Past exam of the options I course of the University of Oxford / 2007 / E (Topology)

A sequence converges if there is $x\in X$ such that, for every $ϵ>0$, some $N$ satisfies $d(x_n,x)<ϵ$ whenever $n\ge N$. The limit must belong to $X$, not merely to a larger ambient space.

A metric space is sequentially compact if every sequence in it has a subsequence converging to a point of the same space.

The sequence $x_n=1/(n+1)$ lies in the open interval and converges in the ambient real line to zero. Every subsequence also converges to zero, and uniqueness of limits prevents it converging to any point of the interval. Thus the interval is not sequentially compact.

For a sequence $(x_n,y_n)$ in the product, sequential compactness of $X$ supplies indices $n_j$ with $x_{n_j}\to x\in X$. Apply sequential compactness of $Y$ to the corresponding $y_{n_j}$, extracting a further subsequence $y_{n_{j_{\ell }}}\to y\in Y$. The first coordinate still tends to $x$. For any $ϵ>0$, both coordinate distances are eventually less than $ϵ/2$, so their sum is less than $ϵ$. Hence the same subsequence converges in the given product metric to $(x,y)$. This proves the product is sequentially compact.

The subspace topology isThese sets contain the empty set and $A$, are closed under arbitrary unions, and are closed under finite intersections because intersections distribute over the common factor $A$. Thus they form a topology on $A$.

For $U\in \mathcal{T}$, the inverse image under inclusion is $i^{-1}(U)=A\cap U$, which is open in the subspace topology by definition. The inverse-image criterion for continuity therefore proves the inclusion is continuous.

Every subspace-open set has the form $A\cap U=i^{-1}(U)$ for some $U\in \mathcal{T}$. Continuity of $i:(A,{\mathcal{T}}^{\prime })\to (X,\mathcal{T})$ makes every such inverse image open in ${\mathcal{T}}^{\prime }$. Hence $\boxed{{\mathcal{T}}_A\subseteq {\mathcal{T}}^{\prime }}$.

If $A\cap U$ is subspace-open, thenThe right side is open in $Z$ by the assumed continuity of the composition. This holds for every subspace-open set, proving $f$ is continuous into the subspace topology.

Choose $Z=(A,{\mathcal{T}}_A)$ and let $f$ be the identity function on $A$. The composite $i\circ f$ is continuous by part (i). The stipulated universal implication therefore makes $f:(A,{\mathcal{T}}_A)\to (A,{\mathcal{T}}^{\prime })$ continuous. Its inverse image of any ${\mathcal{T}}^{\prime }$-open set is that set itself, so it must lie in ${\mathcal{T}}_A$. Thus $\boxed{{\mathcal{T}}^{\prime }\subseteq {\mathcal{T}}_A}$.