Topological Subspaces#

Theorem: Topological Subspace

Let \((X, \tau_X)\) be a topological space.

For every non-empty subset \(S\) of \(X\), the collection

\[ \tau_S \overset{\text{def}}{=} \{O \cap S \mid O \in \tau_X\} \]

is a topology on \(S\).

Proof

We need to prove three things:

(I) \(S \in \tau_S\) and \(\varnothing \in \tau_S\).
(II) \(\tau_S\) is closed under arbitrary unions.
(III) \(\tau_S\) is closed under finite intersections.

Proof of (I):

This follows from the fact that \(\varnothing = S \cap \varnothing\) and \(S = S \cap X\).

Proof of (II):

Let \(\mathcal{U}\) be an arbitrary subset of \(\tau_S\). By definition, for each \(U \in \mathcal{U}\) there exists some \(O_U \in \tau_X\) such that \(U = O_U \cap S\). Then

\[ \bigcup \mathcal{U} = \bigcup_{U \in \mathcal{U}} (O_U \cap S) = \left(\bigcup_{U \in \mathcal{U}} O_U\right) \cap S \]

Since \(\bigcup_{U \in \mathcal{U}} O_U\) is a union of open sets, it is itself open, i.e. \(\bigcup_{U \in \mathcal{U}} O_U \in \tau_X\). Therefore \(\left(\bigcup_{U \in \mathcal{U}} O_U\right) \cap S\) is in \(\tau_S\).

Proof of (III):

Consider the intersection \(U_1 \cap \cdots \cap U_n\) where \(U_1,\cdots, U_n \in \tau_S\). By definition, for each \(U_k\) there exists some \(O_k \in \tau_X\) such that \(U_k = O_k \cap S\). This means that

\[ U_1 \cap \cdots \cap U_n = (O_1 \cap S) \cap \cdots \cap (O_n \cap S) = (O_1 \cap \cdots \cap O_n) \cap S \]

Since \((O_1 \cap \cdots \cap O_n)\) is an intersection of finitely many open sets, it is itself open. Therefore, \(U_1 \cap \cdots \cap U_n \in \tau_S\).

Definition: Topological Subspace

The topological space \((S, \tau_S)\) is known as a subspace of \((X,\tau_X)\).

Properties#

Theorem: Openness in Topological Subspaces

Let \((S, \tau_S)\) be a subspace of a topological space \((X, \tau_X)\) and let \(U\) be a subset of \(S\).

If \(U\) is open in \((S, \tau_S)\) and \(S\) is open \((X, \tau_X)\), then \(U\) is also open in \((X, \tau_X)\).

Proof

TODO

Theorem: Closedness in Topological Subspaces (I)

Let \((S, \tau_S)\) be a subspace of a topological space \((X, \tau_X)\).

A set \(A\) is closed in \((S, \tau_S)\) if and only if it is the intersection of a closed set of \((X, \tau_X)\) with \(S\).

Proof

We need to prove two things:

(I) If \(A\) is a closed set of \((S, \tau_S)\), then there exists a closed set \(C\) of \((X, \tau_X)\) such that \(A = C \cap S\).
(II) If there exists a closed set \(C\) of \((X, \tau_X)\) such that \(A = C \cap S\), then \(A\) is closed in \((S, \tau_S)\).

Proof of (I):

Suppose that \(A\) is closed in \((S, \tau_S)\). Then its Sets \(S \setminus A\) is open in \((S, \tau_S)\) and, by definition, there exists an open set \(O_A\) of \((X, \tau_X)\) such that \(S \setminus A = O_A \cap S\). Furthermore, \(X \setminus O_A\) is closed in \((X, \tau_X)\), since \(O_A\) is open in \((X, \tau_X)\).

Proof of (II):

TODO

Theorem: Closedness in Topological Subspaces (II)

Let \((S, \tau_S)\) be a subspace of a topological space \((X, \tau_X)\) and let \(C\) be a subset of \(S\).

If \(C\) is closed in \((S, \tau_S)\) and \(S\) is closed \((X, \tau_X)\), then \(C\) is also closed in \((X, \tau_X)\).

Proof

TODO

Theorem: Base for Topological Subspaces

Let \((S, \tau_S)\) be a subspace of a topological space \((X, \tau_X)\).

If \(\mathcal{B}\) is a base for \((X, \tau_X)\), then the collection

\[ \mathcal{B}_S \overset{\text{def}}{=} \{ B \cap S \mid B \in \mathcal{B}\} \]

is a base for \((S, \tau_S)\).

Proof

TODO

Compactness of Subspaces#

Theorem: Compactness of Subspaces

A subspace \((S, \tau_S)\) of a topological space \((X, \tau_X)\) is compact if and only if every cover of \(S\) by open subsets in \((X, \tau_X)\) contains a finite subcollection which is also a cover of \((S, \tau_S)\).

Proof

We need to prove two things:

(I) If \((S, \tau_S)\) is compact, then every cover of \(S\) by open subsets in \((X, \tau_X)\) contains a finite subcollection which is also a cover of \((S, \tau_S)\).
(II) If every cover of \(S\) by open subsets in \((X, \tau_X)\) contains a finite subcollection which is also a cover of \((S, \tau_S)\), then \((S, \tau_S)\) is compact.

Proof of (I):

Suppose that \((S, \tau_S)\) is compact and let \(\mathcal{C}\) be a cover of \(S\) consisting of open subset of \((X, \tau_X)\).

Theorem

Let \((S, \tau_S)\) be a subspace of a topological space \((X, \tau_X)\).

If \((X, \tau_X)\) is compact and \(S\) is closed in \((X, \tau_X)\), then \((S, \tau_S)\) is also compact.

Proof

TODO