# Help in Munkres Topology first curse problem

hope you are doing awesome.
I would really appreciate if you could give me a hint instead of solving the problem for me, thanks.
I'm having struggles with this problem :
Let X be a regular space with a countable basis; Let U be open in X. Show that U equals a countable union of closed sets.
So far I have tried two things :

1. Let $U$ and open set, and $x\in U$, then because $X$ is $T_3$ there is and open set $U_1$ such that $x_1 \in U_1 \subset \overline U_1 \subset U$, let $x_2 \in U \setminus \overline U_1$, since this last one is open there is set $U_2$ open such that $x_2 \in U_2 \subset \overline U_2 \subset U\setminus \overline U_1$, let $x_3 \in U \setminus \left(\overline U_1 \cup \overline U_2 \right)$ then there is an open set $U_3$ such that $x_3 \in U_3 \subset \overline U_3 \subset U\setminus \left(\overline U_1 \cup \overline U_2 \right)$. Now I'll proceed by induction, suppose that for $n \in N$ there are $x_1,...,x_n \in X$ and $U_1,..., U_n$ such that $x_i \in U_i \subset \overline U_i \subset U\setminus \left(\cup_{i=1|}^{i-1} \overline U_i \right)$. Let $x_{n+1} \in U \setminus \left( \cup_{i=1}^{n} \overline U_i \right)$, this set is open because the last union is a finite union of closed sets which is a closed set, therefore, it exists an open set $U_{n+1}$ such that $x_{n+1} \in U_{n+1} \subset \overline U_{n+1} \subset U\setminus \left(\cup_{i=0}^{n} \overline U_i \right)$, then the $U_n$ sets are defined for every $n \in N$. Lets prove that $U = \left(\cup_{i=1}^\infty \overline U_i \right)$. This is where I'm stucked.
2. This is my second attempt. Show that a close set A equals a countable intersection of open sets of X. Let $x_1 \in X\setminus A$ then there are open sets $V_1, U_1$ such that $x_1 \in V_1 and A \subset U_1$, let $x_2 \in U_1\setminus A$ then there are open sets $V_2, U_2$ such that $x_2 \in V_1 and A \subset U_2$. Now I'll proceed by induction, suppose that for $n \in N$ there are $x_1,...,x_n \in X$ and $U_1,..., U_n, V_1, ..., V_n$ open sets such that $x_i \in V_i and A \subset U_i$ for every $i \le n$. Let $x_{n+1} \in U_n\setminus A$ then there are open sets $V_{n+1}, U_{n+1}$ such that $x_{n+1} \in V_{n+1} and A \subset U_{n+1}$, then the $U_n$ sets are defined for every $n \in N$. Lets prove that $U = \left(\cup_{i=1}^\infty U_i \right)$. As well I'm stocked here. Any hint or suggestion is welcome. Thanks is advanced
• It that a pun in your title ? Its hilarious. Commented Feb 23, 2017 at 12:10

1 is the way to go: (Some $B_n, n \in \mathbb{N}$ for a base)

For every $x \in U$ pick a basic element $B_{n(x)}$ with $x \in B_{n(x)} \subseteq \overline{B_{n(x)}} \subseteq U$ by regularity and the fact the $B_n$ form a base.

Collect all the indices we've used: $N(U) = \{n(x): x \in X\} \subseteq \mathbb{N}$. This is clearly countable.

Then show that $U = \cup\{\overline{B_n}: n \in N(U)\}$ proving $U$ to be a countable union of closed sets.

This is about not trying to be subtle or minimal, just throw "brute force" (AC) at it...

• Thanks a lot, it was so easy and I was trying to do something more complicated. Really, thanks a lot, hope you have a great week.
– GBes
Commented Feb 23, 2017 at 7:26