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Math 490 Midterm 3 (Solutions); due Friday, December 6, 2013
Problem 1.
Consider the set X = [0, 1] with the topology
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T = {∅} ∪ {U ⊆ [0, 1] | [0, ] ⊆ U }
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(you do not need to prove that T is a topology on X).
Determine whether or not the topological space (X, T ) is compact
and give a proof justifying your answer.
Solution.
The space (X, T ) is not compact. Indeed, for every x ∈ ( 21 , 1] put
Ux = [0, 12 ] ∪ {x}. Then Ux is open in (X, T ) and ∪x∈( 1 ,1] Ux = [0, 1],
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so that (Ux )x∈( 1 ,1] ) is an open cover of X. Clearly, this cover does not
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admit a finite subcover. Indeed, for any n ≥ 1, x1 , . . . , xn ∈ ( 21 , 1] we
have ∪ni=1 Uxi = [0, 12 ] ∪ {x1 , . . . , xn } =
6 [0, 1].
Thus (X, T ) is not compact, as claimed.
Problem 2.[10 points]
Let (X, T ) be as in Problem 1.
(a) Determine whether or not the space (X, T ) is connected and give
a proof justifying your answer.
(b) Determine whether or not there exists a metric d on X such that
the topology T is exactly the metric topology on X corresponding to
d. Give a proof justifying your answer.
Solution.
(a) Yes, (X, T ) is connected. Indeed, if U, V ⊆ X are nonempty open
subsets then [0, 21 ] ⊆ U ∩ V . Therefore there do not exist nonempty
open subsets U, V ⊆ X such that U ∩ V = ∅ and U ∪ V = X. Thus
X is indeed connected, as claimed.
(b) No, there does not exist a metric d on X such that that the
topology T is exactly the metric topology on X corresponding to d.
Indeed, if such a metric did exist, then (X, T ) would be a Hausdorff
topological space since all metric spaces are Howsdorff.
However, (X, T ) is not Hausdorff. Indeed, for any open subsets U, V
of X such that 0 ∈ U and 1 ∈ V we have [0, 21 ] ⊆ U ∩ V so that
U ∩ V 6= ∅.
Problem 3.
Determine whether or not the following statement is true:
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If (X, d) is a bounded metric space and A, B ⊆ X are nonempty
closed subsets then there exist a0 ∈ A and b0 ∈ B such that d(a0 , b0 ) =
inf{d(a, b)|a ∈ A, b ∈ B}.
If the statement is true, prove it, and if the statement is false, give
a counter-example.
Solution.
The statement is false.
Let X = [−1, 0) ∪ (0, 1] with d(x, y) = |x − y| for x, y ∈ X.
Put A = [−1, 0) and B = (0, 1]. Then A, B are closed subsets of
X since A = X ∩ (−∞, 0] and B = X ∩ [0, ∞). We have d(A, B) =
inf{d(a, b)|a ∈ A, b ∈ B} = 0. However A ∩ B = ∅ and so there do not
exist a0 ∈ A, b0 ∈ B such that d(a0 , b0 ) = 0.
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Problem 4.[10 points]
(a) Give an example of a topological space X and a nonempty subset
Y ⊆ X, such that Y is compact (with the subspace topology) but that
Y is not closed in X. Justify that your example has the required
properties.
(b) A metric space (X, d) is called totally bounded if for every > 0
there exist n ≥ 1, x1 , . . . , xn ∈ X such that X = ∪ni−1 B(xi , ).
Give an example of a totally bounded metric space (X, d) such that
(X, d) is not compact. Justify that your example has the required
properties.
Solution.
(a) Let X = R and T = {X, ∅} be the trivial topology on R. Then
(X, T ) is compact since any open cover contains X as one of the open
sets in this cover.
Put Y = {1}. Then Y , with the subspace topology, is compact, since
all finite topological spaces are compact. However, Y is not closed in
X since X \ Y = (−∞, 1) ∪ (1, ∞) 6∈ T .
(b) Let X = (0, 1] with the standard metric d(x, y) = |x − y|. The
metric space (X, d) is bounded but not complete. Indeed, the sequence
( n1 )n≥1 is a Cauchy sequence in X which does not converge in X.
The space (X, d) is totally bounded. Indeed, for any > 0 we can
choose an integer n ≥ 1 such that n1 < . Put xi = ni , for i = 1, . . . , n.
Then ∪ni=1 B(xi , n1 ) = X and therefore ∪ni−1 B(xi , ) = X.
Problem 5.[10 points]
Let X = [0, 1] with the subspace topology Tu coming from the upper
limit topology on R. Let Te be the Euclidean topology on X = [0, 1].
Give an example of a continuous function f : (X, Tu ) → (X, Te ) such
that f (0) = 0, f (1) = 1 and that there exists c ∈ (0, 1) such that
c 6∈ f (X). Justify that your example has the required properties.
Solution.
Consider the function f : X → X defined as
(
0 for 0 ≤ x ≤ 21 ,
f (x) =
1 for 21 < x ≤ 1.
