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We know that continuous image of a compact set is closed and bounded. But, compactness is not necessary to claim that. I want to know is there any theorem that says that continuous image of some special kind of non-compact set is also closed and bounded.

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  • $\begingroup$ Work up some examples. $\endgroup$ – Lubin Oct 9 '15 at 4:07
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No, there can't be any such theorem. Here's a negative result.

Proposition. Let $(X,d)$ be a metric space, and suppose $E \subset X$ is not compact. Then there exists a continuous function $f : X \to \mathbb{R}$ such that the image $f(E)$ is not closed.

Proof. By the Bolzano-Weierstrass theorem, there is a sequence $\{x_n\} \subset E$ which does not have a limit point in $E$. (Without loss of generality, assume all the points $x_n$ are distinct.) This sequence may or may not have a limit point in $X$.

If $\{x_n\}$ does have a limit point in $X \setminus E$, call it $p$, then let $f(x) = d(p,x)$. Since $p \notin E$, we have $0 \notin f(E)$. But since $x$ is a limit point of $\{x_n\}$, we have $f(x_n) = d(p, x_n) \to 0$, so $0$ is a limit point of $f(E)$. Thus $f(E)$ is not closed.

If $\{x_n\}$ does not have a limit point in $X$, then $D := \{x_n\}$ is a closed discrete set. So any function on $D$ is continuous; in particular, consider the function $g : D \to \mathbb{R}$ defined by $g(x_n) = n$. By the Tietze extension theorem, $g$ has a continuous extension $\bar{g} : X \to \mathbb{R}$, so that $\bar{g}(x_n) = n$ as well. Set $f(x) = 1/(1 + \bar{g}(x)^2)$. Clearly $0 \notin f(X)$ so $0 \notin f(E)$. But $f(x_n) = 1/(1+n^2) \to 0$, so $0$ is a limit point of $f(E)$. Thus $f(E)$ is not closed. $\Box$

Getting a function which is not bounded could be a little trickier. For instance, if $E$ is not compact but is contained in a compact set $F$ (such an $E$ is sometimes called "precompact" or "relatively compact"), then for every metric space $Y$ and every continuous $f : X \to Y$, $f(E)$ will be bounded (since it is a subset of $f(F)$ which is compact and thus bounded).

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