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Definition. Let $X$,$Y$ be metric spaces.Then a map $T:X\to Y$ is an homeomorphism if $T$ is continuous, open and bijective. I don't find a counterexample of such maps, may someone give me at least one example where I can understand how this is done to show $T$ is homeomorphic. Regards!

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The question is unclear ... –  Martin Brandenburg Apr 14 '12 at 16:08
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How about $X = \mathbb R$ with the discrete topology (discrete metric) and $Y = \mathbb R$ with the usual topology (induced by the Euclidean metric)? Then take the identity map $id: X \to Y$. It's bijective and continuous but its inverse $id: Y \to X$ isn't: $id: Y \to X$ is open and bijective but not continuous.

As pointed out by Brian in the comment, if you take the absolute value map $|\cdot | : \mathbb Z \to \mathbb N $ and endow $\mathbb Z$ and $\mathbb N$ with the discrete topology then $|\cdot|$ is continuous and open but not bijective.

I hope I understood your question correctly. Otherwise ping me with a comment and I'll adjust my answer.

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You might add to your answer that the identity map from $Y$ to $X$ is an open bijection that isn’t continuous. Finally, if you give $\Bbb Z$ and $\Bbb N$ the discrete topology, the absolute value map is a continuous, open surjection that isn’t a bijection. Thus, the three conditions really are independent. –  Brian M. Scott Apr 15 '12 at 0:45
    
@BrianM.Scott Thank you very much! Added. : ) –  Matt N. Apr 15 '12 at 6:11
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Are you asking for an example of a homeomorphism between two metric spaces? Consider a function $f: X \rightarrow Y$, and consider $X$ and the graph of $f$ which is simply $ \{(x, f(x)) \in X \times Y\}$. Then $X$ is homeomorphic to the graph of $f$ under the usual projection map $\pi : X \times Y \longrightarrow X$.

$\textbf{Edit:}$ Please edit your question and make it clear what exactly you are asking. I can edit my answer to suit this.

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Let $X$ be the discrete space of $N$ points, and let $Y$ be any finite space with $N$ points. Then any bijection between $X$ and $Y$ is homeomorphic.

Let $X$ be $\mathbb{R}$, and let $Y=(0,1)$. Then there are continuous bijective $T:X\to Y$ (such as $x\mapsto \tan \pi (x-0.5)$)

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