homeomorphism

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August 12th 2009, 10:41 AM
Kat-M

homeomorphism

Let $U \subseteq R^n$ be an open set. Suppose that the map $h:U \rightarrow R^n$ is a $homeomorphism$ $\underline{onto}$ $R^n$ and is $uniformly$ $continuous$ .
Prove that $U=R^n$ .
August 12th 2009, 11:16 AM
siclar

Quote:

Originally Posted by Kat-M

Let $U \subseteq R^n$ be an open set. Suppose that the map $h:U \rightarrow R^n$ is a $homeomorphism$ $\underline{onto}$ $R^n$ and is $uniformly$ $continuous$ .
Prove that $U=R^n$ .

A homeomorphism is by definition a bijection, thus one to one and onto, so the assumption that the function be onto $\mathbb{R}^n$ is redundant.
August 12th 2009, 11:39 AM
Kat-M

homeomorphism

i thought if $f$ is homeomorphism, then $f$ is $one-to-one$ and $onto$ on $f(U)\subseteq R^n$ but not nessesarily entire $R^n$ and the inverse exists on $f(U)$ . so i thought the question emphasized that $f(U)=R^n$ . i dont know if that is right but that is what i thought.
August 12th 2009, 11:52 AM
siclar

Quote:

Originally Posted by Kat-M

i thought if $f$ is homeomorphism, then $f$ is $one-to-one$ and $onto$ on $f(U)\subseteq R^n$ but not nessesarily entire $R^n$ and the inverse exists on $f(U)$ . so i thought the question emphasized that $f(U)=R^n$ . i dont know if that is right but that is what i thought.

Well typically if the function is defined to be $U\rightarrow \mathbb{R}^n$ and is onto, it is usually assumed to be onto the given range $\mathbb{R}^n$ in my experience. Emphasis never hurts though(Rock)
August 12th 2009, 11:56 AM
Amer

Quote:

Originally Posted by Kat-M

Let $U \subseteq R^n$ be an open set. Suppose that the map $h:U \rightarrow R^n$ is a $homeomorphism$ $\underline{onto}$ $R^n$ and is $uniformly$ $continuous$ .
Prove that $U=R^n$ .

$U\subseteq R^n$ every $x\in U$ then $x\in R^n$

f is one-one and onto so every $y\in R^n$ is an image so in set theory $Card(U)=Card(R^n)$ and $U\subseteq R^n$ so $U=R^n$
August 12th 2009, 12:05 PM
Kat-M

homemorphism

the question says $f$ is uniformly continuous. so if we are to prove, i think we have to use this property otherwise there should be a counter example when $f$ is just continuous.
August 12th 2009, 12:07 PM
siclar

Quote:

Originally Posted by Amer

$U\subseteq R^n$ every $x\in U$ then $x\in R^n$

f is one-one and onto so every $y\in R^n$ is an image so in set theory $Card(U)=Card(R^n)$ and $U\subseteq R^n$ so $U=R^n$

Not true! For example, there exists a bijection between $(-1,1)$ and $\mathbb{R}$ (and in fact a homeomorphism). Infinite subsets of infinite sets can often be put into bijective correspondance with their supersets.

The crucial piece of information here is that the function is in fact uniformly continuous. I'm still working on figuring out what that entails.
August 12th 2009, 01:28 PM
siclar

I think I have it. Suppose $U\neq \mathbb{R}^n$ . Then $U$ is not closed (since only closed and open sets are the empty and whole sets). In particular there exists a $x\in \mathbb{R}^n\setminus U$ such that there exists a sequence $u_n\rightarrow x$ with each $u_n\in U$ . Then since the sequence is convergent, it is Cauchy, and since the function is uniformly continuous the sequence of images $f(u_n)$ is Cauchy. Since $\mathbb{R}^n$ is complete, there exists $y$ such that $v_n=f(u_n)\rightarrow y$ . Since the function was onto, there exists u such that $f(u)=y$ .

Then since $f^{-1}$ is continuous, $v_n\rightarrow y$ implies $u_n=f^{-1}(v_n)\rightarrow f^{-1}(y)=u$ . But $u_n\rightarrow x$ so by uniqueness of limits $x=u$ . Since we assumed $x\notin U$ but $u\in U$ , we have our contradiction!
August 12th 2009, 01:43 PM
Kat-M

homeomorphism

how do you know that the convergent sequence ${u_n}$ is Cauchy? i know that every Cauchy sequence converges in $R^n$ but does the converse holds too?
August 12th 2009, 01:51 PM
siclar

Quote:

Originally Posted by Kat-M

how do you know that the convergent sequence ${u_n}$ is Cauchy? i know that every Cauchy sequence converges in $R^n$ but does the converse holds too?

Yes! The proof is easy too. Suppose $x_n\rightarrow x$ . Let $\epsilon>0$ be given. Then by the definition of convergence there exists a $N$ such that for all $n>N$ , $|x_n-x|<\dfrac{\epsilon}{2}$ . Then for all $n,m>N$ , $|x_n-x_m|\leq |x_n-x|+|x_m-x|<\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilo n$ .

Convergence is a stronger condition than Cauchy since to be convergent we must identify the element the sequence converges to, whereas with cauchy we only know the sequence elements get closer together.