Download Math 525 More notes about compactness (sec 26

Math 525 More notes about compactness (sec 26-29)
Be sure to check out Thm 26.7 and Lemma 26.8 in Munkres.
Recall that in Munkres it is proved that a totally ordered set X with the order topology is
connected iff it satisfies:
(1) Every subset bounded above has a least upper bound;
(2) If x < y, there is a z such that x < z < y.
Furthermore, if X is connected, then any interval in X is also connected. In particular, this
theorem implies that every interval in R is connected, and we’ve showed that any non-interval
in R is disconnected. Thm. 27.1 in Munkres is a related result pertianing to compactness in
totally ordered sets. We’ll state only the result here, since the proof is given in Munkres.
Thm 27.1 If X is a totally ordered set with the order topology which has the l.u.b property
(see (1) above), then every closed interval [a, b] in X is compact.
Note: For a closed interval [a, b] in a totally ordered set, conditions (1) and (2) above are
necessary and sufficient for connectedness, whereas only (1) is necessary and sufficient for
compactness.
Heine-Borel Theorem A subset of (R, τu ) is compact iff it is closed and bounded.
Proof : By Thm 27.1, any closed, bounded interval [a, b] in R is compact. Any closed,
bounded set in R is a closed subset of some interval [a, b], and hence compact by Prop
A52(a). Conversely, any compact set A in R is closed by Prop A52(b). Also, A must
be bounded since an unbounded set in any metric space can be covered by a collection of
bounded open balls, and such a cover has no finite subcover.
¥
Generalized H-B Theorem A subset of (Rn , τu ) is compact iff it is closed and bounded.
Proof : A compact set in any metric space must be closed and bounded, as shown in the
preceding proof. Conversely, any closed, bounded set in Rn is a closed subset of a product
of closed, bounded intervals in R, and the latter is compact by the Tychanov Theorem. ¥
A standard example: the Long Line
Let SΩ × [0, 1) have the dictionary order topology, and let L be the subspace obtained by
deleting the least element (0, 0). We may visualize L as an uncountable number of vertical
columns, each (except for the first) a copy of [0, 1), indexed by the countable ordinals.
L is a linear continuum, and hence connected. How is L related to (R, τu )?
One may show that L is ”locally like” R in the sense that each point in L has a nbhd basis
of sets homeomorphic to open intervals in R (this is not so obvious for points of the form
(0, λ) ∈ L, where λ is a limit ordinal). Thus L, like (R, τu ), is first countable. However L
is not second countable, since L contains uncountably many disjoint open sets (choose an
open interval in each column). Thus (R, τu ) is homeomorphic to a subspace of L (indeed, to
the first column), but L is not homeomorphic to a subspace of R, since second countability
is hereditary. Also, L is not metrizable, since it contains a subspace homeomorphic to SΩ ,
which is not metrizable (we don’t have the tools to prove this).
Compactness in Metric Spaces
For metric spaces, there are several ways to characterize compactness which are non-equivalent
when generalized to topological spaces. Three such alternative versions of compactness
(which are equivalent in metric spaces, but not in topological spaces) are defined as follows:
Def A topological space (X, τ ) is:
(1) limit-point compact iff each infinite subset of X has a limit point;
(2) countably compact iff each countable open cover of X has a finite subcover;
(3) sequentially compact iff every sequence in X has a convergent subsequence.
Prop A55 In any top. space, (compact) ⇒ (countably compact) ⇒ (lim-pt compact). (cf.
Munkres Thm 28.1)
For a T1 space, (limit-point compact) ⇔ (countably compact).
Proof : Clearly, (compact) ⇒ (countably compact). The rest is left to you.
¥
Examples Example 2 in §28 of Munkres shows that SΩ (in the order topology) is limit-point
compact, and hence countably compact, since it is T2 . One may also show that SΩ is sequentially compact, since every sequence in SΩ has a subsequence which is either constant or
increasing, and in either case convergent. Since SΩ is not compact, none of these properties
implies compactness. Furthermore, compactness does not imply sequential compactness,
since [0, 1][0,1] with the product topology is compact by the Tychanov theorem, but it can
be shown not to be sequentially compact.
More results about compactness:
Prop A56 ((X, τ ) sequentially compact) ⇒ ((X, τ ) countably compact).
Prop A57 If (X, τ ) is T1 and first-countable, then ((X, τ ) limit-point compact) ⇒ ((X, τ )
sequentially compact).
Theorem A58 In a metric space (X, τ ), (compact) ⇔ (sequentially compact) ⇔ (limit-point
compact) ⇔ (countably compact). (cf. Munkres Thm 28.2)
Note: The proof of Thm A58 requires the use of the following well-known lemma, which is
proved in Munkres as Lemma 27.5:
Lebesque Covering Lemma
If (X, d) is a metric space which is sequentially compact and
U is an open cover of X, then there exists a δ > 0 (called the Lebesque number for U) such
¯
def
that if A ⊆ X and and diam A = sup{d(x, y) ¯ x, y ∈ A} < δ, then there is a U ∈ U such
that A ⊆ U .
Recall that a function f : (X, d) → (Y, ρ) between two metric spaces is said to be uniformly
continuous iff for every ǫ > 0, there exists a δ > 0 such that d(x, y) < δ ⇒ ρ(f (x), f (y)) < ǫ.
Another result (Munkres Thm 27.6) whose proof uses the Lebesque Covering Lemma is:
Uniform Continuity Theorem: If f : (X, τ ) → (Y, µ) is continuous and (X, τ ) is compact,
then f is uniformly continuous.
Example An interesting property of the ordinal line SΩ is that any continuous map f : SΩ →
(R, τu ) is ”eventually constant”, meaning that there exists λ ∈ SΩ such that f (µ) = f (λ) for
all µ > λ. The proof of this result is based on the following facts:
(1) SΩ is countably compact;
(2) Countable compactness is preserved under continuous maps;
(3) f (SΩ ) is compact in (R, τu ) by Thm A58.
We’ll omit the remaining details.
Prop A59 A compact, metrizable space is separable.
Proof : Let (X, τ ) be compact and metrizable, with d being a metric which induces τ . By
compactness, for any n ∈ N, the set of all open balls of radius
1
n
has a finite subcover
Bn = {B(x1n , n1 ), B(x2n , n1 , . . . , B(xkn , n1 )}. Note that for each n ∈ N, there is a different,
finite set of center points {x1n , x2n , . . . , xkn }. Let D be the countable collection of all these
sets of center points:
[
D=
{x1n , x2n , . . . , xkn }.
n∈N
The claim now is that D is a dense subset of X. To see this, let x ∈ X and U a nbhd of x.
Then there exists an ǫ > 0 such that B(x, ǫ) ⊆ U . Choose m ∈ N such that
1
m
< ǫ. Since
{B(x1m , m1 ), B(x2m , m1 , . . . , B(xkm , m1 )} is a cover of X, we know that x ∈ B(xjm , m1 ) for some
xjm . But this implies that d(x, xjm ) < ǫ, so B(x, ǫ) ∩ D 6= φ, so U ∩ D 6= φ.
¥
Corollary For metrizable spaces: (compact) ⇔ (seq. compact) ⇔ (lim. pt. compact) ⇔
(countably compact) ⇒ (separable) ⇔ (2nd countable).