SEPARATION AXIOMS BETWEEN To AND T1

Definition

Let (X,\mathcal{T}) be a topological space, and A,B be subsets of X.

Then A is said to be weakly separated from B if there exist an open set G such that A \subset G and G \cap B = \varnothing.

This definition of weakly separated sets

Definition

Let S = (X,\mathcal{T}) be a topological space, and \mathscr{A} be a cover of X. Then we call \mathscr{A} an open cover if

\mathscr{A} \subset \mathcal{T}.

Thus, \mathscr{A} is an open cover if it is made up of open sets.

Compactness

Now we define the notion of a space being compact.

Definition

A space X = (S,\mathcal{T}) is compact if every open cover \mathscr{A} of S has a finite subcover that also covers S.

We next prove that subspaces are compact by the same definition.

Lemma

Let X be a space and Y be a subspace of X. Then Y is compact if and only if every cover of Y by open sets in X has a finite subcover.

(\leftarrow) Suppose Y is compact, and \mathscr{A} = \{A_\alpha\}_{\alpha \in I} is a cover of Y by sets open in X. Then the collection

\{A_\alpha \cap Y \mid \alpha \in I\}

is a cover of Y by sets open in Y, making it an open cover of Y. Since we assumed Y is compact, we have that there exists a finite subcover

\{A_{\alpha_1} \cap Y, \dots, A_{\alpha_n} \cap Y\}.

Then \{A_{\alpha_1}, \dots, A_{\alpha_n}\} is a finite subcover of Y by sets open in X.

(\rightarrow) Suppose every cover of Y by open sets in X has a finite subcover. Let \mathscr{B} = \{B_{\alpha}\}_{\alpha \in I} be an open cover of Y. Then, for each \alpha \in J, choose a set A_\alpha open in X such that

A_\alpha = B_\alpha \cap Y.

(To-DO: Prove such a set exists) Then the collection \mathscr{A} = \{A_\alpha\}_{\alpha \in I}. is a cover of Y by sets open in X. Thus there exists a finite subcover \{A_{\alpha_1}, \dots, A_{\alpha_n}\}. Then \{B_{\alpha_1}, \dots, B_{\alpha_n}\} is a finite subcover of Y, and Y is compact.

Next we prove many properties of compact spaces

Theorem

Every closed subspace of a compact space is compact.

Let X be a compact topological space and Y be a closed subspace of X. Then we have that Y is closed in X, implying X \backslash Y is open in X by definition.

Let \mathscr{A} be a cover of Y by sets open in X. Then

\mathscr{B} = \mathscr{A} \cup X \backslash Y

is an open cover of X. Since X is compact, there exists a finite subcover \mathscr{B}^\prime which covers Y. We may safely remove X\backslash Y from \mathscr{B}^\prime to obtain a finite subcover of \mathscr{A}. Thus Y is compact.

Theorem

Every compact subspace of a Hausdorff space is closed.

Let Y be a compact subspace of a Hausdorff space X. We want to prove that X\backslash Y is open, and thus Y is closed.

Let x_0 be a point in X \backslash Y.

Theorem

The image of a compact space under a continuous map is compact.

Let f\colon X \to Y be continuous, and X be a compact space. Let \mathscr{A} be a covering of the set f(X) by sets open in Y. Then the collection

\{f^{-1}(A) \mid A \in \mathscr{A}\}

Is a collection of sets covering X, that are open in X since f is continuous. Since X is compact, there exists a finite subcover

\{f(A_1),\dots,f(A_n)\}

that covers X. Then the sets A_1,\dots,A_n form a finite subcover of f(X), and thus f(X) is compact.

Theorem

The product of fintely many compact spaces is compact.

Let f\colon X \to Y be continuous, and X be a compact space. Let \mathscr{A} be a covering of the set f(X) by sets open in Y. Then the collection

\{f^{-1}(A) \mid A \in \mathscr{A}\}

Is a collection of sets covering X, that are open in X since f is continuous. Since X is compact, there exists a finite subcover

\{f(A_1),\dots,f(A_n)\}

that covers X. Then the sets A_1,\dots,A_n form a finite subcover of f(X), and thus f(X) is compact.