How to choose a continuous function which vanishes **only** on the closed set

We are reading John Roe's book Lectures on Coarse Geometry. We come across a question in P27 line 9: Suppose $X$ is a paracompact and locally compact Hausdorff space, $\bar{X}$ is a compactification of $X$, how to use Urysohn's Lemma to choose continuous functions $f,g:\bar{X}\times\bar{X}\to \mathbb{R}^+$ such that $f$ vanishes only on the diagonal of $\bar{X}\times\bar{X}$, and $g$ vanishes only at infinity (that is, on $\bar{X}\times\bar{X}\backslash X\times X$)?

If $\bar{X}\times\bar{X}$ is perfectly normal, or the diagonal and the infinity are both $G_\delta$ sets, we can have that according to P213 in Munkres's book Topology. But we fail to verify the condition(we are not familiar with paracompact Hausdorff space).

Or any other way?

• This question is not fit for MathOverflow, but would find a very good home on Math.SE. Flagged it for migration. – Alex M. Oct 20 '17 at 9:04
• A note to potential voters: despite the wording, the offending text in Roe’s book is not posed as a question or exercise or some such, it is in fact in the middle of a proof. – Emil Jeřábek supports Monica Oct 20 '17 at 11:27

1. The function $f:\bar X\times \bar X\to\mathbb R$ exists if and only if the compactification $\bar X$ of $X$ is metrizable, which implies that the locally compact space $X$ is metrizable and separable.

This follows from a well-known metrizability theorem saying that a compact topological space $K$ is metrizable if and only if the diagonal is a closed $G_\delta$-set in $K\times K$.

1. The function $g:\bar X\times \bar X\to \mathbb R$ exists if and only if the set $X\times X$ is of type $F_\sigma$ in $\bar X\times \bar X$, which happens if and only if the locally compact space $X$ is $\sigma$-compact if and only if $X$ is Lindelof.

If there is a continuous real-valued function that vanishes only on the diagonal, then that easily implies that the diagonal is $G_\delta$ in $\overline{X}^2$. We can then intersect with $\overline{X}\times\{a\}$ to see that $\{a\}$ is $G_\delta$ in $\overline{X}$, for every $x\in\overline{X}$. There are certainly examples where this is not true. For example, we can let $\omega_1$ denote the first uncountable ordinal, and take $\overline{X}=[0,\omega_1]$ with the order topology. This is the one-point compactification of the space $X=[1,\omega_1]$, which is homeomorphic to $\overline{X}$. We then find that $X$ is compact Hausdorff (and so is certainly locally compact Hausdorff and paracompact) but $\{\omega_1\}$ is not $G_\delta$.

• Why is the intersection with $\bar{X}\times \{a\}$ $G_\delta$? Note that $\bar{X}\times\{a\}$ is not open. – C.Ding Oct 20 '17 at 9:46
• I did not claim that $\overline{X}\times\{a\}$ is $G_\delta$ in $\overline{X}\times\overline{X}$. Please think about the argument more carefully. – Neil Strickland Oct 20 '17 at 10:00
• You say the intersection is $G_\delta$, right? – C.Ding Oct 20 '17 at 10:07
• No, I did not say that. Please think about the argument more carefully. – Neil Strickland Oct 20 '17 at 10:40
• In my book, a compactification is required to have the property that the original space is dense in it. – Emil Jeřábek supports Monica Oct 20 '17 at 10:58