I mean internal, literal and external as in here: Is there a notion that $0 + \mathbb Ri$ or $\mathbb R + 0i$ has an 'internal complexification' equal to 'external complexification' of $\mathbb R$?
According to Wiki, 'every disjoint union of discrete spaces is discrete'. This refers to 'external' disjoint unions like the disjoint union of topological spaces without (necessarily) some common ambient topological space.
2 questions:
- For 'internal' but not 'literal' disjoint unions:
Let $\{A_i\}_{i \in I}$ now be topological spaces that are topological subspaces of a common ambient topological space $S$. Then disjoint union is the subset of $S \times I$ given by $A := \bigcup_{i \in I} A_i \times \{i\}$. I've seen this definition in geometry/topology texts where $I$ is actually a manifold such that $I$ has a topological structure.
For all $A_i$ discrete, what conditions (eg on cardinality or topology of $I$, or topology of $S$) make $A$ discrete?
- For 'internal' and 'literal' disjoint unions:
Consider $S$ as the 'literal' and 'internal' union of $S = \bigcup_{i \in I} \{A_i\}$ of disjoint subspaces. (Disjoint is the original definition that each pair is disjoint: $A_i \cap A_j = \emptyset$.)
For all $A_i$ discrete, what conditions (eg on cardinality or topology of $I$, or topology of $S$) make $S$ discrete?
- Context for Question 2: I notice that for non-constant holomorphic map of connected Riemann surfaces $F: X \to Y$ ($X$ and $Y$ need not be compact, and $F$ need not be surjective), that $F$ is 'discrete' meaning that each fibre of $F$ is discrete. In this case,
$$X = \bigcup_{y \in Y} F^{-1}(y),$$
such that $X$ is a 'disjoint union' of discrete subspaces. However, this is not quite the same 'disjoint union' as talked about in wiki.
Best Answer
If by disjoint union of topological spaces, you are talking about the union of disjoint subsets of a topological space, the answer is no. Here is a counter example.
In $\mathbb{R}$, with the usual topology, every singleton $\{x\}$ is discrete, but $\mathbb{R} = \bigcup_{x\in \mathbb{R}}\{x\}$ is not discrete.