Download Notes for 1/21/15

In Rn we have the notion of (Euclidean) distance de…ned by
deuc (x; y) = jx
yj :
We have the following properties; see Theorem 1.37 on p. 16 (universal quanti…ers are implicit):
(a) deuc (x; y) > 0 if x 6= y; deuc (x; x) = 0;
(b) deuc (x; y) = deuc (y; x);
(c) deuc (x; y)
deuc (x; z) + deuc (z; y) (triangle inequality).
This is a motivation for the De…nition 2.15 of metric space (X; d). From
now on we assume that we are on a metric space (X; d).
A neighborhood of p 2 X is a set Nr (p) consisting of all q 2 X such that
d(p; q) < r, for some r > 0. That is,
Nr (p) = fq 2 X j d(p; q) < rg:
We also say that Nr (p) is the ball of radius r centered at p. It is often denoted
by Br (p), but we shall follow Rudin’s notation.
Remark. Let p; q 2 X. De…ne r = 21 d (p; q). Then p 2 Nr (p), q 2 Nr (q),
and Nr (p) \ Nr (q) = ? is the empty set. Exercise: prove this.
A subset E X is open if for each p 2 E there exists an r > 0 such that
Nr (p) E.
Theorem 2.19 For any p 2 X and r > 0, Nr (p) is an open set.
Proof. Let q 2 Nr (p). Then d(p; q) < r. Let s = r d(p; q), which is
positive. Then, for any x 2 Ns (q) we have d (x; q) < s and hence (using the
triangle inequality)
d (x; p)
d (x; q) + d (q; p) < s + d (p; q) = r:
Thus x 2 Nr (p). We have proved for any q 2 Nr (p) that Ns (q)
Nr (p).
Therefore Nr (p) is an open set.
A point p is a limit point of E if each neighborhood N of p contains a point
q 6= p such that q 2 E.
A subset E X is closed if each limit point of E is a point of E.
Theorem 2.20 If p is a limit point of a set E, then each neighborhood of
p contains in…nitely many points of E.
Proof. I suggest reading the proof on pp. 32–33. The following proof may
be slightly more intuitive, but it is less rigorous.
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We construct a sequence of points getting closer and closer to p.
Let Nr1 (p) be a neighborhood of p. Since p is a limit point of E, there exists
q1 2 Nr1 (p) \ E with q1 6= p. Let r2 = d (q1 ; p). Then 0 < r2 < r1 . Since p is
a limit point of E, there exists q2 2 Nr2 (p) \ E with q2 6= p. Let r3 = d (q2 ; p).
Continue this way to get an in…nite sequence q1 ; q2 ; q3 ; : : : in E with
r1 > d (q1 ; p) > d (q2 ; p) >
:
In particular, each point qi 6= p is distinct and in Nr1 (p) \ E:
Theorem 2.23 A set E is open if and only if its complement is closed.
Proof. Again, read the proof on p. 34; below is just slightly di¤ erent.
(1) Suppose E is open. Let x 2 E. Since E is open, there exists a neighborhood N of x with N
E, that is, N \ E c = ?. This implies x is not a limit
c
point of E .
We have proved : If x 2 E, then x is not a limit point of E c .
So the contrapositive of this is true: If x is a limit point of E c , then x 2 E c .
Thus E c is a closed set.
(2) Suppose E c is closed. Let x 2 E. Then x 2
= E c . Since E c is closed, x
c
is not a limit point of E . Thus there exists a neighborhood N of x such that
(N fxg) \ E c = ?. That is, N fxg E. Since x 2 E, this implies N E.
We have proved : If x 2 E, then there exists a neighborhood N of x such
that N E.
Thus E is an open set.
2.24 Theorem
(a) Any union of open sets is open.
(b) Any intersection of closed sets is closed.
(c) Any …nite intersection of open sets is open.
(d) Any …nite union of closed sets is closed.
The T
word ‘…nite’in (c) and (d) is necessary as seen by the following examples:
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1 1
(c’) i=1
;
= f0g is not an open set.
S1 1 n n
(d’) i=1 n ; 1 = (0; 1] is not a closed set.
More generally, we have (http://en.wikipedia.org/wiki/Topological_space ):
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.
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