Q:

What does this theorem statement mean?

This is from Royden's Real Analysis book (4th edition). On page 17, Proposition 9 says:

Every nonempty open set is the disjoint union of a countable collection of open intervals.

At first, I thought it meant that:

if $X$ is a non-empty open set then there is a set of intervals $I=\{I_1,I_2,I_3,..\}$ such that: (a) $X=\cup_j(I_j)$; (b) if $j \neq k$ then $I_j \cap I_k = \emptyset$.

But from the proof, this doesn't seem right. So I looked up "disjoint union" (https://mathworld.wolfram.com/DisjointUnion.html) but it looks like it gives you back a set of tuple - so I'm not sure how to interpret this.

A:

Your understanding is correct.
The definition of disjoint union in your link is perhaps a little misleading: it is best to see it as defining a binary operator DisjointUnion$(A,B)$ that takes two arbitrary sets as input, and outputs a set that consists of an image of $A$ and an image of $B$ under a simple mapping, such that these two images are disjoint.
But when we say that a set $C$ is the disjoint union of $A$ and $B$, it usually means that (i) $C=A\cup B$ and (ii) $A\cap B=\emptyset$. Unfortunately this usage clashes with $C=$ DisjointUnion$(A,B)$ in the link.

