If X is Hausdorff and K⊂X is compact then the contraction X/K is Hausdorff

Question

I found this claim as an answer to one problem. I tried to prove it in the following way: $X$ is Hausdorff so for every $x$ that does not lie in $K$ compact i can find two disjont open set A,B such that $x∈A$ and $K⊂B$. [As the space is $T2$ both the point and the compact subspace are closed in $X$]. $A$ and $B$ are saturated so their images through the projection are the disjoint open sets that let $X/K$ be Hausdorff. Is this proof correct? If not where does it fail? Thanks in advance

Answer 1

The proof is good to show that a point $x\in X\setminus K$ and $\hat{K}$ (the contraction of $K$) are separated in $X/K$.

However this doesn't fully suffice. If $x,y\in X\setminus K$ and $x\ne y$ you have to prove that they can be separated by disjoint open sets.

Choose $A$ and $B$ as before for $x$, but also $A'$ and $B'$ for $y$ in the same way. Since $X$ is Hausdorff, you can conclude. Fill in the details.

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How do you prove the key fact? For every $z\in K$ choose $U_z$ and $V_z$ disjoint open sets containing $x$ and $z$ respectively. Then $(V_z)_{z\in K}$ is an open cover of $K$ and so there are $z_1,\dots,z_n$ such that $B=V_{z_1}\cup\dots\cup V_{z_n}\supseteq K$. Take $A=U_{z_1}\cap\dots\cap U_{z_n}$.