Past exam of the probability, statistics and options I course of the University of Oxford / 2007 / E / 2

The subspace topology is ${\mathcal{T}}_A=\{A\cap U:U\in \mathcal{T}\}$. For every open $U\subseteq X$,The inverse-image definition of continuity therefore gives the inclusion is continuous from $(A,{\mathcal{T}}_A)$ to $(X,\mathcal{T})$.

If the inclusion is continuous from $(A,{\mathcal{T}}^{\prime })$, then $i^{-1}(U)=A\cap U$ belongs to ${\mathcal{T}}^{\prime }$ for every $U\in \mathcal{T}$. These intersections are precisely all members of ${\mathcal{T}}_A$, soThus the subspace topology is the coarsest topology on $A$ making the inclusion continuous.

Every open set of the subspace is $A\cap U$ for some $U\in \mathcal{T}$. Since $f$ takes values in $A$,The right side is open in $Z$ by the assumed continuity of the composition. Hence preimages of every ${\mathcal{T}}_A$-open set are open, proving $f:Z\to (A,{\mathcal{T}}_A)$ is continuous.

Choose $Z=(A,{\mathcal{T}}_A)$ and let $f$ be the identity set map $A\to A$. The composition $i\circ f$ is the inclusion and is continuous by part (i). The hypothesized universal implication therefore says that the identity mapis continuous. Its preimage of a set $V\in {\mathcal{T}}^{\prime }$ is $V$ itself, so every such $V$ lies in ${\mathcal{T}}_A$. ThereforeCombining this with part (ii), when both properties hold, characterizes the subspace topology uniquely.