continuous image of non-compact sets 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.
 A: 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).
