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Can you please help me to prove that if a space has a countable base, then it has countable dense subset?

I know this happens in the real line, i.e.: it has a countable base as $\{(a,b):a,b \in\Bbb Q\}$ and $\Bbb Q$ is dense in $\Bbb R$.

But how can I generalize this result as mentioned in the problem?

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up vote 6 down vote accepted

We (commonly) say a topological space that has a countable base is second countable. And we also (commonly) say a topological space that has a countable dense subset to be separable.

Here, we would like to show that a topological space $(X,\tau)$ that is second countable implies it is separable.

Let ${\scr B}=\{B_n\}_{n\in{\bf N}} $ be a countable base for $X$. We need a candidate set that could possibly be a countable dense subset of $X$. Since we only have information about ${\scr B}$ is countable, we use it to construct a set.

Suppose without loss that every base set in the collection $\scr B$ is not empty. Now, for each $B_n\in {\scr B}$, pick any element $x_n\in B_n$. This is possible since each $B_n$ is not empty. (And this infinitary choice is possible by Axiom of Choice.)

Write $D=\{x_n\}_{n\in{\bf N}}$, the collection of $x_n$ we just constructed. This set $D$ is our candidate set that is (i) countable, and (ii) dense in $X$.

We know $D$ is countable because we picked an $x_n$ for each $B_n$ from $\scr B$, that is, $D$ is bijective to our countable base $\scr B$. Hence $D$ is (i) countable.

Now we would like to show $D$ is dense in $X$. Recall the definition of a dense set $E$: We say $E$ is dense in $X$ if every nonempty open set $U$ of $X$ intersects $E$.

Now, also recall a property of a base $\scr B$: If $\scr B$ is a base then for every open set $U$ of $X$, and each point $x\in U$, there exists $B\in{\scr B}$ such that $x\in B\subset U$.

Try to show (ii) by using the definition of denseness and definition of a base.

Hint: To show (ii), take any nonempty open set $U$ of $X$. We need to show $U$ intersects our set $D$. Note for any $x\in U$, there exists a base $B_k$ such that $x\in B_k\subset U$. But $B_k$ contains $x_k\in D$. We have claimed.

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Pick one point from every member of the countable base.

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