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I have two questions regarding metric spaces which I can't solve.These are not homework problems.These came in my mind quite naturally. Here's these :

(1) We know that every open (resp. closed) set in the discrete metric space $(X,d)$ is $F_{\sigma}$ as well as $G_{\delta}$.Now my first question is based on whether the converse of the above will hold good or not. i.e. If $(X,d)$ be a metric space such that every open (resp. closed) set is $F_{\sigma}$ as well as $G_{\delta}$ then is it true that $(X,d)$ is the discrete metric space?

(2) We also know that any metric on a finite space is equivalent to the discrete metric.Now my second question is :

Does the converse of the above hold good or not? i.e. If any metric on a space is equivalent to the discrete metric then is it the true fact that the space should be finite?

Please anybody help me in finding proper answers to the above mentioned questions.

Thank you in advance.

Arnab Chattopadhyay.
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  • For the first one,have you tried taking the real line with the usual topology? Every closed is naturally the union of itself,also its the enumerable intersection of open intervals containing it,for the open sets notice that R is separable meaning every open is G-delta and to finish you can use a very similar argument to prove every open set is F-sigma,but such topology on R is not the trivial. – AHandsomeAlien May 04 '17 at 06:31
  • I have trouble understanding the first question. You "know that every open set is $F_\sigma$ as well as $G_\delta$". You know this for every metric space, right? Then doesn't that automatically give a negative answer to your question "if $(X,d)$ be a metric space such that every open set is $F\sigma$ as well as $G_\delta$ then is it true that $(X,d)$ is the discrete metric space?" According to what you say you know, any non-discrete metric space is a counterexample, no? By the way, how is the statement "if ... then $(X,d)$ is discrete space" the converse of a statement with no "discrete"? – bof May 04 '17 at 08:31
  • Wouldn't the "converse" of the statement "every open (resp. closed) set is $F_\sigma$ as well as $G_\delta$ be the statement that "every set which is $F_\sigma$ as well as $G_\delta$ is open (resp. closed)"? – bof May 04 '17 at 08:33
  • I only want the answers to the following questions.

    (1) If $(X,d)$ be such a metric space that every open (resp. closed) set of it is both $F_{\sigma}$ and $G_{\delta}$ then is $(X,d)$ always the discrete metric space.

    (2) If any metric on a non-empty set $X$ is equivalent to the discrete metric.Then whether $X$ is always finite or not.

    These are only my queries.Any help will be appreciated.

     – Arnab Chattopadhyay. May 04 '17 at 10:55

1 Answers1

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(1). In any metric space $(X,d)$, any closed set $Y$ is a $G_{\delta}$ set.

Let $Y_n=\cup \{B_d(p,1/n):p\in Y\}$ for each $n\in \mathbb N.$ Obviously $Y$ is a subset of the $G_{\delta}$ set $Z=\cap_{n\in \mathbb N}Y_n.$

And if $q\in Z,$ then for every $n\in \mathbb N$ there exists $p\in Y$ with $d(p,q)<1/n$,...( because $q\in Y_n\implies \exists p\in Y\;(q\in B_d(p,1/n)$),... so $q\in Z\implies q\in \bar Y=Y.$

BTW if $Y$ is any subset of $X$ then $Z$ (as defined above) is $\bar Y$.

A normal space in which every closed set is a $G_{\delta}$ set is called a perfectly normal space. (It is not good to call it perfect-normal because among metric spaces, a perfect space means something different.)

(2). Two metrics on the same SET are called equivalent iff they generate the same topology. Be careful to distinguish between "set" and "space". I think that what you meant was:

If every metric on the set $X$ is equivalent to "the" discrete metric (i.e. the metric $d$ where $p\ne q\implies d(p,q)=1$), then is $X$ finite?

The answer to this is Yes. There are many ways to prove this. One way is to construct a non-discrete metric space $(X,d)$ for any infinite set $X$, as follows:

Let $f:\mathbb N\cup \{0\}\to Y$ be a bijection with $Y\subset X.$

For $p\in X$ \ $Y$ and $p\ne q\in X$ let $d(p,q)=1.$

Let $d(f(m),f(n))=1/m-1/n$ for $0<n<m.$

Let $d(f(0), f(n))=1/n$ for $0<n.$

Then $f(0)\in Cl(\;Y)$ \ $\{f(0)\}\;)$ so $(X,d)$ is not a discrete space.

The idea is to identify some $Y\subset X$ with the real subspace $\{0\}\cup \{1/n: n\in \mathbb N\}$ and to let any point in $X$ \ $Y$ be at a distance $1$ from every other point.

DanielWainfleet
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