Let $\Omega$ be an open, bounded from $\mathbb{R}^n$ and $f: \overline{\Omega} \rightarrow \overline{\Omega}$ a contiuous function such that $f(x)=x, \forall x \in \partial \Omega$.

Prove that $f(\overline{\Omega})=\overline{\Omega}$.

how to solve it ? any idea please ?

  • 1
    $\begingroup$ Did you mean that $\Omega\subset\mathbb{R}$? $\endgroup$
    – Marra
    Feb 23, 2013 at 14:44
  • 1
    $\begingroup$ $\Omega \subset \mathbb{R}^n$ sorry $\endgroup$
    – Vrouvrou
    Feb 23, 2013 at 14:57
  • $\begingroup$ If we divide $\overline{\Omega}$ in it's connected components, I think we just have to show that if $U$ is an component, then $f(int(U))= U$ where $int$ is interior. Is this true? $\endgroup$
    – Tomás
    Feb 23, 2013 at 15:18
  • $\begingroup$ @Tomás Yes, but you probably want to look at the components of $\Omega$. Otherwise $int(U)$ doesn't need to be connected. I think one can show fairly easily that it suffices to prove the result for a connected set. $\endgroup$
    – JSchlather
    Feb 23, 2013 at 15:50
  • $\begingroup$ For the sufficiency it is enough to prove that $f(int (U))\subset U$. $\endgroup$
    – Tomás
    Feb 23, 2013 at 15:59

3 Answers 3


The problem can be solved using degree theory.

Suppose otherwise, that $f(\overline{\Omega})\subset\overline{\Omega}$ is strict. Since $f(\partial\Omega)=\partial{\Omega}\subset f(\overline{\Omega})$, then there is some $p\in\Omega$ such that $p\notin f(\Omega)$.

Since $p\notin\partial\Omega=f(\partial\Omega)=\mathbb{I}(\partial\Omega)$, then $\text{deg}(f,\Omega,p)=\text{deg}(\mathbb{I},\Omega,p)=1$ by the Poincare-Bohl theorem.

But the basic properties of degree give that $\text{deg}(f,\Omega,p)\neq0\Longrightarrow\exists x\in\Omega$ such that $f(x)=p$, so we get a contradiction.

  • $\begingroup$ What happens if $p\in f(\bar\Omega)\setminus\bar\Omega$ instead? $\endgroup$
    – Asaf Karagila
    Mar 2, 2013 at 2:41
  • $\begingroup$ $f(\overline{\Omega})\backslash\overline{\Omega}=\emptyset$ since $f:\overline{\Omega}\longrightarrow\overline{\Omega}$. $\endgroup$
    – mdg
    Mar 2, 2013 at 2:48
  • $\begingroup$ D'oh, thanks! :-) $\endgroup$
    – Asaf Karagila
    Mar 2, 2013 at 2:51
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    $\begingroup$ The various techniques which constitute the area of nonlinear (functional) analysis are all equally impressive and useful, and I believe deserve more attention and study than they get...! $\endgroup$
    – mdg
    Mar 2, 2013 at 20:09

Note that $\Omega$ can be written as a countable union of disjoint open intervals, i.e. $\Omega=\cup I_n$, where $I_n=(a_n,b_n)$. Then $f(a_n)=a_n$ and $f(b_n)=b_n$ and because $f$ is continous, $f$ satisfies the intermediate value property, hence, for each $u\in [a_n,b_n]$ with $f(a_n)\leq u\leq f(b_n)$, you can find $v\in [a_n,b_n]$ such that $f(v)=u$. This implies that $\overline{\Omega}\subset f(\overline{\Omega})$

Edit: As Ayman pointed out, I just proved that $\overline{\Omega}\subset f(\overline{\Omega})$ and because I think this might be helpful to someone, I will not delete the answer.

Remark: When I answered the question, $\Omega$ was a subset of $\mathbb{R}$.

