A continuous function maps a compact set to a compact set. Is the converse of this true? That is, is a function that maps every compact set to a compact set necessarily continuous?
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2Every function with a finite range obviously satisfies the condition, and you should be able to produce examples of this which are not continuous. – Mariano Suárez-Álvarez Dec 19 '15 at 06:11
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No. Consider the discontinuous function $$f(x) = \begin{cases} 1 & x \geq 0\\ -1 & x <0.\end{cases}$$ Then for any subset $A \subseteq \mathbb{R}$ (not just compact subsets), $f(A)$ is either $\emptyset$, $\{-1\}$, $\{1\}$, or $\{-1, 1\}$, all of which are compact.
Michael Albanese
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Wow. This even works if you replace "continuous" with "locally continuous everywhere". – John Dvorak Dec 19 '15 at 07:14
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1My mistake. A function that's locally continuous everywhere is continuous. OTOH there is a difference between locally monotonous everywhere and monotonous globally. There is also monotonous on every defined interval, which is again slightly different. – John Dvorak Dec 19 '15 at 10:00
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The characteristic function of the rationals maps every set, in particular every compact set, to a compact set. But it is discontinuous at every point.
zhw.
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The most bare-bones counterexample is probably the identity function from a two element set equipped with the trivial topology to a two element set equipped with a nontrivial topology.
Alex Provost
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