈ Vβ . This gives that
y ∈ Pβ−1 (Vβ ) ∩ A. Hence Pβ−1 (Vβ ) ∩ A 6= φ for all A ∈ B implies Pβ−1 (Vβ ) ∈ B. Again if
B is a basic open set containing x in the product space (X, J ) then B = Pβ−1
(Vβ1 ) ∩
1
Pβ−1
(Vβ2 ) ∩ · · · ∩ Pβ−1
(Vβn ) for some Vβi ∈ Jβi , i = 1, 2, 3, . . . , n. We have proved that
n
2
each Pβ−1
(Vβi ) ∈ B and hence B ∈ B. Hence whenever B is a basic open set containing
i
x, then B ∩ A 6= φ (A ∈ B) implies x ∈ A, for all A ∈ B implies x ∈ ∩ A 6= φ. Now
A∈B
96
A ⊆ B implies ∩ A 6= φ. That is, whenever A is a collection of closed subsets of
A∈A
the product space (X, J ) and further A has f.i.p then ∩ A 6= φ. This proves that
A∈A
(X, J ) is a compact topological space.
Now let us introduce the notion of a generalized sequence, known as net and
convergence of a net in a topological space.
Let (X, ≤) be a partially ordered set. Further suppose for α, β ∈ X there exist
γ ∈ X such that α ≤ γ and β ≤ γ. Then we say that (X, ≤) is a directed set.
(In the above case if α ≤ γ then we also say γ ≥ α.)
Definition 4.4.7. Let X be a nonempty set and (D, ≤) be a directed set. Then any
function f : D → X is called a net in X. For each α ∈ D, f (α) = xα ∈ X and we
say that {xα }α∈D is a net in X.
Example 4.4.8. Let D = N and ≤ be the usual relation on N. Then (N, ≤) is
a directed set. If X is a nonempty set and f : N → X then for each n ∈ N,
f (n) = xn ∈ X. Hence our net {xn }n∈N is the well known concept namely sequence
in X. In this sense we say that every sequence is a net. Now take D = [0, 1]. Then
(D, ≤) is also a directed set. Define f : [0, 1] → R as f (α) = α + 3, ∀ α ∈ [0, 1]. Here
f = {f (α)}α∈D = {α + 3}α∈[0,1] is a net (generalized sequence) in R.
It is intuitively clear that the net {α + 3}α∈[0,1] approaches to 4. What do we
mean by saying that the net {xα }α∈D approaches to 4? Can we also say that the net
{xα }α∈D approaches 3? Well, in R consider a sequence {xn }n∈N = {xn }∞
n=1 . We know
that lim xn = x (ı.e xn → x as n → ∞) if and only if for each > 0 there exists
n→∞
n0 ∈ N such that xn ∈ (x − , x + ) for all n ≥ n0 . Note that xn → x as n → ∞ if
and only if for each open set U containing x there exists n0 ∈ N such that xn ∈ U for
all n ≥ n0 .
97
Keeping this in mind, we define:
Let (X, J ) be a topological space and {xα }α∈D be a net in X. Then we say that
the net {xα }α∈D converges to an element x ∈ X if and only if for each open set U
containing x there exists α0 ∈ D such that xα ∈ U , ∀ α ≥ α0 (that is α ∈ D with
α0 ≤ α). If {xα }α∈D converges to x then we write xα → x.
In a metric space (X, d) we know that a sequence {xn }∞
n=1 in X converges to
at most one element x in X.
What about in a topological space ? Whether a net {xα }α∈D in a topological
space converges to at most one element in X. Obviously the answer is no. For
example, let X be any set containing at least two elements and J = {φ, X}. Take
x1 , x2 ∈ X, x1 6= x2 . Now with usual ≤, (N, ≤) is a directed set.
Define f : N → X as

 x1 when n is odd
f (n) =
 x when n is even
2
Here our net is {x1 , x2 , x1 , x2 , . . .} that is our net is a sequence in X. Let x = x1 .
