Subspace topology

The subspace topology is a natural way of reframing a subspace as a whole space. Let $(𝑋, T 𝑋)$ be a topological space, and $𝑌 \subseteq 𝑋$ be any subset with the canonical inclusion $𝑖 : 𝑌 ↣ 𝑋$ . The subspace topology on $𝑌$ is the coarsest topology for which the canonical inclusion is continuous.¹ #m/def/topology

T 𝑌 = {𝑖 - 1 𝑈 : 𝑈 \in T 𝑋}

More generally, if $𝑓 : 𝑆 ↣ 𝑋$ is an injective map, then the subspace topology induced by $𝑓$ is the coarsest topology for which $𝑠$ is continuous. In this case $𝑓$ is an embedding.

Further characterisations

Explicit

Let $(𝑋, T)$ be a topological space and $𝑌 \subseteq 𝑋$ be a subset. A subset $𝑉 \subseteq 𝑌$ is then open relative to $𝑌$ iff there exists an open subset $𝑈$ (relative to $𝑋$ ) such that $𝑉 = 𝑈 \cap 𝑌$ .² The system $T 𝑌$ of all subsets open relative to $𝑌$ is called the subspace topology induced by $𝑋$ , and $(𝑌, T 𝑋)$ forms a topological space.

Universal property

For every topological space $(𝑍, T 𝑋)$ and every map $𝑓 : 𝑍 \to 𝑌$ , then $𝑓$ is continuous iff $𝑖 𝑓 : 𝑍 \to 𝑋$ is continuous. #m/thm/topology

Proof

First we will prove that the subspace topology as characterised above satisfies the universal property. Let $(𝑋, T 𝑋)$ be a topological space and let $𝑌 \subseteq 𝑋$ be a subset endowed with the subspace topology $T 𝑌$ . Let $(𝑍, T 𝑍)$ be some topological space, and $𝑓 : 𝑍 \to 𝑌$ be a function. If $𝑓$ is continuous, then so is the composition $𝑖 𝑓$ of continuous functions. Now suppose $𝑖 𝑓 : 𝑍 \to 𝑋$ is continuous, and let $𝑈 \in T 𝑌$ . Then $𝑈 = 𝑖 - 1 𝑉$ for some $𝑉 \in T 𝑋$ . Since $𝑖 𝑓$ is continuous, $𝑓 - 1 𝑈 = (𝑖 𝑓) - 1 𝑉 \in T 𝑍$ , thus $𝑓$ is continuous. Therefore $𝑓$ is continuous iff $𝑖 𝑓$ is continuous.

Now let $T'$ be a topology on $𝑌$ satisfying the universal property. In particular, let $(𝑍, T 𝑍) = (𝑌, T 𝑌)$ and $𝑓 = i d 𝑌 : 𝑦 \mapsto 𝑦$ . Then since $𝑖 i d 𝑌 = 𝑖$ is continuous so is $i d 𝑌$ , wherefore $T'$ is coarser than $T 𝑌$ Now let $(𝑍, T 𝑍) = (𝑌, T')$ with $𝑓 = i d 𝑌$ . Since $i d 𝑌$ is continuous, so too is $𝑖 i d 𝑌 = 𝑖$ . But $T 𝑌$ is the coarsest topology on $𝑌$ for which $𝑖$ is continuous, therefore $T' = T 𝑌$ .

Properties

The subspace is closed iff $𝑖$ is a closed map
The subspace is open iff $𝑖$ is an open map

#state/tidy| #lang/en | #SemBr

2020, Topology: A categorical approach, p. 25–26 ↩
2010, Algebraische Topologie, p. 9 (Definition 1.2) ↩

Subspace topology

Further characterisations

Explicit

Universal property

Properties

Footnotes