Let $A \subseteq \mathbb R^n $ such that for every continuous function $f : A \to \mathbb R$ , $f(A)$ is closed in $\mathbb R$ ; then I know that $A$ is bounded ; my question is , is $A$ closed in $\mathbb R^n$ ? ( If we changed the co-domain from $\mathbb R$ to $\mathbb R^n$ , the answer would be trivially yes , but I don't know what happens when the co-domain is real line ) . Please help . Thanks in advance
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I don't think A is closed in general. There's no easy way I can think of to prove it, but there's some good stuff here under the properties of continuous functions: their images are compact if the pre-image is compact. Images under continuous functions are connected if their pre-image is connected; closedness is not in general preserved under continuous maps, only compactness. – Bourque Jul 13 '16 at 05:53
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@Bourque : I know very well closed-ness is not preserved , and I am infact asking the converse question than what you think or linked . I am trying to conclude about domain from given structure of image whereas what you are saying is concluding about image from given structure of domain .. – Jul 13 '16 at 05:58
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Perhaps a search for counterexamples would be warranted. Take for example the closure of the graph of $y=\sin\left(\frac{1}{x}\right)$ but minus the point $(0,0)$ for $A$. Can it be continuously mapped to a non-closed set in $\mathbb{R}$? – John Wayland Bales Jul 13 '16 at 06:05
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Assume $A$ is not closed. Then there is some $x_0 $ in the closure of $A$, but not in $A$.
This implies that the continuous function $f(x) := |x - x_0| $ assumes only positive values on $A$, but the closure of the image contains $0$, which contradicts closeness of $f(A)$.
Hence, $A$ is closed.
PhoemueX
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