Then the only open set U containing x1 is X and hence n0 = 1 ∈ N. Then for all
n ≥ n0 , xn ∈ X = U . Hence xn → x for any x ∈ X. Also nothing special about the
net {x1 , x2 , x1 , x2 , . . .}. In fact if D is a directed set and {xα }α∈D is an arbitrary net
in X then for each x ∈ X, xα → x.
Example 4.4.9. Now consider X = R and Jf , the cofinite topology on R. D = R
and ≤ is our usual relation. Then (D, ≤) is a directed set. Define f : D → R as
f (α) = α for α ∈ D = R. Then {α}α∈R is a net in R. Fix an element say x ∈ R
Whether xα → x ? How to start? Start with an open set U containing x in our
98
topological space (R, J ). Now U ∈ Jf , x ∈ U (that is U 6= φ) implies U c is a finite
subset of R.
Case (i). U c = φ (⇒ U = X).
Case (ii). U c 6= φ.
That is U c is a nonempty finite subset of R. Hence there exists n0 ∈ N and
x1 , x2 , . . . , xn0 ∈ R = D such that U c = {α1 , α2 , . . . , αn0 }. Now take a real number
say α0 such that α0 > αi for all i = 1, 2, . . . , n0 . This α0 ∈ D is such that xα = α ∈ U
∀ α ≥ α0 , (α ≥ α0 , α0 > αi ⇒ α > αi ⇒ α ∈
/ U c ⇒ α ∈ U ).
Conclusion: We started with an open set U containing x and we could get an α0 ∈ D
(α0 depends on U ) such that xα ∈ U , ∀α ≥ α0 . Hence by our definition xα → x.
That this net {xα } = {α}α∈D converges to every element x of the given topological
space (R, J ).
(iii) D = {1, 2, . . . , p} and ≤ is our usual relation. (D, ≤) is a directed set (check).
What about {xα }α∈D . Here D = {1, 2, . . . , 10} implies {xα }α∈D = {1, 2, . . . , 10}.
Now for any open set U containing 10 there exists α0 = 10 ∈ D is such that α ∈ D,
α ≥ α0 = 10 ⇒ α = 10 and xα = α = 10 ∈ U . Hence {xα }α∈D → 10.
Theorem 4.4.10. In a Hausdorff topological space (X, J ) a net {xα }α∈D in X cannot
converge to more than one element.
Proof. Suppose a net {xα }α∈D converge to say x, y ∈ X, where x 6= y. Now x 6= y,
(X, J ) is a Hausdorff topological space implies there exist open sets U, V in X such
that (i) x ∈ U, y ∈ U , (ii) U ∩ V = φ. Now xα → x, U is an open set containing x
implies
there exists α1 ∈ D such that xα ∈ U for all α ≥ α1 .
99
(4.5)
Also yα → y, V is an open set containing y implies
there exist α2 ∈ D such that yα ∈ V for all α ≥ α2 .
(4.6)
Note that D with a relation ≤ is a directed set and hence for α1 , α2 ∈ D there exists
α0 ∈ D such that α0 ≥ α1 and α0 ≥ α2 (that is α1 ≤ α0 and α2 ≤ α0 ). Now α0 ≥ α1
implies xα0 ∈ U from Eq. (4.5) and α0 ≥ α2 implies xα0 ∈ V from Eq. (4.6). Hence
xα0 ∈ U ∩ V , a contradiction to U ∩ V = φ. We arrived at this contradiction by
assuming xα → x, xα → y and x 6= y. This means {xα }α∈D cannot converge to more
than one element.
Note. In a Hausdorff topological space a net {xα }α∈D may not converge. If a net
converges then it converges to a unique limit.
?
Theorem 4.4.11. Let (X, J ) be a topological space and A ⊆ X. Then an element x
of X is in A if and only if there exists a net {xα }α∈D in A such that xα → x.
