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I was reading a proof [Theorm 1.2], there says the continuity of the function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ implies the closedness of the set $\{x \mid f(x) \le \alpha\}$. Could someone refresh me on how such a claim is achieved?

Updates: The context is: a function $f$ is coercive if and only if for every $\alpha \in \mathbb{R}$ the set $\{x \mid f(x) \leq \alpha \}$ is compact.

Logan
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    What does "closeness of the set ${x}$" mean? – arkeet Oct 10 '16 at 23:15
  • What space to start with? – user251257 Oct 10 '16 at 23:16
  • Edited and added more contexts. Hope this would clarify some of the ambiguity. – Logan Oct 11 '16 at 04:19
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    No, it's still rather unclear. What does the "coercive" definition have to do with the rest of the question? Can you show us the proof you were looking at? (Either a reference or an image) – arkeet Oct 11 '16 at 07:22
  • @arkeet Added the link for your reference. – Logan Oct 12 '16 at 08:42
  • OK, that entirely changes the meaning of the question. (Closeness and closedness are rather different, and ${x}$ is a one-element set, which is not what's in your link.) I'll go ahead and edit to make the question clearer. – arkeet Oct 12 '16 at 16:59
  • A function is continuous iff the preimage of an open set is always open, or equivalently if the preimage of a closed set is always closed (proving it's equivalent is almost immediate given that the complement of a closed set is open and vice versa). That's the definition. Now, what is ${x,|,f(x)\leq \alpha}$, in terms of preimage? – Jean-Claude Arbaut Oct 12 '16 at 17:07

2 Answers2

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The preimage of any closed set is closed under a continuous map.

Nitin
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The easiest way to see that the set $\{x \mid f(x) \le \alpha\}$ is closed is that it is the preimage by $f$ of the closed subset $(-\infty,\alpha]$ of $\mathbb{R}$. (Remember that in general, if $f \colon X \to Y$ is continuous and $A \subseteq Y$ is closed, then $f^{-1}(A)$ is closed.)

Alternatively, you could show that for any convergent sequence $(x^\nu)$ in $\mathbb{R}^n$ with $f(x^\nu) \le \alpha$ for all $\nu$, you would see that $f(\lim_{\nu\to\infty} x^\nu) = \lim_{\nu\to\infty} f(x^\nu) \le \alpha$ (where the equality holds by continuity of $f$).

arkeet
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  • But what about this question tells you that $f(x)$ is closed, though? If you know that $f(x)$ is closed, then I could understa?d why, from the theorem you cited would allow us to say that ${x|f(x) \le \alpha }$ is closed. But otherwise, how can you draw that conclusion? – makansij Apr 21 '18 at 23:04