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Let $X=[0,1)\times[0,1)$, $\tau$ its topology with base $$\beta = \{ [a,b)\times[c,d): 0 \leq a < b \leq 1, 0 \leq c < d \leq 1 \}\;.$$ Please help me prove, that it is regular, but not a normal topological space.

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    Is this a homework assignment? And what have you tried so far?2012-05-15

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HINTS: For regularity, first prove that $\beta$ is a base for $\tau$ consisting of clopen (i.e., both closed and open) sets; regularity of $X$ follows immediately. For non-normality, let $D=\{\langle x,1-x\rangle:x\in[0,1)\}$; you can easily prove that $D$ is a closed, discrete subset of $X$. Let $H=\{\langle x,1-x\rangle\in D:x\in\Bbb Q\}$, and let $K=D\setminus H$; then $H$ and $K$ are disjoint closed subsets of $X$ that cannot be separated by disjoint open sets. For this part you’ll probably want the Baire category theorem.

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    I believe one could also apply Jone's Lemma to show non-normality: If $X$ contains a dense set $D$ and a closed, relatively discrete subspace $S$ with $|S|\ge 2^{|D|}$, then $X$ is not normal. c.f Willard *General Topology*. pg. 100.2012-05-15
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    @David: True. (For some reason I always forget about Jones’ Lemma.)2012-05-15
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    Sorry, but can you please explain why X is regular. Thanks in advance2012-05-16
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    @user31497: Which part are you having trouble with? Seeing that each member of $\beta$ is closed as well as open? Or seeing why that implies regularity?2012-05-16
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    I think the second.2012-05-16
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    @user31497: Let $x\in X$, and let $F$ be a closed set in $X$ not containing $x$. Then $X\setminus F$ is an open set containing $x$, so there is some $B\in\beta$ such that $x\in B\subseteq X\setminus F$. $B$ is closed as well as open, so $B$ and $X\setminus B$ are disjoint open sets with $x\in B$ and $F\subseteq X\setminus B$. Does that help?2012-05-16
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    Yes, thank you.2012-05-16
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    @Brian M. Scott Why $D$ is discrete?2018-09-13