# Manifolds and Disjoint unions

How do I prove the following proposition:

Let $$(A_j)$$ be an indexed collection of non-empty countable sets. If $$\bigcup_{j\in J}A_j$$ is countable then $$J$$ is countable.

May I have hints on this, please?

I'm trying to construct a map $$f:J \rightarrow \bigcup_{j \in J}A_j$$ that is injective, but I'm unable to do so. Do I need to use choice?

The reason I'm asking this is because I'm trying to prove the following:

If $$(X_i)_{j \in J}$$ is a collection of $$n$$-manifolds and their disjoint union is an $$n$$-manifold then $$J$$ is countable.

My attempt constitutes of the idea that I cover the disjoint union by a countable basis and then show that there exists a mapping between the index sets, so I conjectured the above.

• Unless the $A_j$ are pairwise disjoint there might not be such a map. What if $A_j = A$ for all $j$ and $A$ is, moreover, finite? – Thomas Jan 17 at 19:24
• (Btw, what is the relation to manifolds, indicated in the title?) – Thomas Jan 17 at 19:26
• @Thomas I just wrote why, sorry. – topologicalorientablesurface Jan 17 at 19:29
• If each $A_j \neq \emptyset$ and you want a map $f\colon J \to \cup_{j\in J} A_j$ such that $f(j) \in A_j$ for arbitrary $J$, then this is the axiom of choice. If the $A_j$'s are moreover disjoint then a choice function will be injective, and from this is follows that if $J$ is uncountable then the union will be as well. – William Jan 17 at 20:11

You don't actually need choice for this, assuming the $$A_j$$'s are disjoint. Note that the proposition you want to prove is actually false in general (see Thomas' comment on your question or Paul K's comment on my answer) but it is true if you add the disjointness assumption.

Disjointness implies that we can define a function $$f\colon\cup_{j \in J} A_j \to J$$ without having to make any choices by sending $$x$$ to $$j$$ if $$x\in A_j$$, and moreover $$f$$ is surjective iff each $$A_j$$ is non-empty. Therefore if the union is countable then $$J$$ must be as well, since it is the image of a countable set.

You could then apply this to your manifolds problem as follows. An arbitrary disjoint union of manifolds will still be Hausdorff and locally Euclidean, so we we only need to be concerned with second countability. What I will actually show is

Suppose $$X = \cup_{j\in J} X_j$$ where each $$X_j$$ is non-empty and $$X_i\cap X_j =\emptyset$$ for every $$i\neq j$$. If $$X$$ has a countable basis then $$J$$ is countable.

Suppose $$A$$ is a countable basis for $$X$$. One property of a basis is that any open set of $$X$$ can be written as a union of elements of $$A$$. In particular, if we have an open set $$U\subset X_j$$ for some $$j$$, then $$U$$ will be the union of elements in $$A$$, which will necessarily be subspaces of $$X_j$$. Therefore the following is also a basis, and will still be countable since it is a subset of $$A$$:

$$A' = \{ U \in A\ |\ U\subset X_j\text{ for some }j\}.$$ If we let $$A_j = \{ U \in A\ |\ U\subset X_j\}$$, this will be a basis for $$X_j$$ and therefore non-empty since $$X_j$$ is non-empty, and $$A_i \cap A_j = \emptyset$$ for $$i\neq j$$. Then since $$A' = \cup_{j\in J} A_j$$ it follows from the above that $$J$$ must be countable.

Note that this argument didn't depend on the $$X_j$$'s being manifolds, the only input was second countability of $$X$$, and the non-empty/disjointess.

• I'm more interested in the case where $A_j$ are not necessarily disjoint. – topologicalorientablesurface Jan 17 at 20:32
• In that case it's not necessarily true, like Thomas points out their comment. For your problem with manifolds you can take $A_j$ to be a countable basis for $X_j$, but if $X_i\cap X_j = \emptyset$ then $A_i \cap A_j = \emptyset$ so the disjointness condition would be satisfied. – William Jan 17 at 20:35
• @topologicalmagician but it is wrong if $A_j$ are not necessarily disjoint. Take an arbitrary set $J$ and the disjoint union $\bigcup_{j \in J} A$ for some set $A$. – Paul K Jan 17 at 20:35
• I'm concerned that people have down-voted my answer into a negative score, but no one has said there's anything wrong with it, and it's even the accepted answer. Why is it bad? – William Jan 17 at 23:00
• @William I just upvoted, I didn’t realize it was dowmvoted until now! – topologicalorientablesurface Jan 18 at 10:05