# Regular but not normal space

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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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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Thank you for the simple and elegant solution :) –  hattenn Jan 19 '11 at 16:41

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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Thank you so much for the answer. –  hattenn Jan 19 '11 at 16:40

In case of finite topological spaces regularity and normality are equivalent.

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But Why is that so ? –  mick Oct 12 '12 at 15:16