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Open and Closed Sets
Adrian Down 16779577
August 2, 2005
1 Analogy with R
Properties that make analysis on (a, b) very different than analysis on [a, b]: i) Given x โ (a, b), โr > 0 such that (x โ r, x + r) โ (a, b). [a, b] does not have this property. Take x = a โ no r > 0 works ii) Given (sn) โ [a, b] and limnโโ sn = s, then s โ [a, b]. (a, b) does not have this property. Consider sn = (^1) n โ (0, 1). Then limnโโ sn = 0 โ/ (0, 1). Goal is to generalize these ideas to metric spaces.
2 Open sets
Definition: Given a metric space (S, d), a subset E is open if
โx โ E, โr > 0 such that Br(x) โ E (1) Br(x) = {y โ S|d(x, y) < r} (2)
In general, given any set E, a point x โ E satisfying (1) is called an interior point of E and E^0 = {x โ E|x is an interior point }. E^0 is called the interior of E. A set is open if and only if every point of E is interior to E, or E is open if and only if E = E^0. Can also be shown that E^0 is always open, for any set E. Would have to consider a point in the interior, then find some positive radius so that the open ball is contained in the interior. Does not follow immediately from the definition.
3 Examples
- (a, b) is open in R. In general, any open ball Br(x) is open in (S, d).
- โ
is always open for any S.
- If d is the discrete metric, so d(x, y) = 1, โx 6 = y, then any E โ S is open.
B 12 (x) = {x} โ E, โx โ E
- [a, b] is not open in R. (1) fails at x = a. Given any r > 0, Br(a) = (aโr, a+r) which is not a subset of [a, b]. You can still compute the interior, and get that [a, b]^0 = (a, b).
4 Propositions
Proposition: if (Uฮฑ)ฮฑโA is any collection of open sets, then
U =
ฮฑโA
Uฮฑ
is open. Means you donโt need a countable or finite union. Proof: let x โ U. Then x โ Uฮฑ for some
ฮฑ โ A โ โr > 0 such that Br(x) โ Uฮฑ โ U
So U is open. Proposition: if U 1 ,... , Un are open sets, then
U =
โ^ โ
i=
Ui
is open Proof: let x โ U. So x โ Ui, โi โ โri > 0 such that Bri (x) โ Ui. Let r = min{r 1 ,... , rn} > 0 and
Br(x) โ Bri (x)โi โ Ui
So Br(x) โ U The finiteness of the set of U โฒs is used to get that the minimum of the radius is greater than 0.
d(xn, x) < (^1) n , โn โ (xn) converges to x, but x /โ E. So E is not closed, a contradiction. Assume S/E is open. Assume E is not closed โ โ(sn), sn โ E such that lim sn = s /โ E. s โ S/E โ โr > 0 such that Br(s) โ S/E. โN such that n > N โ d(sn, s) < r, i.e. sn โ Br(s) โ S/E, a contradiction.
6 Compactness
Closed is not the exact analogy to the closed interval, since we can make closed sets like [a, โ). Weโre interested in the Bolzano Weirstrass property: given any sequence (sn) โ [a, b], [a, b] bounded โ โ a convergent subsequence (snk ) and limnโโ snk = s โ [a, b] Definition: Given a metric space (S, d), a set E โ S is compact if: given any sequence (sn) โ E, โ a convergent subsequence (snk ) and lim snk = s โ E. Notes:
- if E is compact then E is closed. Proof: if (sn) is convergent sequence in E, โ a convergent subsequence (snk ), lim snk = s โ E. But (sn) converging โ lim sn = lim snk = s โ E. So E = Eยฏ.
- Not every closed set is compact. For example, E = R is not compact, but [a, b] is compact. Theorem (Heine-Borel): In the space Rk^ (with the standard metric), a set E is compact if and only if E is closed and bounded. Proof: assume E is compact โ E is closed by the previous theorem. Assume for the sake of contradiction that E is unbounded. Fix x 0 โ S. So E/ โ Bn(x 0 ) far any n โ N. So โn โ N, โxn โ E but d(xn, x 0 ) โฅ n. So E compact โ โ a convergent subsequence (xnj ) and d(xnj , x 0 ) โฅ nj , โj. However, (xnj ) convergent โ (xnj ) is bounded. Contradiction, since nj goes to โ as j goes to โ. Assume E is closed and bounded. Given any sequence (sn) โ E, E is bounded โ (sn) is a bounded sequence, so by the Bolzano-Weirstrass theorem (on Rk), โ a convergent subsequence (snk ) โ E and limkโโ snk = s โ E because E is closed. So E is compact. Note that the Bolzano-Weirstrass theorem only applies to Rk. Thus the Heine-Borel theorem is false for metric spaces other than Rk.
Challenge: classify the compact sets in l^1 (N) and C[0, 1] = {f continuous on [0, 1]}, d(f, g) =
|f (x) โ d(x)|dx. These two spaces are similar. For C[0, 1], use the idea of an equicontinuous family of functions.
