5

Being $F : \mathbb R^n\to \mathbb R^n$ a continuous function such that for every $x,y\in \mathbb R^n$, $$\|x − y\| \leq\|F(x) − F(y)\|\,.$$ I'd like to prove that the image of $F$ is a closed set in $\mathbb R^n$.

So far I have tried to use that if $x_n$ is a sequence in $\mathbb R^n$ such that $F(x_n)$ is a Cauchy sequence, $x_n$ must be Cauchy as well, in addition to the fact that any closed set has to contain all the limits of its sequences, but with this approach I haven't been able of find a proof.
Any other ideas on how to solve this proof?

Hanno
  • 6,302
takama245
  • 51
  • The answer here applies to this case also with an obvious modifcation: https://math.stackexchange.com/questions/2657549/if-f-mathbbrn-rightarrow-mathbbr-such-that-alpha-x-y-leq-fx?rq=1 – Kavi Rama Murthy Nov 28 '20 at 12:24

2 Answers2

3

A set $C$ in $R^n$ is closed exactly iff all the Cauchy sequences in $C$ converge to an element inside $C$. You can carry a Cauchy sequence in $F(C) $ to a cauchysequence in $C$ with the preimage. This Cauchy sequence converges in $C$. Using continuity of $F$ we get that the sequence in $F(C) $ converges as well to an element of $F(C) $.

Dude1662
  • 101
0

Let $\{y_n\}_{n\in \mathbb{N}}$ be a convergent sequence in $Im(F)$, since $\mathbb{R}^n$ is a complete space, $y_n$ is a Cauchy sequence, so, for a given $\epsilon >0;~\exists n_0\in \mathbb{N}$ such that $\lVert y_n-y_m\rVert <\epsilon ~\forall n,m\geq n_0$.

Now, construct a sequence $\{x_n\}_{n\in \mathbb{N}}$ putting $x_n$ as any element of $f^{-1}(y_n)$. With this you have that for $n,m\geq n_0; \lVert x_n-x_m\rVert \leq\lVert y_n-y_m\rVert <\epsilon$. That is, $x_n$ is Cauchy, so is convergent.

Let $x=lim(xn)$, using that $F$ is continuous we have that $lim(y_n)=lim(F(x_n))=F(lim(x_n))=F(x)$. That is, $y_n$ converge to a point in $Im(F)$, so, this is a closed set.

PrV
  • 151