Topological Subspaces

Theorem: Topological Subspace

Let $(X,{\tau }_X)$ be a topological space.

For every non-empty subset $S$ of $X$, the collection

$${\tau }_S\stackrel{\text{def}}{=}\{O\cap S\mid O\in {\tau }_X\}$$

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 complement $S\setminus A$ 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

$${\mathcal{B}}_S\stackrel{\text{def}}{=}\{B\cap S\mid B\in \mathcal{B}\}$$

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