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Question:

Given two distinct points $x_1$, $x_2$ of normal space $X$, show that there exists continuous function $f:X\to\mathbb R$ such that $f(x_1)\neq f(x_2)$.

My attempt:

I thought last process of proof is using Urysohn's lemma, so I tried hard to take two closed sets $C$, $D$, which contains $x_1, x_2$. So long time I can't solved. Please help!

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  • $\begingroup$ Have you tried $\{x_1\}$ and $\{x_2\}$? $\endgroup$ – Kavi Rama Murthy Jun 6 '18 at 10:25
  • $\begingroup$ I missed that simple idea... thank you $\endgroup$ – 박윤수 Jun 6 '18 at 10:30
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Your idea is good. Take $C=\{x_1\}$ and $D=\{x_2\}$, which are closed sets since, by definition, every normal space is $T_1$.

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