We claim that f : (X, Tu ) → (X, Te ) is continuous. First, observe that
the sets [0, 21 ] and ( 12 , 1] are open in Tu . Indeed, the sets (−∞, 21 ] =
1
1
1
1
∞
∪∞
n=1 (−n, 2 ] and ( 2 , ∞) = ∪n=1 ( 2 + n , n] are open in the upper limit
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topology on R. Therefore the sets [0, 12 ] = X ∩ (−∞, 12 ] and ( 12 , 1] =
X ∩ ( 12 , ∞) are open in Tu .
Let U ⊆ X be an open subset in Te .
If 0 6∈ U and 1 6∈ U then f −1 (U ) = ∅ is open in (X, Tu ).
If 0 ∈ U but 1 6∈ U then f −1 (U ) = [0, 21 ] is open in (X, Tu ).
If 0 6∈ U but 1 ∈ U then f −1 (U ) = ( 21 , 1] is open in (X, Tu ).
If 0, 1 ∈ U then f −1 (U ) = X is open in (X, Tu ).
Thus for every open subset U of (X, Te ) the set f −1 (U ) is open in
(X, Tu ). Therefore f is a continuous function.
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However, by construction, we have 100
6∈ f (X).
Problem 6.[10 points]
Let X = F (R, R) be the set of all bounded functions from R to R.
[A function f : R → R is called bounded if there exists C ∈ R, C > 0
such that for every x ∈ R we have |f (x)| ≤ C.]
For f, g ∈ X define d(f, g) := supx∈R |f (x) − g(x)|.
(a) Prove that (X, d) is a metric space.
(b) Determine whether or not the metric space (X, d) is complete.
Give a proof justifying your answer.
Solution.
(a) First, if f, g ∈ F (R, R) then there exists C > 0 such that |f (x)| ≤
C and |g(x)| ≤ C for every x ∈ R. Therefore for every x ∈ R we have
|f (x) − g(x)| ≤ |f (x)| + |g(x)| ≤ 2C
and hence d(f, g) = supx∈R |f (x) − g(x)| ≤ 2C < ∞. Thus d is indeed
a function d : X × X → [0, ∞). By construction, for every f, g ∈ X we
have d(f, g) = d(g, f ) since |f (x) − g(x)| = |g(x) − f (x)|.
If f ∈ X then d(f, f ) = supx∈R |f (x)−f (x)| = 0. On the other hand,
if f, g ∈ X and d(f, g) = supx∈R |f (x)−g(x)| = 0 then |f (x)−g(x)| = 0
for all x ∈ R, so that f (x) = g(x) for all x ∈ R and therefore f = g.
Finally, if f, g, h ∈ X, then
d(f, h) = sup |f (x) − h(x)| ≤ sup (|f (x) − g(x)| + |g(x) − h(x)|) ≤
x∈R
x∈R
sup |f (x) − g(x)| + sup |g(x) − h(x)| = d(f, g) + d(g, h).
x∈R
x∈R
Thus (X, d) is a metric space, as required.
(b) Yes, the metric space (X, d) is complete. Indeed, suppose that
(fn )n≥1 is a Cauchy sequence in (X, d).
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Thus for every > 0 there exists an integer N ≥ 1 such that for all
m, n ≥ N we have
(*)
d(fn , fm ) = sup |fn (x) − fm (x)| < .
x∈R
Therefore for every x ∈ R and every > 0 for all m, n ≥ N we
have |fn (x) − fm (x)| < . This means that for every x ∈ R the sequence (fn (x))n≥1 is a Cauchy sequence of real numbers. Since R is a
complete metric space, for every x ∈ R the sequence (fn (x))n≥1 has a
limit in R, denote this limit by f (x). Thus for every x ∈ R we have
limn→∞ fn (x) = f (x) ∈ R.
We have thus constructed a function f : R → R.
We claim that the function f is bounded. Indeed, for = 1 and N =
N1 , we know that for every n ≥ N we have supx∈R |fn (x) − fN (x)| < 1.
Since fN ∈ X, there exists C > 0 such that for every x ∈ R we
have |fN (x)| ≤ C. Hence for every x ∈ R and every n ≥ N we have
|fn (x)| ≤ C + 1. Therefore for every x ∈ R we have
|f (x)| = | lim fn (x)| ≤ C + 1.
n→∞
Thus the function f is bounded, so that f ∈ X.
We now claim that limn→∞ fn = f in (X, d)
Indeed, let > 0 be arbitrary and let N ≥ 1 be as in (*) above.
Fix an arbitrary x ∈ R. Then for every n, m ≥ N we have |fn (x) −
fm (x)| < . In particular, for every n ≥ N we have fn (x) ∈ [fN (x) −
, fN (x) + ]. Since limn→∞ fn (x) = f (x), it follows that f (x) ∈
[fN (x) − , fN (x) + ]. Therefore for every n ≥ N we have |fn (x) −
f (x)| < 2.
Since x ∈ R was arbitrary, this means that for every n ≥ N we have
d(fn , f ) < 2. Hence, by definition of convergence in a metric space,
we have limn→∞ fn = f in (X, d).
Thus every Cauchy sequence in (X, d) has a limit in X, so that the
metric space (X, d) is complete.