If $\sqcup_{i\in A} X_i$ is second countable and $X_i$ is second countable, then $A$ is countable. Is this true?

Question

This is basically a general statement adapted from Lee's Topological Manifold Ex.3.44 in the text.

$\{X_i\vert i\in A\}$ is a set of $n-$manifolds. $\sqcup_{i\in A} X_i$ is $n-$manifold endowed with disjoint union topology if and only if $A$ is countable.

In order to show the reverse direction by contradiction, I realize that I need the hausdorffness of $\sqcup X_i$ and then pick the subset of countable basis of $\sqcup X_i$ with the property that $x_i\in X_i$ such that $x_i\in B_i\subset X_i$. This will show the contradiction.

So my question is that if I drop hausdorffness, do I still get $A$ countable. In other words, I wonder the following statement is true.

If $\sqcup_{i\in A} X_i$ is second countable and $X_i$ is second countable, then $A$ is countable. Is this true?

Answer 1

If $\bigsqcup_{i\in A} X_i$ is second countable, it's Lindelöf, and $\{ X_i : i \in A \}$ is an open cover. Therefore, there exists a countable subcover; but assuming each $X_i \neq \emptyset$, the subcover must be everything, so $A$ is countable.

Answer 2

Each $X_i$ $i \in A$ is open in the disjoint union and non-empty. So for each them we need some $B_i$ from any base $\mathcal{B}$ for that union, such that $B_i \subseteq X_i$, so for any base for the disjoint union $|\mathcal{B}| \ge |A|$, as these $B_i$ are disjoint (as the $X_i$ are). So $A$ should be at most countable if we want to have countable base.

In fact, in terms of cardinal functions: $w(\sqcup_{i\in I}X_i)=|I|\sup\{w(X_i): i \in I\}$

Answer 3

The cellularity of X, denoted $c(X),$ is the least infinite cardinal $k$ such that the cardinal of a family of pair-wise disjoint open sets (a.k.a. a discrete open family) is at most $k.$ The density of $X,$ denoted $d(X)$, is the least infinite cardinal $k'$ such that $X$ has a dense subset of cardinal $k'$ or less. The weight of $X,$ denoted $w(X),$ is the least infinite cardinal $k''$ such that $X$ has a base of cardinal $k''$ or less.

It is easy to show that $c(X)\leq d(X)\leq w(X)$ for any $X.$

In the topology of the disjoint union $X$ of $\{X_i:I\in A\}$, (a.k.a the topological sum), each $X_i$ is open. So $F=\{X_i:i\in A\}$ is a discrete open family in $X,$ and if each $X_i\ne \phi$, the cardinal of $F$ is the cardinal of $A$.

On the other hand, if $B_i$ is a base for $X_i$ the the disjoint union of $\{B_i:I\in A\}$ is a base for $X.$