Proof. Let us assume that x ∈ A. Our tasks are the following: (i) using the fact
that x ∈ A construct a suitable directed set, (D, ≤), (ii) and then define a net
{xα }α∈D that converges to x. Now x ∈ A implies for each open set U containing x,
U ∩ A 6= φ. (If our topology J is induced by a metric d on X then J = Jd . In this
case B(x, n1 ) ∩ A 6= φ for each n ∈ N. So take xn ∈ B(x, n1 ) ∩ A. Then d(xn , x) <
and
1
n
1
n
→ 0 as n → ∞. Hence xn → x). Take D = Nx = {U ∈ J : x ∈ U } that is Nx
is the collection of all open sets containing x. For U, V ∈ Nx , define U ≤ V if and
only if V ⊆ U (reverse set inclusion is our relation ≤). Now define f : Nx → X as
f (U ) = xU ∈ U ∩ A (U ∩ A 6= φ for each U ∈ Nx implies by axiom of choice such a
function exists). Now we have a net {xU }U ∈Nx .
100
Claim: xU → x.
Take an open set U0 containing x, then such an U0 ∈ Nx implies f (U0 ) ∈ U0 ∩A.
Now U ∈ Nx (our directed set) and U ≥ U0 implies U ⊆ U0 implies xU ∈ U ⊆ U0 .
Now U ≥ U0 implies xU ∈ U0 . Hence by definition of convergence of a net, xU → x.
Conversely, assume that there is a net say {xα }α∈D in A such that xα → x.
Now we will have to prove that x ∈ A. So start with an open set U containing x.
Hence xα → x implies
there exists α0 ∈ D such that xα ∈ U for all α ≥ α0 .
(4.7)
(D is a directed set means (D, ≤) is a directed set). In particular when α = α0 ,
α ≥ α0 and therefore from Eq. (4.7), xα0 ∈ U . Also xα0 ∈ A. Hence xα0 ∈ U ∩ A.
That is for each open set U containing x, U ∩ A 6= φ. This implies x ∈ A.
Theorem 4.4.12. Let X, Y be topological spaces and f: X → Y . Then f is continuous
if and only if for every net {xα }α∈J converging to an element x ∈ X the net
{f (xα )}α∈J converges to f(x).
Proof. Assume that f : X → Y is a continuous function. Now let {xα }α∈J be a net
in X such that xα → x for some x ∈ X. We will have to prove that f (xα ) → f (x).
Let V be an open set containing f (x) in Y . Now V is an open set containing f (x)
and f : X → Y is a continuous function implies
there exists an open set U containing x such that f (U ) ⊆ V .
(4.8)
Now U is an open set containing x and xα → x implies there exists an α0 ∈ J such
that xα ∈ U for all α ≥ α0 . Hence from Eq. (4.8), f (xα ) ∈ V for all α ≥ α0 . That is,
101
for each open set V containing f (x) there exists α0 ∈ J such that f (xα ) ∈ V for all
α ≥ α0 . This in turn implies f (xα ) → f (x).
Now let us assume that whenever a net {xα }α∈J converges to an element x
in X then f (xα ) → f (x) in Y . In this case we will have to prove that f : X → Y
continuous. We know that f is continuous if and only if f (A) ⊆ f (A) for all A ⊆ X.
(An element z of X is closer to A, that is if z ∈ A then the image f (z) is closer to
f (A).) So start with A ⊆ X and an element y ∈ f (A) (f (A) = φ ⇒ f (A) ⊆ f (A)).
Now y ∈ f (A) implies there exists x ∈ A such that y = f (x). Hence x ∈ A implies
there exists a net {xα }α∈J in A such that xα → x (refer the previous theorem) this
implies by our assumption, f (xα ) → f (x). Now f (xα ) ∈ f (A) and f (xα ) → f (x)
implies f (x) ∈ f (A) (again refer the previous theorem). So we have proved that
f (A) ⊆ f (A) whenever A ⊆ X. This implies f : X → Y is a continuous function. Alternate proof of theorem 4.4.12.
