f is continuous, show that f(closure) is a subset of closure of f

Question

If $f:X\rightarrow Y$ is continuous and $E\subset X$, prove that $f(\overline{E})\subset \overline{f(E)}$. Provide an example to show that the inclusion does not have to be equality.

So far what I have is that the preimage of a closed set in $Y$ is closed in $X$. So $f^{-1}(\overline{f(E)})$ is closed and contains $E$.

Let $x\in \overline E$. So there is a net $x_a\to x$. But since $f$ is continuous, we have $f(x_a)\to f(x)$ i.e., $f(x)\in \overline {f (E)}$. Hence $f(\overline E)\subset \overline {f(E)}$ — Jose Antonio, Nov 05 '14 at 02:06
For some counterexamples, see also: Example of a continuous function s.t. $f(\overline{A}) \subsetneq \overline{f(A)}$. — Martin Sleziak, Sep 27 '21 at 15:17

Przemysław Scherwentke · Answer 1 · 2014-11-05T02:09:07.300

8

A counterexample to the equality: $f(x)=\arctan(x)$, $X={\bf R}$.

edited Nov 05 '14 at 02:09

answered Nov 05 '14 at 02:03

Przemysław Scherwentke

13,836

score 4 · Accepted Answer · answered Nov 05 '14 at 02:04

You're off to a great start. $f^{-1}(\overline{f(E)})$ is a closed set containing $E$. Then, since $\overline{E}$ is by one definition $E \cup E'$, where $E'$ is the set of limit points of $E$, then it must be the case that $E' \subseteq f^{-1}(\overline{f(E)})$ (why?). But then that means $\overline{E} \subseteq f^{-1}(\overline{f(E)})$, which implies $f(\overline{E}) \subseteq \overline{f(E)}$.

score 4 · Answer 3 · answered Nov 05 '14 at 02:17

You're doing fine. You've shown $E\subseteq f^{-1}(\overline{f(E)})$ and $f^{-1}(\overline{f(E)})$ is closed. Now recall that the closure of $E$ is the smallest closed set containing $E$; so it follows from what you've shown that $\overline E\subseteq f^{-1}(\overline{f(E)})$. Now recall (or prove) that $A\subseteq f^{-1}(B)\iff f(A)\subseteq B$, and apply it with $A=\overline E$ and $B=\overline{f(E)}$.

score 2 · Answer 4 · answered Jul 21 '18 at 03:10

2

A counterexample, consider $X=Y=\mathbb{R}$, $E=\overline{E}=[0,\infty) \subset \mathbb{R} $ and a continuos function $ f(x) = \frac{1}{1+x^2} $

Note that $f(E) = f(\overline{E})=(0,1] \neq \overline{f(E)} = [0,1] $

answered Jul 21 '18 at 03:10

tnt235711

421

Ari Royce Hidayat · Answer 5 · 2023-08-06T10:26:56.763

First, we need this Theorem 1:

$\overline A$ is the smallest closed set containing $A$.

By Theorem 1, we know that: $f(E) \subset \overline{f(E)}$ (of course by assuming $E$ is an open set). As $f^{-1}(f(E)) = E$, we have: $E \subset f^{-1}(\overline{f(E)})$.

Then we need this Theorem 2:

$f$ is continuous if and only if for each closed set $C$ of $Y$, $f^{-1}(C)$ is closed in $X$.

By Theorem 2, we know that $f^{-1}(\overline{f(E)})$ is closed in $X$. By Theorem 1, we know that: $E \subset \overline E \subset f^{-1}(\overline{f(E)})$.

Thus: $f(\overline E) \subset \overline{f(E)}$.

Note that if $E$ is a closed set, then $f(E) = f(\overline E) = \overline{f(E)}$.

aram · Answer 6 · 2014-11-05T02:09:19.643

1

Solution: Let $x \in f(\overline{E})$ there is some $y \in \overline{E}$ such that $f(y) = x$, now because its in the closure we have some $y_n \in E$ with $y_n \rightarrow y$, now by continuity this means that

$$\lim f(y_n) = f(\lim y_n) = f(y) = x $$

So $x \in \overline{f(E)}$, because $(f(y_n)) \in f(E)$ for all $n$ and $f(y_n)_n \rightarrow x$

edited Nov 05 '14 at 02:09

answered Nov 05 '14 at 02:03

aram

2,145

1

I think in this case is better to use nets. – Jose Antonio Nov 05 '14 at 02:07

f is continuous, show that f(closure) is a subset of closure of f

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