Subspace

Definition

Let $X$ be a topological space with topology $\mathcal{T}$ Let $S\subseteq X$

$${\mathcal{T}}_S=\{S\cap U|U\in \mathcal{T}\}$$

We can also write:

$${\mathcal{T}}_S=\{U\subseteq S|U=S\cap V\text{ for some }V\subseteq X\}$$

${\mathcal{T}}_S$ is the smallest topology such that the map $S\to X$ is continuous

reason: we want the identity map ${\iota }_S:S\to X$ to be continuous Assume we have $U\subseteq X$ to be open $⟹$ the preimage ${\iota }_S^{-1}(U)=U\cap S$ must be open (definition continuous) $⟹U\cap S\in {\mathcal{T}}_S$ $⟹\{U\cap S|U\text{ is open in }X\}\subseteq {\mathcal{T}}_S$ We check this is indeed a topology. ($\varnothing =S\cap \varnothing$ and $S=S\cap X$ and closed under finite intersections, and arbitrary unions) So we let ${\mathcal{T}}_S=\{S\cap U|U\in \mathcal{T}\}$, and define this to be a subspace.

Examples

Warning

a subset may be open respect to the subspace topology and closed respect to a larger space

Propositions

Let $U\subseteq S\subseteq X$

• If $U$ is open (closed) in $X$ , then $U$ is open (closed) in $S$.
  • Proof: $U\subseteq X$ is open, $U=U\cap S$ $⟹$ $U$ is open in $S$.
• If $S$ is open (closed) in $X$ and $U$ open (closed) in $S$, then $U$ is open (closed) in $X$.
  • so if $S$ itself is open in $X$, then the converse of the first proposition holds.
  • Proof: $U\subseteq S$ is open $⟹$ $U=S\cap V$ for some $V\subseteq X$ open in $X$. (by definition)
  • Note that both $S$ and $V$ are open in $X$.
  • $⟹U$ is also open in $X$.

Characteristic Property

Let $S\subseteq X$ be a subspace. For any space $Y$, a function $f:Y\to S$ is continuous if and only if ${\iota }_s\circ f:Y\to X$ is continuous:

Corollary

Let $X$,$Y$,$Z$ be spaces and $f:X\to Y$ continuous

• If $S\subseteq X$, then $f_{|S}:S\to Y$ is continuous
• If $T\subseteq Y$ and $f(X)\subseteq T$, then $f:X\to T$ continuous.
• If $Y\subseteq Z$, then $f:X\to Z$ is continuous.

Inheritance

Subspaces preserves:

• Hausdorff property
• First countability
• Second countability