NoteNextra-origin/content/Math4201/Math4201_L30.md

Math4201 Topology I (Lecture 30)

Compactness

Compactness in Metric Spaces

Limit point compactness

A topological space X is limit point compact if every infinite subset of X has a limit point in X.

• Every compact space is limit point compact.

Sequentially compact

A topological space X is sequentially compact if every sequence in X has a convergent subsequence.

Theorem of equivalence of compactness in metrizable spaces

If (X,d) is a metric space then the following are equivalent:

1. X is compact.
2. X is limit point compact.
3. X is sequentially compact.

Proof

(1) \implies (2):

We proceed by contradiction,

Let X be compact and A\subseteq X be an infinite subset of X that doesn't have any limit points.

Then X-A is open because any x\in X-A isn't in the closure of A otherwise it would be a limit point for A, and hence x has an open neighborhood contained in the complement of A.

Next, let x\in A. Since x isn't a limit point of A, there is an open neighborhood U_x of x in X that U_x\cap A=\{x\}. Now consider the open covering of X given as

\{X-A\}\cup \{U_x:x\in A\}

This is an open cover because either x\in X-A or x\in A and in the latter case, x\in U_x since X is compact, this should have a finite subcover. Any such subcover should contain U_x for any x because U_x is the element in the subcover for x.

This implies that our finite cover contains infinite open sets, which is a contradiction.

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(2) \implies (3):

Let \{x_n\}_{n\in\mathbb{N}} be an arbitrary sequence in X.

Since d(z,x_{n_k})\leq \frac{1}{k} the subsequence (x_{n_k}) converges to z.

This completes the proof.

except possibly z.

Now we consider

B_{r_k}(z)\text { with } r_k=\min \left(\frac{1}{k}, d_k\right)

This ball has a point x_{n_k} from \{x_n\} which isn't equal to z.

r_k\leq d_k\implies n_k\geq n_{k-1}.

Since z is a limit point of \{x_n\}, there exists x_{n_k} such that d(z,x_{n_k})<\frac{1}{k}. So x_{n_k}\in B_{r_k}(z).

So, we have a convergent subsequence (x_{n_k}).

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(3) \implies (1):

First wee prove the analogue of Lebesgue number lemma for a sequentially compact space (X,d).

Let \{U_\alpha\}_{\alpha\in I} be an open covering of X. By contradiction, assume that for any \delta>0, there are two points x,x' with d(x,x')<\delta don't belong to the same open set in the covering.

Take \delta=\frac{1}{n}, and let x_n,x_n' be the points as above, then d(x_n,x_n')<\frac{1}{n}.

x_n,x_n' don't belong to the same open set in \{U_\alpha\}_{\alpha\in I}.

By assumption \{x_n\} is convergent after passing to a subsequence

\{x_{n_k}\}_i

Let y be the limit of this subsequence and U_\alpha be an element of the open covering containing y. There is \epsilon>0 such that B_\epsilon(y)\subseteq U_\alpha.

If k is large enough, then x_{n_k}\in B_{\epsilon/2}(y) and d(x_{n_k},x_{n_k}')<\epsilon/2. (take k such that \frac{1}{n_k}<\epsilon/2)

Then d(x_{n_k}',y)<\epsilon/2 this implies that x_{n_k}'\in U_\alpha.

Thus, d(x_{n_k}',y)\leq d(x_{n_k}',x_{n_k})+d(x_{n_k},y)<\epsilon/2+\epsilon/2=\epsilon.

So, x_{n_k}'\in B_\epsilon(y)\subseteq U_\alpha.

This is a contradiction.

Next we show that for any \epsilon, there are

y_1,y_2,\cdots,y_k

such that X=\bigcup_{i=1}^k B_{\epsilon}(y_i).

Let's assume that it's not true and construct a sequence of points inductively in the following way:

• Pick y_1 be arbitrary point in X.
• In the $k$-th step, if X\neq B_{\epsilon}(y_1)\cup \cdots \cup B_{\epsilon}(y_k), then pick y_{k+1}\notin B_{\epsilon}(y_1)\cup \cdots \cup B_{\epsilon}(y_k).
• In particular, d(y_{k+1},y_j)\geq \epsilon for all j<k.

By iteration, this process we obtain a sequence such such that the distance between any two elements is at most \epsilon.

This sequence cannot have a converging subsequence which is a contradiction.

To prove the compactness of X, take an open covering \{U_\alpha\}_{\alpha\in I} of X and let be \delta>0 such that any set with diameter at least \delta is one of the $U_\alpha$'s. Let also y_1,y_2,\cdots,y_k\in X be chosen such that B_{\frac{\delta}{2}}(y_1)\cup \cdots \cup B_{\frac{\delta}{2}}(y_k)=X.

Since diameter of B_{\frac{\delta}{2}}(y_i) is less than \delta, it belongs to U_\alpha for some \alpha\in I.

Then \{U_{\alpha_i}\}_{i=1}^k is a finite subcover of X.

This completes the proof.