Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 1 / 10F (Analysis and Topology)

Assume inverse images of open sets are open. If $x_n\to a$ and $V$ is any open neighbourhood of $f(a)$, then $f^{-1}(V)$ is an open neighbourhood of $a$. By the definition of convergence in a metric space, $x_n\in f^{-1}(V)$ eventually, and hence $f(x_n)\in V$ eventually. Thus $f(x_n)\to f(a)$.

Assume the sequential condition, and let $A\subseteq Y$ be open. If $f^{-1}(A)$ were not open, some $a\in f^{-1}(A)$ would have no ball contained in $f^{-1}(A)$. For each $n$, chooseThen $x_n\to a$, so $f(x_n)\to f(a)\in A$. Because $A$ is open, this forces $f(x_n)\in A$ eventually, a contradiction. Therefore $f^{-1}(A)$ is open. This proves the sequential characterization of continuity in metric spaces.

This is always true: it is the Heine-Cantor theorem. If uniform continuity failed, there would be an $\epsilon >0$ and sequences $x_n,y_n\in X$ such thatBy sequential compactness of a compact metric space, some subsequence $x_{n_k}$ converges to $x\in X$. The triangle inequality gives $y_{n_k}\to x$. Continuity then makes both image subsequences converge to $f(x)$, contradicting their separation by $\epsilon$.

This may be false because continuity only forces the continuous image of a compact space $f(X)$ to be compact, not the whole codomain. For example, let $X=\{0\}$, let $Y=\mathbb{R}$, and set $f(0)=0$. The domain is compact and $f$ is continuous, but $\mathbb{R}$ is not compact.

This is always true. The continuous image $f(X)$ of a connected space is connected. Its closure is also connected: if the closure were separated into disjoint nonempty relatively open sets, connectedness would put $f(X)$ inside one of them, preventing its closure from meeting the other. Since $f(X)$ is dense in $Y$, its closure is $Y$, so $Y$ is connected.

This is always true for metric spaces. Let $x_n\to x$. If $f(x_n)$ did not converge to $f(x)$, some subsequence would remain at least $\epsilon >0$ from $f(x)$. Compactness of $Y$ gives a further subsequenceThen $(x_{n_k},f(x_{n_k}))\to (x,y)$. The graph is closed, so $(x,y)$ belongs to it and $y=f(x)$, contradicting the $\epsilon$ separation. Thus $f(x_n)\to f(x)$ for every convergent sequence, and the equivalence proved above makes $f$ continuous. This is the closed-graph criterion with compact codomain.