NoteNextra-origin/content/Math4201/Math4201_L29.md

Math4201 Topology I (Lecture 29)

Compact and connected spaces

Compact spaces

Theorem of uncountable compact Hausdorff spaces without isolated points

Any non-empty compact Hausdorff space X without any isolated points is uncountable.

Proof

By contradiction, let X=\{x_n\}_{n\in\mathbb{N}} be a countable set.

We construct inductively a sequence of open non-empty subspaces \{V_i\} of X such that

\overline{V_1}\supseteq \overline{V_2}\supseteq \overline{V_3}\supseteq \dots

where x_i not in \overline{V_i}.

This could imply that for any j\leq i, x_j is not in \overline{V_j} but \overline{V_j}\supseteq \overline{V_i}. This contradicts the fact that x_j not in \overline{V_i}, that is x_1, ..., x_i are not in \overline{V_i}.

This is a contradiction because \{\overline{V_i}\} satisfies the finite intersection property.

\bigcap_{i=1}^\infty \overline{V_i} \neq \emptyset

But in this case, \bigcap_{i=1}^\infty \overline{V_i} = \emptyset because x_i not in \overline{V_i}.

To construct such \{V_i\}, we can start with V_1,\dots, V_{k-1} are constructed, then there is a point y_k in V_{k-1} which isn't same as x_k.

So x_k not in \overline{V_k} because U_k is an open neighborhood of x_k that don't intersect with V_k.

Since X is Hausdorff, there exists an open neighborhood U_k of x_k and U_y of y_k such that U_k\cap U_y=\emptyset.

Let x_k\in U_k and y_k\in W_k\subseteq X that is open, and U_k\cap W_k=\emptyset.

Let V_k=W_k\cap V_{k-1}. Then this is open and is contained in V_{k-1} since \overline{V_k}\subseteq \overline{V_{k-1}}.

Case 1: x_k\notin V_{k-1}. There is such y_k because V_{k-1} is not empty.

Case 2: x_k\in V_{k-1}. Since x_k is not isolated points, any open neighborhood of x_k including V_{k-1} contains another point.

Therefore, X is uncountable.

Definition of limit point compact

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

That is, \forall A\subseteq X and A is infinite, there exists a point x\in X such that x\in U, \forall U\in \mathcal{T} containing x, (U-\{x\})\cap A\neq \emptyset.

This property also holds for finite sets, for example, any finite set with discrete topology is limit point compact. (since you cannot find a infinite subset of a finite set that has a limit point)

Definition of sequentially compact

A space X is sequentially compact if any sequence has a convergent subsequence. i.e. If \{x_n\}_{n\in\mathbb{N}} is a sequence in X, then there are n_1<n_2<\dots<n_k<\dots such that \{y_i=x_{n_i}\}_{i\in\mathbb{N}} is convergent.

Theorem of limit point compact 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.

Example of limit point compact spaces but not compact

Let X'=\{a,b\} with trivial topology, and X=\mathbb{N}\times X' with the product topology where we use the discrete topology on \mathbb{N}.

X isn't compact because \{\{i\}\times X':i\in\mathbb{N}\} is an open cover of X that doesn't have a finite subcover.

because these open sets are disjoint, X is limit point compact.

Let A\subseteq X be an infinite subset of X. In particular, it contains a point of the form (i,a) or (i,b) for i\in\mathbb{N}. Let (i,a)\in A. Then (i,b) is a limit point of A, since any open neighborhood (\{i\}\times X') of (i,b) contains a point of the form (i,a) or (i,b).

X is not sequentially compact because the sequence \{(n,a)\}_{n\in\mathbb{N}} has no convergent subsequence.

Proof

First, we show that 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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Continue with the proof that 2. implies 3. next time.

Corollary of compact spaces

1. If X is a compact topological space, then it is limit point compact.
2. If X is a sequentially compact topological space, then it is limit point compact.

Proof

Proof of 1. follows from the theorem of limit point compact spaces.

That means, sequentially compact is a stronger property than limit point compact, and compact is the stronger property than limit point compact.

Warning

Hope you will not use it soon for your exams but here are some interesting examples.

There exists spaces that are sequentially compact but not compact.

link to spaces

S000035

There exists spaces that are compact but not sequentially compact.

link to spaces

Consider the space of functions f:[0,1]\to [0,1] with the topology of pointwise convergence. This space is compact I^I but not sequentially compact (You can always find a sequence of functions that does not converge to any function in the space, when there is uncountable many functions in the space).

S000103