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Let $X$ and $Y$ be topological spaces, $X$ compact, $f : X \to Y$ continuous. Then the preimage of each compact subset of $Y$ is compact.

With the stipulation that $X$ and $Y$ are metric spaces, this is a theorem in Pugh’s *Real Mathematical Analysis*. The proof uses sequential compactness. Is this theorem true in general (*i.e.* can it be proved with covering compactness alone)?

- A compact Hausdorff space that is not metrizable
- Compact metrizable space has a countable basis (Munkres Topology)
- Prove that there exists a Cauchy sequence, compact metric space, topology of pointwise convergence
- Stone-Čech compactification. A completely regular topological space is locally compact iff it is open in its Stone-Čech compactification.
- What's going on with “compact implies sequentially compact”?
- Construct a metric for the topology of compact convergence on $Y^{X}$ so that $Y^{X}$ is complete when $(Y,d)$ is complete and $X$ is $\sigma$-compact

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- $\inf_{x\in}f(x)=\inf_{x\in\cap\mathbb{Q}}f(x)$ for a continuous function $f:\to\mathbb{R}$
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- Continuous function positive at a point is positive in a neighborhood of that point
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- Does $\sqrt{x}$ have a limit for $x \to 0$?
- A continuous injective function and its inverse
- Showing that $\Omega$ is of class $C^1$
- Countable compact spaces as ordinals

The claim is true if $Y$ is Hausdorff: if $C\subseteq Y$ is compact, then it is closed; therefore $f^{-1}(C)$ is closed in $X,$ hence compact.

For a counterexample, take $Y=X$ with the indiscrete topology and $f$ the identity map. Then every subset of $Y$ is compact, which can be easily arranged for $X$ not to.

This is not true in general.

Let $X=Y=[0,1]$. Take $X$ with the usual topology. For $Y$, take the topology

$$\tau=\left\{\varnothing,Y,(1/2,1]\right\}.$$

Then $id:x\in X\mapsto x\in Y$ is continuous, but $(1/2,1]=id^{-1}(1/2,1]$ is not compact, although $(1/2,1]$ is compact in $Y$.

On the other hand, if $Y$ is Hausdorff, then every compact of $Y$ is closed, so the inverse image of compact sets is closed, hence compact.

A map $f:X\to Y$ is called *proper* if the preimage of every compact subset is compact. It is called *closed* if the image of every closed subset is closed.

If $X$ is a compact space and $Y$ is a Hausdorff space, then every continuous $f:X\to Y$ is closed and proper.

Here are some examples where $f$ is not proper:

- With $X$ compact: Let $X=[0,1]$ and $f=\text{Id}:(X,\tau)\to(X,\sigma)$ where $\tau$ is the Euclidean topology and $\sigma$ is the cofinite topology.
- With $Y$ Hausdorff: Let $f:\Bbb R\to\{*\}$ be the constant map.

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