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Let $$ f:X\rightarrow Y $$

If $f$ is defined on $X$ such that every compact set in $X$ is mapped to a compact set in $Y$. Is $f$ a continuous map?

MrDi
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2 Answers2

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Not necessarily. Let

$$f:\Bbb R\to\Bbb R:x\mapsto\begin{cases} 0,&\text{if }x\le 0\\ 1,&\text{if }x>0\;. \end{cases}$$

clearly $f[A]$ is compact for every $A\subseteq\Bbb R$, but $f$ is certainly not continuous.

Brian M. Scott
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  • If $A=[-1,1]$ then is $f[A]=[0,1]$ or ${0,1}$? – MrDi Jan 17 '16 at 01:04
  • @MrDi: The range of $f$ is the two-point set ${0,1}$; there is no $x\in\Bbb R$ such that $f(x)=\frac12$, for instance. Thus, $f[A]={0,1}$. – Brian M. Scott Jan 17 '16 at 01:05
  • Is ${0,1}$ closed set? – MrDi Jan 17 '16 at 01:05
  • @MrDi: Of course: every finite subset of $\Bbb R$ is closed. – Brian M. Scott Jan 17 '16 at 01:06
  • If $f$ is continuous but $X$ is not compact then is it true that $f$ can't be uniformly continuous? – MrDi Jan 17 '16 at 01:20
  • @MrDi: No: a constant function, for instance, is uniformly continuous. – Brian M. Scott Jan 17 '16 at 01:20
  • @BrianM.Scott - after reading several articles (and noticed the OP used this vernacular), is reading $f : X \to Y$ as “$f$ is a $Y$ valued map on a set $X$” an effective way of communicating the data of a map? e.g., $f$ is a real-valued map on a set $M$ means $f : M \to \mathbb{R}$. – Taylor Rendon Nov 13 '20 at 21:18
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    @TaylorRendon: In the case of real-valued, integer-valued, complex-valued, and perhaps two-valued it’s pretty standard, but $Y$-valued sounds a bit odd to me, though I expect that it would be understood. – Brian M. Scott Nov 13 '20 at 21:59
  • @BrianM.Scott - gotcha, that’s what I was thinking as well! Regarding the domain of the map $f : X \to Y$, do you think it would be better to say “$f$ is a map defined on $X$” or “$f$ is a map on $X$“? I’ve seen both, but I’d assume the former would be more precise. – Taylor Rendon Nov 13 '20 at 23:04
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    @TaylorRendon: I would normally say simply that $f$ is a function (or occasionally a map) from $X$ to $Y$, – Brian M. Scott Nov 13 '20 at 23:06
  • @BrianM.Scott - wonderful, thank you for your time answering my questions! (-: – Taylor Rendon Nov 13 '20 at 23:39
  • @TaylorRendon: You’re welcome! – Brian M. Scott Nov 13 '20 at 23:40
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Let $X$ be some topological space and $Y$ some finite topological space. Then every subset of $Y$ is compact, so every map $f \colon X \to Y$ maps compact sets to compact sets. But not every function into a finite topological space is continuous.

Jendrik Stelzner
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