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!

  • $\begingroup$ 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

Your idea is good. Take $C=\{x_1\}$ and $D=\{x_2\}$, which are closed sets since, by definition, every normal space is $T_1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.