Proof. Assume that whenever a net {xα }α∈J converges to an element x ∈ X then
f (xα ) → f (x) in Y. Now suppose f is not continuous at x. Then there exists an open
set V containing x such that f (U ) * V for every U ∈ Nx . Then for each U ∈ Nx
there exists xU ∈ U such that f (xU ) ∈
/ V. Now observe (refer the proof of the theorem
4.4.11) that the net {xU }U ∈Nx such that xU → x in X but f (xU ) 9 f (x) in Y.
Now let us define a concept which generalize the concept of a subsequence.
Recall that if X is a nonempty set {xn }∞
n=1 = (xn )n∈N is a sequence in X if and only
if there exists a function f : N → X satisfying the condition that f (xn ) = xn . Here
we have a net {xα }α∈J in X. Hence in place of N we have a directed set (J, ≤) and
a function f : J → X satisfying the condition that f (α) = xα for all α ∈ J.
102
What do we mean by saying that {xnk }∞
k=1 is a subsequence of {xn }? We have
a subset {nk }∞
k=1 of natural numbers satisfying n1 < n2 < · · · < nk < nk+1 < · · · .
Let D = {nk : k ∈ N} and we have f : N → X such that f (n) = xn for all n ∈ N.
That is essentially f : N → X is sequence in X. Now we have another function say
g : D → N satisfying (i) g(k) = nk , (ii) n1 < n2 < · · · < nk < nk+1 < · · · that is
k < l ⇒ nk < nl that is k < l ⇒ g(k) < g(l). Also note that for each n0 ∈ N there
exists k0 ∈ N such that f (k0 ) = nk0 > n0 . So keeping this motivation in mind we
define the concept of subnet of a given net in X. It is given that f : J → X is a net
in X ( so it is understood that (J, ≤) is a directed set). Suppose D with a relation
≤ is a directed set (need not be the same relation as given in J. But for the sake
of simplicity we use same notation ≤ for both sets). Suppose g : D → J such that
i, j ∈ D, i ≤ j implies g(i) ≤ g(j) and for each α ∈ J there exists γ ∈ D such that
g(γ) ≥ α (this is like saying that, when J = N = D, for each n0 ∈ N there exists
k ∈ N such that g(k) = nk ≥ n0 ). In such a case f ◦ g : D → X is called subnet of
X. ((f ◦ g)(k) = f (g(k)) = f (nk ) = xnk ).
Definition 4.4.13. Let (X, J ) be a topological space and {xα }α∈J = (xα )α∈J be a
net in X. An element x ∈ X is said to be an accumulation point of the given net
(xα )α∈J if and only if for each open set U containing x, the set KU = {α ∈ J : xα ∈ U }
is cofinal in J. Now KU is cofinal in J means for each α ∈ J there exists β ∈ KU
such that β ≥ α (it is like saying that k → ∞ implies nk → ∞).
Now let us prove:
Theorem 4.4.14. Let (xα )α∈J be a net in a topological space. Then a point x in X
is an accumulation point of the given net (xα )α∈J if and only if (xα )α∈J has a subnet
and that subnet converges to x.
103
Proof. ⇒ Assume that x is an accumulation point of (xα )α∈J . By the definition of
accumulation point of a net we have for each open set U containing x
KU = {α ∈ J : xα ∈ U } is cofinal in J.
(4.9)
Let K = {(α, U ) ∈ J × Nx : xα ∈ U }, where Nx is the collection of all open sets
containing x. From Eq. (4.9), KU 6= φ. (Fix α ∈ J. Now KU is cofinal in J implies
there exists β ∈ KU such that β ≥ α.) For (α, U ), (β, V ) ∈ K define (α, U ) ≤ (β, V )
if and only if α ≤ β and V ⊆ U (reverse set inclusion). It is easy to see that (K, ≤)
is a directed set. It is given that (xα )α∈J is a net in X. Hence (J, ≤) is a directed set
and f : J → X is such that f (α) = xα . Now define g : K → J as g(α, U ) = α (refer
Eq. (4.9)).