  • $\begingroup$ The question says $f : \overline{\Omega} \to \overline{\Omega}$. So equality holds after all. $\endgroup$ Feb 23, 2013 at 15:00
  • $\begingroup$ All you needed to prove was that $\overline{\Omega}\subseteq f\left(\overline{\Omega}\right),$ since we have $f:\overline{\Omega}\to\overline{\Omega}.$ However, the question has been edited so that $\Omega\subseteq\Bbb R^n$, so you'll need to take a different approach. $\endgroup$ Feb 23, 2013 at 15:02
  • $\begingroup$ It is true, but now I have to change it. $\endgroup$
    – Tomás
    Feb 23, 2013 at 15:04
  • $\begingroup$ so you say that we have only $\overline{\Omega}\subset f(\overline{\Omega})$ ? $\endgroup$
    – Vrouvrou
    Feb 23, 2013 at 16:15
  • 1
    $\begingroup$ @KarimaMht You have $f(\overline{\Omega})\subseteq \overline{\Omega}$ by assumption. $\endgroup$
    – Julien
    Feb 23, 2013 at 16:21

Wrong Solution

If $\Omega = \emptyset$, then we are done.

The plan for nonempty $\Omega$ (as hinted at by JSchlather and Tomás) is to show that $f(\overline{U}) = \overline{U}$ for any connected component $U$ of $\Omega$. The result follows from this, since the connected components exhaust $\Omega$.

We already know $f(\overline{U}) \subset \overline{\Omega}$.

Now, $\overline{U}$ is connected (since $U$ is connected), so $f(\overline{U})$ is connected, since the continuous image of a connected set is connected.

Consequently, $f(\overline{U})$ lies entirely within one connected component of $\Omega$.

Since $f(x) = x$ on $\partial U \subset \overline{U}$, it must be that $f(\overline{U}) \subseteq \overline{U}$.

Therefore, all we need to show now is that $f(\overline{U}) \supseteq \overline{U}$.

Let $B \supset \overline{U}$ be a closed ball neighborhood of $\overline{U}$. Extend $f$ to a function $F: B \to B$ by the identity map on $B\smallsetminus\overline{U}$. Then $F$ fixes the boundary of $B$ (in addition to lots of other elements of $B$), and is continuous. Also $F$ takes $B\smallsetminus\overline{U}$ to itself, and $\overline{U}$ to itself. So it will suffice to show that $F$ is surjective for any map $F: B \to B$ from a ball to itself fixing the boundary.

Let $N_1 \supseteq N_2 \supseteq \cdots$ be a decreasing sequence of closed balls whose intersection is $x$. Let $K_i = F^{-1}(N_i)$. Then $K_1 \supseteq K_2 \supseteq \cdots$ is a decreasing sequence of closed sets such that $F^{-1}(\mbox{int }N_i) \subset K_i$; in particular, $K_i$ is nonempty. (The statement in bold is wrong, as pointed out by @G. Smith.)

Since $B$ is compact, the intersection of all the $K_i$ is nonempty; call it $\mathcal{K}$. Then $f(\mathcal{K}) \subseteq N_i$ for all $N_i$. Therefore $f(\mathcal{K}) = x$.

Now, $x$ was arbitrary, so $f$ is surjective, and we are done.

  • $\begingroup$ This is not correct because $K_i$ may be empty. Since $U$ is open and you are assuming $x\in U$, then eventually $\text{int}N_i\subset U$. But there is no guarantee that $F^{-1}(\text{int}N_i)\neq\emptyset$, which is exactly what we are trying to prove, i.e. there may be no points in $U$ that map to $\text{int}N_i$. $\endgroup$
    – mdg
    Mar 2, 2013 at 7:36
  • 1
    $\begingroup$ -blush- Well, I suppose I ought to leave this nonanswer up as a warning to those attempting a solution without using more heavy machinery. It's so tempting. $\endgroup$ Mar 6, 2013 at 20:17
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    $\begingroup$ It would be nice to be able to prove such a simple statement directly. Using degree theory seems to be shooting an ant with an atom bomb... $\endgroup$
    – mdg
    Mar 6, 2013 at 23:18

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