How to show that any separable space is CCC

Question

I thought I had the proof of this in my head, but it doesn't sound right on paper. Can someone see if my argument could be improved.

Let $(X,\tau)$ be a topological space that is separable, then it contains a countable dense subset $\{D_n\}_{n \in \mathbb{N}}$. Then $X$ is CCC if it does not contain any uncountable collection of disjoint open sets.

My argument:

By contradiction, suppose $X$ is separable but is not CCC.

Since a countable dense subset could only intersection countably many open sets, there exists some open sets $U$ that $D_n$ fails to intersection.

Therefore, $X$ is not separable as claimed.

My problem with this statement is that: Since a countable dense subset could only intersection countably many open sets...

Is there to make the middle paragraph rigorous? The whole proof feels wrong, please assist!

Answer 1

You have the right basic idea, but there’s no need to argue by contradiction. Let $D$ be a countable dense subset of $X$, and suppose that $\mathscr{U}$ is a family of pairwise disjoint non-empty open sets. For each $U\in\mathscr{U}$ there is an $x_U\in D\cap U$. If $U,V\in\mathscr{U}$, and $U\ne V$, then $U\cap V=\varnothing$, so $x_U\ne x_V$. Thus, the function $\mathscr{U}\to D:U\mapsto x_U$ is injective, and it follows immediately that $|\mathscr{U}|\le|D|$, i.e., that $\mathscr{U}$ is countable.

Answer 2

Your argument is essentially correct, but can be made more formal:

Suppose that $\{ U_i: i \in I \}$ is a family of non-empty pairwise disjoint sets.

Then define for $i \in I$, $f(i) = \min \{n: d_n \in U_i \}$, where $d_n$ is an enumeration of the dense set by $\mathbb{N}$. As $U_i$ is non-empty and open, the set of such $n$ is empty, and for definiteness we choose its minimum.

Then for $i \neq j$ we have $f(i) \neq f(j)$, as otherwise the common value $n$ would be the index of a $d_n$ that is in both $U_i$ and $U_j$ which cannot be. So $f$ is 1-1, and $I$ is at most countable.

Answer 3

Let $\{ U_{\alpha}\}_{\alpha \in I}$ be a pairwise disjoint collection of open sets in $(X, \tau)$. The index set $I$, at the outset, may or may not be countable. We will show that it is

Because $X$ is separable, it has a dense countable subset $D$. Therefore, for each $\alpha \in I$, we can choose an element $u_{\alpha} \in D \cap U_{\alpha}$. Since the $U_{\alpha}$'s are pairwise disjoint, all the $u_{\alpha}$'s are distinct, and on the other hand, all the $u_{\alpha}$'s lie in $D$. Therefore, the $u_{\alpha}$'s constitute a countable set, which is in 1-to-1 correspondence with the collection $I$.