# Hausdorff Property

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?

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

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