# characterization of subspace topology

###### Theorem.

Let $X$ be a topological space and $Y\subset X$ any subset. The subspace topology on $Y$ is the weakest topology making the inclusion map continuous.

###### Proof.

Let $\mathcal{S}$ denote the subspace topology on $Y$ and $j\colon Y\hookrightarrow X$ denote the inclusion map.

Suppose $\{\mathcal{T}_{\alpha}\mid\alpha\in J\}$ is a family of topologies on $Y$ such that each inclusion map $j_{\alpha}\colon(Y,\mathcal{T}_{\alpha})\hookrightarrow X$ is continuous. Let $\mathcal{T}$ be the intersection $\bigcap_{\alpha\in J}\mathcal{T}_{\alpha}$. Observe that $\mathcal{T}$ is also a topology on $Y$. Let $U$ be open in $X$. By continuity of $j_{\alpha}$, the set $j_{\alpha}^{-1}(U)=j^{-1}(U)$ is open in each $\mathcal{T}_{\alpha}$; consequently, $j^{-1}(U)$ is also in $\mathcal{T}$. This shows that there is a weakest topology on $Y$ making inclusion continuous.

We claim that any topology strictly weaker than $\mathcal{S}$ fails to make the inclusion map continuous. To see this, suppose $\mathcal{S}_{0}\subsetneq\mathcal{S}$ is a topology on $Y$. Let $V$ be a set open in $\mathcal{S}$ but not in $\mathcal{S}_{0}$. By the definition of subspace topology, $V=U\cap Y$ for some open set $U$ in $X$. But $j^{-1}(U)=V$, which was specifically chosen not to be in $\mathcal{S_{0}}$. Hence $\mathcal{S_{0}}$ does not make the inclusion map continuous. This completes the proof. ∎

Title characterization of subspace topology CharacterizationOfSubspaceTopology 2013-03-22 15:40:32 2013-03-22 15:40:32 mps (409) mps (409) 6 mps (409) Theorem msc 54B05