Claim: g(K) is cofinal in J.
So take α ∈ J. Now KU is cofinal in J (refer Eq. (4.9)) there exists β ∈ KU
such that β ≥ α. Now β ∈ KU implies xβ ∈ U that is (β, U ) ∈ K is such that
g(β, U ) = β ≥ α implies g(K) is cofinal in J. Also (α, U ), (β, V ) ∈ K, (α, U ) ≤ (β, V )
implies g(α, U ) = α ≤ β = g(β, V ). Hence f ◦ g : K → X is a subnet of f (or say
f (α) = (xα )). Now let us prove that this subnet converges to x. So take an open set
U containing x. This implies KU is cofinal in J. Fix (α0 , U ) ∈ K. Now α0 ∈ J, KU
is cofinal in J implies β0 ∈ KU such that β0 ≥ α0 . Hence (α, V ) ∈ K, (α, V ) ≥
(α0 , U ) implies (f ◦ g)(α, V ) = f (α) = xα ∈ V ⊆ U . That is for each open set U
containing x there exists (α0 , U ) ∈ K such that (α, V ) ∈ K (α, V ) ≥ (α0 , U ) implies
(f ◦ g)(α, V ) ∈ U . This proves that f ◦ g → x.
Conversely, suppose there is a subnet of (f (α))α∈J = (xα )α∈J which converge
to an element x ∈ X. A subnet of f converges to x means there exists a directed set
104
say (K, ≤) and a function say g : K → J such that i, j ∈ K, i ≤ j implies g(i) ≤ g(j),
g(K) is cofinal in J, and (f ◦ g)(i) = f (g(i)) → x. Now let us prove that x is an
accumulation point of the net f . So take an open set U containing x.
Claim: {α ∈ J : f (α) = xα ∈ U } is cofinal in J.
Let α0 ∈ J. Now f ◦ g : K → X is a subnet such that f ◦ g → x. Hence for
this given α0 ∈ J there exists β ∈ K such that g(β) ≥ α0 (note g(K) is cofinal in J).
Now f ◦ g → x, U is an open set containing x implies there exists β0 ∈ K such that
α ∈ K, α ≥ β0 ⇒ f (g(α)) ∈ U , β ∈ J is such that g(β) ≥ α0 . Take γ0 ∈ K such that
α0 ≥ β, β0 . Then (f ◦ g)(γ0 ) ∈ U and g(γ0 ) ≥ g(β) ≥ α0 . That is for α0 ∈ J, there
exists g(γ0 ) ∈ J such that f (g(γ0 )) ∈ U implies {α ∈ J : f (α) = xα ∈ U } is cofinal
in J. Hence x is an accumulation point.
Recall that a metric space (X, d) is a compact metric space if and only if every
∞
sequence {xn }∞
n=1 in X has a subsequence {xnk }k=1 that converges to an element
in X. It is to be noted that this result is not true for an arbitrary topological space.
For a topological space we have the following theorem.
Theorem 4.4.15. A topological space (X, J ) is compact if and only if every net in
X has a subnet that converges to an element in X.
Proof. Assume that (X, J ) is a compact topological space and f : J → X is a net
in X. We will have to prove that f has a subnet that converges to an element in X.
So it is enough to prove that f has an accumulation point.
For each α ∈ J, let Aα = {xβ : α ≤ β} (note: f : J → X is a net means with
respect to a relation ≤, (J, ≤) is directed set). Now {Aα }α∈J is a collection of sets
which has finite intersection property. For Aα1 , Aα2 , . . . , Aαk if we take α ∈ J such
105
that α ≥ αj for all j = 1, 2, . . . , k, that is αj ≤ α, then xα ∈ Aαj , ∀ j = 1, 2, . . . , k and
k
hence x ∈ ∩ Aαj . Now (X, J ) is a compact topological space {Aα }α∈J is a collection
j=1
of closed subsets of X which has finite intersection property implies ∩ Aαj 6= φ. Let
α∈J
x ∈ ∩ A αj .
