Munkres states explicitly that:
Theorem 30.3: Suppose that $X$ has a countable basis. Then: (b) There exists a countable subset of $X$ that is dense in $X$.
The following proof he gives is relatively short.
From each nonempty basis element $B_n$, choose a point $x_n$. Let $D$ be the set consisting of the points $x_n$. Then $D$ is dense in $X$. Given any point $x \in X$, every basis element containing $x$ intersects $D$, so $x$ belongs to $\overline{D}. \hspace{0.8cm} \blacksquare$
I have a few confusions with the proof.
$(1)$ This seems to make the idea of a space having a dense subset less colorful, as Munkres has just constructed a completely arbitrary dense set $D$ in $X$, which leads me to my next confusion
$(2)$ Munkres didn't take advantage of second countability in his proof. Or if he did, I cannot see where having a countable base necessarily implies separability here.
I think what would be most helpful would be if someone could explain to me what exactly would happen if we only had an uncountable base. Would we still have a dense subset, albeit an uncountable one?
Best Answer
For (2): Munkres most definitely uses the countable basis, otherwise, picking one element from each basis set would not necessarily result in $D$ countable.
Not sure what to make of your complaint in (1). Second countable doesn’t give you much in the way to be more specific than this, and this argument works.
I suppose it still requires, if not the full axiom of choice, at least the axiom countable choice. That make the description of $D$ non-constructive. But in specific cases, we can often make it constructive.
For example, on the real line we have a basis $(a,b)$ where $a<b\in \mathbb Q$ and we can pick $\frac{a+b}2,$ which gives $D=\mathbb Q.$
Or in any metric space where the center of a ball is always unique, any countable basis of balls we can choose the centers of those balls to get a dense subspace.
So, the proof is non-constructive, because we don’t know anything specific about our space other than the countable basis.
The reality is that the construction usually goes the other way around. Given a countable sense set, we use it to construct a countable basis.