Show that for any $E\subseteq X$ we have $f(\overline{E})$ $\subseteq$ $\overline{f(E)}$.

Question

Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces and $f : X → Y$

(a) Assume that $f$ is continuous. Show that for any $E\subseteq X$ we have $f(\overline{E})$ $\subseteq$ $\overline{f(E)}$.

(b) Assume that $f$ is continuous. Show that for any $B\subseteq X$ we have $\overline{f^{−1}(B)}$ $\subseteq$ $f^{−1}( \overline B)$

I am having a difficulty understading the differences between these closures. Can I pleasure get further explanation as to what the following differ from one another?

$f(\overline{E})$ vs $\overline{f(E)}$

and

$\overline{f^{−1}(B)}$ vs $f^{−1}( \overline B)$

Answer 1

When we apply successive operations on mathematical objects, the order that we apply them in is important. It makes a difference : if I multiply a number by $3$ and subtract $2$, I would most likely get a different number from the one obtained if I subtracted by $2$ and then multiplied by $3$, isn't it?

In this case, image and closure are the two operations. $f(\overline E)$ is the image, of the closure of $E$, while $\overline {f(E)}$ is the closure, of the image of $E$. Note the difference in the order of application : this tells you that most likely, these operations are to differ in their results.

Now, given $f(\overline E)$, you want to show that it lies in $\overline {f(E)}$. So let $x \in f(\overline E)$. This means, that $x = f(y)$ for some $y \in \overline E$, therefore there is a sequence $y_n \in E$ which converges to $y$. By continuity, $f(y_n) \in f(E)$ converges to $f(y) = x$, but then this implies that $x \in \overline {f( E)}$, since we found a sequence converging to $x$.

A similar argument applies for the second question : use definitions, these are the key for beginners in topology.

EDIT : Of course, the counterexample is vitally important. We'll do better : we will try to find the problem in the converse statement.

Let $x \in \overline {f(E)}$. Then, there is a sequence of points $x_n \in f(E)$ such that $x_n$ converge to $x$. But, $x_n$ are in the image of $f_n(E)$, so there exist $y_n \in E$ so that $f(y_n)$ converges to $x$.

We wanted to prove that $x$ is in the image of the closure of $E$. But there is a clear problem here : $y_n$ need not be convergent! That is, we can't really do anything with this sequence $y_n$, it won't help give us the point whose image we would like to be $x$.

A nice example of this is given by taking $\mathbb N$ with the standard topology, and taking the function $f : \mathbb N \to \mathbb R$ given by $f(x) = \frac 1x$. Take $E = \mathbb N$. Clearly, $\overline E = E$, so $f(\overline E) = f(\mathbb N) = (0,1]$, while $\overline {f(E)}$ is just the closure of $(0,1]$, which is $[0,1]$. The two sets are clearly different.

Aside : If it turns out that for all $E$, we have equality above, if and only if, $f$ is a continuous bijection whose inverse is also continuous i.e. $f$ is a homeomorphism.

Answer 2

$f(x)=x$ for $x>0$, $f(0)=-1$, $f(x)=x-2$ for $x<0$, then with $E=(0,1)$, $\overline{E}=[0,1]$, so $f(\overline{E})=\{-1\}\cup(0,1]$. But $f(E)=(0,1)$ and hence $\overline{f(E)}=[0,1]$, so $f(\overline{E})$ is not contained in $\overline{f(E)}$, one sees from graph that $f$ is not continuous.

Answer 3

To show that the inclusion could be strict : take $f = Arctan$ and $E = \Bbb R$. Then $f(\overline{E}) = (-\frac \pi2, \frac \pi 2) $ while $\overline{f(E)} = [-\frac \pi2, \frac \pi 2]$.