characterization of subspace topology

Theorem.

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

Proof.

Let $𝒮$ denote the subspace topology on $Y$ and $j:Y\hookrightarrow X$ denote the inclusion map.

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

We claim that any topology strictly weaker than $𝒮$ fails to make the inclusion map continuous. To see this, suppose ${𝒮}_0⊊𝒮$ is a topology on $Y$. Let $V$ be a set open in $𝒮$ but not in ${𝒮}_0$. 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 ${𝒮}_{\mathcal{0}}$. Hence ${𝒮}_{\mathcal{0}}$ does not make the inclusion map continuous. This completes the proof.∎