My textbook says "Every $T_{4}$-space is a $T_{3}$-space."


A $T_{4}$-space is a normal $T_{1}$-space. For every two disjoint closed sets, there are two disjoint open sets containing one each. A $T_{3}$-space is a regular $T_{1}$ space. For any closed set and a point outside it, there are two disjoint sets containing one each.

For every $T_{4}$-space to be a $T_{3}$-space, for every point $x$ and a closed set $E$ not containing it, there will have to be closed set $A$ such that $x\in A$, $E\cap A=\emptyset$. How can we ensure this?

Please note that $A$ can't be the (closed) complement of the open set containing $E$, as then the open set containing $A$ will intersect with the open set containing $E$,unless $A$ is a clopen set. We want disjoint open sets.

Thanks in advance!

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    $\begingroup$ In a $T_1$-space, points are closed. So apply the $T_4$ property to $E$ and $\{x\}$ to get the $T_3$-property. $\endgroup$ – martini Jun 18 '13 at 10:15
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    $\begingroup$ @martini Considering that your comment is effectively an answer to OP, you may want to promote it to one. $\endgroup$ – Lord_Farin Jun 18 '13 at 11:00
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    $\begingroup$ @martini: Seconded. $\endgroup$ – Cameron Buie Jun 18 '13 at 11:31
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    $\begingroup$ @Lord_Farin, Cameron Buie: Done. $\endgroup$ – martini Jun 18 '13 at 15:46

In a $T_1$-space, points are closed: Let $X$ be $T_1$ and $x \in X$. For each $y \in X \setminus \{x\}$, there is an open neighbourhood $U_y$ of $y$ with $x \not\in U_y$. As $X \setminus \{x\} = \bigcup_{y\ne x} U_y$, $X \setminus \{x\}$ is open, hence $\{x\}$ is closed.

To see that $T_4$-spaces are $T_3$ suppose $X$ is $T_4$, $E \subseteq X$ is closed and $x \in X \setminus E$. Then $E$ and $\{x\}$ are disjoint closed sets in $X$, hence have disjoint open neighbourhoods.


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