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Let $X$ be a topological space and $C \subseteq X$ closed. Give $X/C$ the quotient topology and suppose $X$ is regular. Then this implies that $X/C$ is Hausdorff.

Approach: Let $p: X \to X/C$ be denoted as the quotient map. Take two points $x,y \in X/C$, we then have to find two disjoint neighborhoods. Then I looked at $p^{-1}(\{x\})$ and $p^{-1}(\{y\})$, which are sets closed in $X$. And then I don't know.

Can anyone give me an hint or a solution? Thanks

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    At least one of $p^{-1}(\{x\})$ and $p^{-1}(\{y\})$ is a singleton. – 2017-01-27
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    Ok. thanks, and why? – 2017-01-27
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    Because in $X/C$, there is only one point whose preimage may not be a single point. – 2017-01-27
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    I still don't understand it. Can you please give some more details about why that is? Thanks. – 2017-01-27
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    Recall what the space $X/C$ is. What is its definition? – 2017-01-27
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    That all elements of $C$ will be identified as one element? So the only point whose preimage is not a single point is that one, correct? – 2017-01-27
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    Yes. (It might be a singleton, if $C = \{z\}$, then $X/C \cong X$.) – 2017-01-27
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    Ok, but that case will be trivial. If I now suppose first that $p^{-1}(\{x\})$ and $p^{-1}(\{y\})$ are singletons. First we can note that $X \backslash C$ is regular as as subspace of a regular space, so one can find disjoint opens around them, their images will then be also open in the quotient topology, because they don't intersect $p(C)$. So in this case it is OK. Is this correct? But if now $p^{-1}(\{y\})=C$, then one can again take disjoint opens around $p^{-1}(\{x\})$ and $C$, say $U_1$ and $U_2$, $p(U_1)$ will be open in $X \backslash C$, but why is $p(U_2)$ open? Thanks – 2017-01-27
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    What is $p^{-1}(p(U_2))$? – 2017-01-27
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    Possibele Duplicate of: http://math.stackexchange.com/questions/487318/x-regular-a-subset-x-closed-rightarrow-x-a-quotient-space-is – 2017-01-27
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    Ah, yes of course, $U_2$, Thanks for helping!! – 2017-01-27
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    Possible duplicate of [$X$ - regular, $A \subset X$ - closed $\Rightarrow \ \ X/A$ - quotient space is Hausdorff](http://math.stackexchange.com/questions/487318/x-regular-a-subset-x-closed-rightarrow-x-a-quotient-space-is) – 2017-01-27

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