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This was an exam question of ours:

Let $\chi$ be the set, $\chi = \left \{ a, b, c, d \right \}$. Create a topology $\tau$ on $\chi$ so that $\left ( \chi ,\tau \right )$ is regular but not normal.

I couldn't solve the question but I'm wondering what the answer is. Is it possible at all? Can anyone help me with this?

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4 Answers 4

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Suppose $\chi$ is a regular space. Suppose $A,B$ are closed disjoint sets. If one of them is a singleton, then the regularity implies that they can be separated by open sets. Otherwise, since the sets are disjoint, then each of them is of size 2.

If this is the case then $A= \chi - B$ is open as a complement of a close set so $A,B$ are open and you can separate A from B, therefore $\chi$ is normal.

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  • $\begingroup$ Thank you for the simple and elegant solution :) $\endgroup$
    – hattenn
    Jan 19, 2011 at 16:41
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No, it is not possible. For example check Munkres's Topology book, p 200:

Theorem 32.1 Every regular space with a countable basis is normal.

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    $\begingroup$ Thank you so much for the answer. $\endgroup$
    – hattenn
    Jan 19, 2011 at 16:40
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In case of finite topological spaces regularity and normality are equivalent.

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  • $\begingroup$ But Why is that so ? $\endgroup$
    – mick
    Oct 12, 2012 at 15:16
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Thats possible. Take $p$-exclusion space on Real Line. Just to assure it does not have countable basis. Given space is Normal but not Regular.

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  • $\begingroup$ I think you need to expand this answer. Given that the cardinality of $\chi$ is 4 I can't see how you're going to create an uncountable basis for $\tau$ from it... $\endgroup$
    – postmortes
    Sep 6, 2017 at 6:26

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