Show with an exemple that the inclusion could be real

Question

Im trying to Solve this problem. I have started to Solve the first part, but need to show it with an example. Thankfull for help.

Answer 1

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = \frac{x^2}{x^2 + 1}$. Let $A = [0,\infty)$. Note that $A$ is already closed so $\bar{A} = A$.

Observe that $f(A) = f(\bar{A}) = [0,1)$. Hence $\overline{f(A)} = [0,1]$.

Hence in this example, $f(\bar{A}) \neq \overline{f(A)}$.

Answer 2

For $Y$ take a metric space such that the topology on it is not the discrete one.

Now let $X$ have the same underlying set and let it be equipped with discrete topology.

Then also $X$ is a metric space. This with metric $d\left(u,v\right)=0$ if $u=v$ and $d\left(u,v\right)=1$ otherwise.

Let $f$ be prescribed by $x\mapsto x$.

Then $f$ is continuous (as any function having the discrete space as its domain).

Let $A$ be some set that is not closed in $Y$. Sets like that exist since the topology on $Y$ is not discrete.

Observe that $f\left(\overline{A}^{X}\right)=f\left(A\right)=A\varsubsetneq\overline{A}^{Y}=\overline{f\left(A\right)}^{Y}$.