How can a continuous function induce a proper inclusion $f(\overline{A})\subsetneq \overline{f(A)}$? Let $f:(X, d_X)\longrightarrow (Y, d_Y)$ be a continuous function between two metric spaces, $A\subseteq X$. We have $f(\overline{A})\subseteq \overline{f(A)}$ from this question. Can you please provide a counter-example to $f(\overline{A})\supseteq \overline{f(A)}$? I.e. in which cases is the forward inclusion $f(\overline{A})\subseteq \overline{f(A)}$ proper?
My thoughts are that $f$ creates a new set of sequences that are not the images $(f(x_n))$ of sequences $(x_n)$ in $X$, with limits in $\overline{f(A)}$ but not in $f(\overline{A})$. I cannot describe those sequences I'm thinking of.
 A: For instance, if $f(x) = e^x$, then $(0, \infty) = f(\overline {\Bbb R}) \subsetneq \overline{f(\Bbb R)} = [0, \infty)$.
If a function is closed, in the sense that it maps closed sets to closed ones, then it satisfies the requirement:
$$\overline A \supset A \implies f(\overline A) \supset f(A)  \implies f(\overline A) = \overline {f(\overline A)} \supset \overline {f(A)}$$
A: For a counter example you have to have a limit point  $y\in\overline f(A)$ that is not in $f(A)$ and $y \ne f(x)$ for any limit point of $A$.  As $f$ is continuous if $a_n \rightarrow x$ then $f(a_n) \rightarrow f(x)$ so for the only way for this to happen that I could see we'd need $a_n$ diverge but $f(a_n)$ converge.  
So my thought was to have a (partially) unbounded $A$ that doesn't have boundery  points wheres $f(A)$ is bounded and does.
My thought was to manipulate $A = \mathbb N$ (because {n} diverges as $n \rightarrow \infty$) and $f(x) = 1/x$ (because {1/n} converges as $n \rightarrow \infty$) so $0 = \lim_{n\rightarrow \infty} 1/n \in \overline {f(A)}$ but $0 \not \in f(\overline A)$ .  
But I like the other posters' answers of $f(x) = e^x$ better which is the same idea.  Namely $\mathbb R$ has no lower bound but $f(\mathbb R)$ does. 
(I think I'd need some tweaking as $f(x) = 1/x$ isn't continuous and f(0) isn't defined if I include 0 in the domain.  I guess this isn't a deal breaker but it didn't feel right to not have 0 in the domain but to have it in the range space.  
(I finally used $f(x) = x$ if $x \le 1$ and $2- 1/x$ if $x > 1$ so $f: \mathbb R \rightarrow (-\infty, 2) \subset \mathbb R$ so for $A = \mathbb N$ $f(A) = f(\overline A) = \{2 - 1/n\}$ whereas $\overline {f(A)} = \{2 - 1/n\}\cup \{2\}$.)
(But I think $f(x) = e^x$ is cleaner and neater.)
A: The continuous image of a closed set is not necessarily closed.
Example 1: Let $X=R^2$ and $Y=R$ with the usual topologies. The projection $f:(x,y)\to x$ to the first co-ordinate is continuous. The set $\{(x,1/x): x>0\}$ is closed in $R^2$ but its image is $(0,\infty)$ which is not closed in $R.$
Example 2: Let X=(0,1] and Y=[0,1] with the usual topologies. Let $f$ be  the identity embedding $(f(x)=x)$ of $X$ into $Y$. Then $X$ is certainly closed in $X$, but its image is not closed in $Y.$
