Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 2 / 4A (Metric and Topological Spaces)

True. Let $\{V_{\alpha }\}$ be an open cover of $f(S)$ in its subspace topology. Write $V_{\alpha }=f(S)\cap W_{\alpha }$ with $W_{\alpha }$ open in $Y$. Since $f$ is continuous, the sets $S\cap f^{-1}(W_{\alpha })$ are open in $S$ and cover $S$. By compactness, finitely many cover $S$. Their corresponding $V_{\alpha }$ then cover $f(S)$, proving that the continuous image of a compact space is compact.

False. Give $X=\mathbb{R}$ its usual topology, let $Y$ be a one-point space, and take the constant continuous map $f:X\to Y$. The set $K=Y$ is compact, but $f^{-1}(K)=\mathbb{R}$ is not compact. For example, the open cover $\{(-n,n):n\ge 1\}$ has no finite subcover. Thus continuity alone does not preserve compactness under inverse images.

False. The real line with its usual metric is complete and is homeomorphic to $(0,1)$ with its usual metric, throughIts inverse is $h^{-1}(u)=\tan (\pi (u-1/2))$, so both maps are continuous. However, $(1/n{)}_{n\ge 2}$ is a Cauchy sequence in $(0,1)$ and has no limit there: its only possible real limit is zero. Hence $(0,1)$ is not a complete metric space. Completeness is a property of the metric, not just of its topology.