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Let $f: \mathbb R^2 \to \mathbb R^2$ be a continuous function ; then is it true that there is a non-empty proper closed subset $A \subseteq \mathbb R^2$ such that $ A \subseteq f(A)$ ?

I can show that if $f: \mathbb R^n \to \mathbb R^n$ is continuous then there is a non-empty proper closed subset $A \subseteq \mathbb R^n$ such that $ f(A) \subseteq A$ ; but I have no idea on what happens if we want a reverse inclusion . Please help . Thanks in advance

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Not true. In fact let me give an example for $\mathbb R$ already. Let $f$ be a strictly increasing function bounded from below by $C$ and let $f(x)>x$ (there are plenty of functions like this). Now towards contradiction let $A$ be one of the subsets you desire. $A$ has a minimal point $p$ as the image of $f$ is bounded from below (and $A$ is closed). Now the point $f(p)$ is the minimal point of the image of $A$ as $f$ is increasing and $f(p)>p$. In particular $p$ is not in $f(A)$.

rschwieb
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Patrik
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  • The domain and range of $f$ is $\mathbb R^2$, not $\mathbb R.$ What do you mean by $f(x)>x$? – DanielWainfleet May 15 '17 at 08:50
  • Well the statement is false for R. It is easy to extend to a counterexample for any dimension (use this f on all coordinates for example) – Patrik May 15 '17 at 08:55
  • I see what you mean now. But in $\mathbb R^2,$ the closed set ${(x,y): 0<x \land 0<y \land xy\geq 1}$ has no minimum first or second co-ordinate – DanielWainfleet May 15 '17 at 09:11
  • Maybe that's not the easiest way to extend the counterexample. Probably easier to first project to the x-axis :) – Patrik May 15 '17 at 09:30
  • For $\Bbb R\to\Bbb R$, we have $e^x$, for instance. How would you generalise that function to $\Bbb R^2\to\Bbb R^2$? – Arthur May 15 '17 at 11:32
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    (x,y) to (e^x,0) works – Patrik May 15 '17 at 12:07
  • Ahh, I think I had something reversed when thinking about this. You're right, it is easy to generalise from $\Bbb R$ to $\Bbb R^2$. – Arthur May 15 '17 at 12:09
  • @Arthur. Yes.$ F(x,y)=(e^x,0)$ works. But the $f:\mathbb R\to \mathbb R$ in the Answer should have an additional condition: $\inf (f(x)-x)>0.$ This takes care of the case where $\phi\ne A\subset \mathbb R^2$ is closed but $\min {x:(x,y)\in A}$ does not exist, because any $(x,y)\in A$ with $x$ close enough to $\inf {x:(x,y)\in A}$ will not be in $f(A)$. – DanielWainfleet May 15 '17 at 19:25
  • (f(x),f(y)) also works and no extra condition is needed (while I realized after posting my first comment it is significantly harder to prove than (f(x),0), it is a fun exercise :) and I might still have missed a very easy way to do it). – Patrik May 16 '17 at 13:42