In a metric space, it is true that every open set can be represented as the union of closed sets (e.g. the union of closed balls).

I am wondering if every open set can be represented as the countable union of closed sets?


Say $U$ is an open set. If $U$ is empty or if it is the whole space $X$ then $U$ is also closed, so we are done. Now assume that $U$ and its complement $U^c=X\setminus U$ are both non-empty.

Let $A_n = \bigcup_{x\in U^c} B(x,\dfrac1n)$. Notice that $A_n$ is open and its complement $A_n^c$ (which is closed) is contained in $U$. To finish off the problem show that $U$ is the union of the $A_n^c$.

A set which is the union of countably many closed sets is called an $F_\sigma$ set. It is known that every open set in a metric space is an $F_\sigma$ set.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.