If $X$ is compact Hausdorff and $p\in X$, then there is a continuous $f:X\to\mathbb{R}$ that vanishes at $p$ and nowhere else

Question

Prove or refute: If $p$ is a point in a compact Hausdorff space $X$ then there exists a continuous real-valued function $f:X\to\mathbb{R}$ that vanishes at $p$ and nowhere else.

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score 3 · Answer 1 · edited Apr 13 '17 at 12:21

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HINT: Consider the point $\omega_1$ in the space $\omega_1+1$ with the order topology. This question should help.

Added: Here’s a simpler example. Let $X$ be an uncountable set, and let $p\in X$. A set $U\subseteq X$ is open if and only if either $p\notin U$, or $X\setminus U$ is finite. Show that $X$ is a compact Hausdorff space, and that if $f:X\to\Bbb R$ is continuous, and $f(p)=0$, then there are uncountable many points $x\in X$ such that $f(x)=0$.

edited Apr 13 '17 at 12:21

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answered Jun 29 '13 at 18:33

Brian M. Scott

616,228

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What is $\omega_1$? – stickyperson Jun 29 '13 at 18:34
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@stickyperson: It’s the first uncountable ordinal. $\omega_1+1$ is the set of all ordinals less than or equal to $\omega_1$. – Brian M. Scott Jun 29 '13 at 18:36

score 1 · Answer 2 · edited Apr 13 '17 at 12:21

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I asked a very closely related question on this site a while back. The accepted answer shows that a compact Hausdorff space has the property that every singleton set is the zero set of a continuous real-valued function iff it is first countable. This gives a very clean answer to your question.

edited Apr 13 '17 at 12:21

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answered Jun 29 '13 at 18:52

Pete L. Clark

97,892

score 0 · Answer 3 · answered Jun 29 '13 at 18:37

Let $X=\{0,1\}^{\mathbb R}$, the uncountable product of a discrete space. The point $0:=(0_i)_{i\in\mathbb R}$ is not the intersection of countably many open sets (otherwise $\{0\}$ would be a so-called $G_\delta$-set), but if there existed such a function, then one could take the preimages of $(-\frac1n,\frac1n)$ and intersect them to obtain $0$.

If $X$ is compact Hausdorff and $p\in X$, then there is a continuous $f:X\to\mathbb{R}$ that vanishes at $p$ and nowhere else

3 Answers3