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How to show that every pseudo metric space $(X,d)$ is completely regular?

My attempt is constructing a function by distance between a point $x$ and a closed set $A$ such that $x\notin A$, but I don't know how to makes the distance to $A$ be $1$ since $d$ is a pseudo metric.

And how to show that the function is continuous.

I just want to get some hints.

Pump Kin
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1 Answers1

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The proof is the same as in metric case.

Take $d(v,A):=\inf \{d(v,a)\ |\ a\in A\}$. It is a simple exercise that $d(\cdot, A)$ is continuous (the linked proof applies to pseudometrics as well).

Then put $f(v)=d(v,A)/d(x,A)$. Note that when $A$ is closed and $x\not\in A$ then $d(x,A)>0$, which I leave as an exercise. And so $f$ is well defined. It is also continuous as a composition of continuous functions. Finally $f(x)=1$ while $f_{|A}=0$.

freakish
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  • What confused me is just how to process the case that $d(x,A)=0$ for $x\notin A$, but now I understand that this case is impossible. Thanks a lot! – Pump Kin May 19 '20 at 08:26