Homeomorphic subset of $\mathbb R^n$ such that... is actually $\mathbb R^n$

Question

Let $ f:A\subset \mathbb R^n \to \mathbb R^n$ be a homeomorphism, such that $f$ is also uniformly continuous, and where $A$ is an open subset of $\mathbb R^n$

Prove that $A=\mathbb R^n$.

I have the intuition that if that happens, then $A$ is also closed, and then by connectedness, $A=\mathbb R^n$ but I'm not sure

Answer 1

Uniformly continuous maps take Cauchy sequences to Cauchy sequences. Take a Cauchy sequence $(a_n)$ in $A$. By the previous remark, $(f(a_n))$ is Cauchy in $\mathbb R^n$, so it's convergent. Since $f$ is a homeomorphism $(a_n)$ must be convergent in $A$, i.e. $A$ is complete. We also know that a subspace of a complete space is closed if and only if it's complete [ProofWiki], so $A$ is closed.

To finish up, use connectedness, as you indicated.

Answer 2

Let $a_n\in A$ such that $\lim_{n\to\infty} a_n=a\in\mathbb{R}^n$. Certainly $a_n$ is Cauchy, thus by uniform continuity of $f$, $f(a_n)$ is Cauchy as well and thus by completeness of $\mathbb{R}^n$, $\lim_{n\to\infty} f(a_n) = b\in\mathbb{R}^n$. Now notice that $f^{-1}(b) = f^{-1}(\lim_{n\to\infty}f(a_n)) = \lim_{n\to\infty}f^{-1}(f(a_n)) = \lim_{n\to\infty} a_n = a$, where we have used the continuity of $f^{-1}$. Hence $a\in\mathrm{im}(f^{-1})=A$ and we conclude that $A$ is closed.