α∈J
Now we aim to prove that x is an accumulation point of f . So, start with an
open set U containing x, and we will have to prove that {α ∈ J : xα ∈ U } is cofinal
in J. Take α0 ∈ J. Now U is an open set containing x, x ∈ Aα0 implies U ∩ Aα0 6= φ.
Hence there exists α ≥ α0 such that xα ∈ U . This proves that {α ∈ J : xα ∈ U } is
cofinal in J. Hence we have proved that x is an accumulation point of the stated net
f . This implies there exists a subnet of f which converges to f .
To prove the converse part let us assume that every net in X has convergent
subnet in X. By assuming this, we aim to prove that (X, J ) is a compact topological
space.
To prove that (X, J ) is a compact topological space, let us prove: if A is a collection
of closed subsets of X which has finite intersection property then ∩ A 6= φ. So, we
A∈A
have a collection A of closed subsets of X which has finite intersection property.
Let B = {A ⊆ X : A = A1 ∩A2 ∩· · ·∩Ak , k ∈ N, A1 , . . . , Ak ∈ A}. That is B is
the collection of finite intersection of members of A. (Note. ∩ A = X and hence we
A∈φ
do not require to consider this case.) For A, B ∈ B define A ≤ B, whenever B ⊆ A.
Then (B, ≤) is a directed set. Now define f : B → X as f (A) = f (A1 ∩A2 ∩· · ·∩Ak ) =
xA , where xA ∈ A1 ∩ A2 ∩ · · · ∩ Ak is fixed (A1 ∩ A2 ∩ · · · ∩ Ak ) may contains more
than one element and in that case first take any one element form A1 ∩ A2 ∩ · · · ∩ Ak .
Hence f = (f (A))A∈B is a net in X. By our assumption this net f will have a subnet
106
that will converge to an element say x in X. So there will exists a directed set K and
a function g : K → B satisfying f ◦ g is a subnet of f and f ◦ g converges to x.
Now we claim that x ∈ A for each A ∈ A. Suppose for some A ∈ A, x ∈
/ A.
Then x ∈ Ac = U , an open set. Since f ◦g → x and U is an open set containing x there
exists α0 ∈ K such that (f ◦ g)(α) ∈ U for all α ≥ α0 . Now α0 ∈ K implies g(α0 ) ∈ B
implies there exists k ∈ N and A1 , A2 , . . . , Ak ∈ A such that g(α0 ) = A1 ∩A2 ∩· · ·∩Ak .
A1 ∩ A2 ∩ · · · ∩ Ak ∈ B is such that A1 ∩ A2 ∩ · · · ∩ Ak ≥ g(α0 ). We have f ◦ g(α0 ) =
f (g(α0 )) ∈ U = Ac . Now K is a directed set and g(K) is cofinal in B implies there
exists α ∈ K such that (i) α ≥ α0 , (ii) g(α) ≥ A1 ∩ A2 ∩ · · · ∩ Ak . Now α ≥ α0 implies
(f ◦g)(α) ∈ U = Ac but by the definition of f, f (g(α)) ∈ g(α) ⊆ A1 ∩A2 ∩· · ·∩Ak ⊆ A.
So we get a contradiction. Therefore x ∈
/ A for some A ∈ A cannot happen. We have
proved that if A is a collection of closed subsets of X which has finite intersection
property then ∩ A 6= φ. Hence (X, J ) is a compact topological space.
A∈A
Definition 4.4.16. A topological property is any property so that if (X, J ), (Y, J 0 )
are topological spaces and f : (X, J ) → (Y, J 0 ) is a homeomorphism (that is (X, J )
is homeomorphic to (Y, J 0 )) then (X, J ) has the property if and only if (Y, J 0 ) has
the same property.
Example 4.4.17. Compactness, connectedness, local compactness are all topological
properties.
107