First countable spaces and metric spaces.

[alemsalem]

Homework Statement

Show That every metric space is first countable. Hence show that every SUBSET of a metric space is the intersection of a countable family of open sets.

Homework Equations

no equation

The Attempt at a Solution

its easy to show that it is first countable, because for every point in the space there is the set of rational open balls which are included in every other open set.
but the second part of the question is confusing:
how can every subset be an intersection of a countable family? we only know that at every point there is a countable family but. there maybe an uncountable number of point..

Thanks for the help, I've been staring at the question for two hours..
by the way this problem is from P. Szekeres chapter 10 problem 10.9

 

[mighty2000]

To show that every subset of a metric space is the intersection of a countable family of open sets, we can use the fact that every metric space is first countable. This means that for every point in the space, there exists a countable family of open sets that contains that point.

Now, let S be a subset of a metric space X. For each point x in S, we can find a countable family of open sets that contains x. Since S is a subset of X, these open sets also contain all the points in S. Therefore, we can take the union of all these countable families of open sets to obtain a countable family of open sets that contains all the points in S.

Finally, we can take the intersection of this countable family of open sets to obtain the set S. This is because every point in S is contained in all the open sets in the countable family. Therefore, S is the intersection of a countable family of open sets, as required.

In conclusion, every subset of a metric space is the intersection of a countable family of open sets, using the fact that every metric space is first countable.