Download Theorem 2.24. (a) For any collection {G α}α∈Aof open sets, G

Theorem 2.24.
S
(a) For any collection fG g 2A of open sets, G =
2A G is open.
T
(b) For any collection fF g 2A of closed sets,
F
is closed.
2A
Tn
(c) For any …nite collection G1 ; : : : ; Gn of open sets, H = i=1 Gi is open.
Sn
(d) For any …nite collection F1 ; : : : ; Fn of closed sets, i=1 Fi is closed.
Proof. (a) Let x 2 G. Then x 2 G for some 2 A: Since G is open, x
is an interior point of G : Since G
G, x is an interior point of G. We have
proved G is open.
T
S
c
(b)
=
F c : For each 2 A, since F is closed, F c is open.
2A F
T
S 2A c
c
, we
Part (a) implies
2A F
2A F is open. Since this equal to
T that
conclude that
F
is
closed.
2A
(c) Let x 2 H. Then x 2 Gi for each i = 1; : : : ; n: Since each Gi is open,
there exists ri > 0 such that Nri (x) Gi . De…ne r = min fr1 ; : : : ; rn g. Since
r ri for each i, we have Nr (x)
Tn Nri (x), which implies Nr (x) Gi . Since
this is true for each i; Nr (x)
i=1 Gi = H. We have proved that H is open.
Sn
Tn
c
c
c
(d) ( i=1
F
)
=
F
:
Since
each Fi is closed,
open. By (c) we
i
i=1 i
Tn
Sn Fi is
c
c
have
that
F
is
open.
Since
this
is
equal
to
(
F
)
,
we
conclude that
i
i=1 i
i=1
Sn
F
is
closed.
i
i=1
Déjà vu:
De…nition A topological space is a set X together with a collection of
subsets of X, called open sets, with the following properties:
(i) ? and X are open sets.
(ii) Any union of open sets is open.
(iii) Any …nite intersection of open sets is open.
A closed set is de…ned to be the complement of an open set.
Given a set E, let E 0 denote the set of limit points of E. The closure is
de…ned by E = E [ E 0 .
Theorem 2.27. If X is a metric space and E X; then
(a) E is closed,
(b) E = E if and only if E is closed,
(c) E F for every closed set F X such that E F:
Proof. (a) We show that (E)c is open. Let x 2 (E)c . Since x 2
= E, we have
that x 2
= E and x is not a limit point of E. Thus there exists a neighborhood
Nr (x) such that Nr (x) \ E = ?.
Claim. Nr (x) \ E 0 = ?.
1
The claim implies that Nr (x) \ E = ?, i.e., Nr (x)
implies that (E)c is open. Part (a) follows.
(E)c . This, in turn,
Proof of the claim: Suppose there is a y 2 Nr (x)\E 0 . Let s = r d (y; x) > 0.
Then Ns (y) Nr (x). Since y 2 E 0 , there exists z 2 Ns (y) \ E; which implies
z 2 Nr (x) \ E: This contradicts Nr (x) \ E = ?.
(b) (i) Suppose E = E. Since E is closed by part (a), we conclude that E
is closed.
(ii) Suppose E is closed. Then, by de…nition, E 0 E. Hence E = E [ E 0 =
E.
(c) Let F
X be a closed set with E
F: Let x 2 E 0 . Since x is a limit
point of E and since E F , x is a limit point of F . Since F is closed, x 2 F .
Therefore E 0 F . Since E F and E = E [ E 0 , we conclude that E F .
2