# Prove the following theorem involving homeomorphic metric spaces.

I want to prove this theorem:

If $(X,d)$ and $(X,d')$ be homeomorphic metric spaces, then they have the same convergence sequences.
However, there exists homeomorphic metric spaces $(X,d), (X,d')$ such that only one of them is complete.

So what we want to prove is that if $d(x_n,x)< \epsilon$ then $d'(x_n,x)<\epsilon$ but the thing is that I don't know facts about homeomorphic spaces, only the definition, this is that they the same topology since we are dealing with the same space $X$ (because the function is the identity, so we have to have the same open set in both metric space so that the identity would be inverse continous).I think that the definition should be enogh since we haven't seen anything fancy about homeomorphism yet (I think the ball version of convergence should help, but I am stuck here). So, Can someone help me to prove this with this basic stuff please?

NOTE: We are working in the same space $X$ :)

HINT FOR 1: Let $h$ be a homeomorphism. Then $h(x_n)$ converges iff $x_n$ converges. This is true since $h$ is invertible, the inverse is also continuous, and convergence is preserved under a continuos function.
HINT FOR 2: $\mathbb{R}$ is complete and homeomorphic to $(0,1)$. But $x_n=1/n$ is Cauchy sequence not converging in $(0,1)$.
COMMENT: your definition of homeomorphism is not correct. A homeomorphism is a bijection that is continuous and whose inverse is also continuous. Continuity of an invertible function does not guarantee that the inverse is continuous. For example. Define $f$ from $[0,2\pi)$ to the unitary circle in $\mathbb{C}$ by setting $f(x)=\exp(ix)$. Then the function is invertible and continuos but the inverse is not continuous.
• Thanks for your answer a lot, let me think in your hints, but who is $1_n$?, Is this the contant sequence only having 1? Commented Aug 19, 2015 at 14:48
• I have a doubt with the first hint, How $h(x_n)$ should be equal to $x_n$? Commented Aug 19, 2015 at 14:59
• The sequence $(h(x_n))$ and the sequence $(x_n)$ live on different spaces. If you have every had a course in analysis, you must have proved that for a continuous function $\lim_{\to\infty} f(x_n) = f(\lim_{n\to\infty} x_n)$. Commented Aug 19, 2015 at 15:36