Example of a continuous function s.t. $f(\overline{A}) \subsetneq \overline{f(A)}$ This question is a subproblem of the A map is continuous if and only if for every set, the image of closure is contained in the closure of image. 
I am not able to come up with any example of a continuous function s.t. $f(\overline{A}) \subsetneq \overline{f(A)}$. Can anyone give such an example?
 A: Hint : consider $f = \arctan$ and $A = \Bbb R$. 
A: Take $f\colon\mathbb{R}\longrightarrow\mathbb R$ define by $f(x)=\frac1{1+x^2}$ and $A=\mathbb R$. Then $\overline A=\mathbb R$, $f(A)=(0,1]$, and $\overline{f(A)}=[0,1]$.Another example is the identity function $f$ from $\mathbb R$ endowed with the discrete metric into $\mathbb R$ endowed with the usual metric. Take $A=\mathbb Q$. Then $\overline A=\mathbb Q$ (with respect to the discrete metric). But $\overline{f(\mathbb{Q})}=\overline{\mathbb{Q}}=\mathbb R$ (with respect to the usual metric). 
A: Let $f:X \to Y$ be any continuous, non-surjective function with dense range in $Y$. Then 
$$f(\overline{X}) = f(X) \neq \overline{f(X)} = Y$$ 
An explicit example is given by $f: \ell^2 \to \ell^2$ defined by $f(x) = (2^{-n} x_n)_{n \geq 1}$ since the space $c_{00}$ is contained in the range of $f$ and hence the range of $f$ is dense but e.g. the sequence $(2^{-n})_{n \geq 1}$ is easily seen to not be in the range of $f$.
A: Let $f:X\to Y$ where $Y$ is equipped with indiscrete topology. 
Then automatically $f$ is continuous and $\overline{f(A)}=Y$ for every non-empty $A\subseteq X$.
For suitable $X$ plenty examples can be found with $\varnothing\neq A\subseteq X$ and$f(\overline{A})\neq Y$.
