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Suppose $f:X\rightarrow Y$ is a homeomorphism. Show that if $X$ is hausdorff then so is $Y$.

My attempt: Let $y_1,y_2\in Y$ be distinct, by bijectivity of $f$, there exists distinct $x_1,x_2 \in X$ such that $f(x_1)=y_1$ and $f(x_2)=y_2$. Since $X$ is hausdorff, there exists disjoint open subsets of $X$, $V_1$ and $V_2$ containing $x_1$ and $x_2$ respectively. Since $f$ is a homeomorphism, it is an open map, hence $f(V_1)\cap f(V_2)$ is the union of two open sets. Since $f$ is injective, $f(V_1 \cap V_2)= f(V_1) \cap f(V_2)= \varnothing$, and $f(V_1),f(V_2)$ contain $y_1,y_2$ respectively. Thus $Y$ is hausdorff.

Is it correct?

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    $\begingroup$ Yep seems right ! $\endgroup$
    – Popyaitte
    Dec 24, 2019 at 15:48
  • $\begingroup$ Just say $f[V_1]$ and $f[V_2]$ are open sets, etc. $\endgroup$ Dec 24, 2019 at 16:36

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There is a little problem with the sentence 'hence $f(V_1)\cap f(V_2)$  is the union of two open sets', because you should write $f(V_1)\cup f(V_2)$ instead of $f(V_1)\cap f(V_2)$, and because you should precise that $f(V_1)$ and $f(V_2)$ are open sets.

One should also precise that $f(V_1)$ and $f(V_2)$ are disjoint, and contain $y_1$ and $y_2$ respectively. But this is done in your text

The rest of your argument is correct.

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Homeomorphisms preserve all topological properties, and the result follows.

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  • $\begingroup$ He has to prove that Hausdorff is a topological property.... $\endgroup$ Dec 24, 2019 at 17:19
  • $\begingroup$ @HennoBrandsma Well it is a statement about the existence of certain types of open sets. $\endgroup$
    – user403337
    Dec 24, 2019 at 17:27

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