# Could someone point me out the error in this 'proof' that every Hausdorff space is regular?

Clearly the statement is not true because counter examples exist. However consider the following argument:

Definition (Hausdorff Space): The topological space $X$ has the Hausdorff property if for each pair of distinct points $x,x'$ of $X$ there exists neighborhoods $U$ of $x$ and $U'$ of $x'$ which are disjoint.

Definition (Regular Space): The topological space $X$ is regular if for each point $x$ of $X$ and each closed set $E$ such that $x \notin E$, there exists a neighbourhood $V$ of $x$ and a neighbourhood $W$ of $E$ which are disjoint.

Now supppose $X$ is a Hausdorff space. Then for every pair $x,x' \in X$ distinct points, There exists open neighbourhoods $U$ of $x$ and $U_{x'}$ of $x'$ that are disjoint. Fix arbitrary closed set $E$ such that $x \notin E$. Then there is an union $$E\subseteq\bigcup_{x' \in E} U_{x'} =:U_{E}$$ which is an open neighbourhood of $E$. Now as $U \cap U_{x'}=\emptyset$ for every $x'$ distinct from $x$ we conclude that $$U \cap E= U \cap \big( \bigcup_{x'\in E} U_{x'} \big) = \bigcup_{x'} \big( U \cap U_{x'} \big) = \emptyset.$$ As $E$ was fixed arbitrarily we conclude that $X$ is regular.

Where have I made my mistake?

• You cannot ensure that there is one $U$ that works for every $x'\in E$. It may happen that for each $x'\in E$ you can separate $x'$ with $U(x')$ from $U(x,x')\ni x$, but then you would want to take $\bigcap_{x'\in E} U(x,x')$, which might not be open. Your proof does show that if $E$ es compact, such separation is possible, for then you need only intersect finitely many of the $U(x,x')$, and this is open. – Pedro Tamaroff Feb 5 '18 at 0:18
• Got it, thanks! If you would copy and paste the above as an answer I would love to mark it as the correct one, if that matters by any chance. – Meagain Feb 5 '18 at 0:31

You cannot ensure that there is one $U$ that works for every $x'\in E$. It may happen that for each $x'\in E$ you can separate $x'$ with $U(x')$ from $U(x,x')\ni x$, but then you would want to take $\bigcap_{x'\in E} U(x,x')$, which might not be open. Your proof does show that if $E$ es compact, such separation is possible, for then you need only intersect finitely many of the $U(x,x')$, and this is open.