TITLE: Topology on the (canonical) union of sets vs disjoint union topology
QUESTION [0 upvotes]: Let $B_{i} \subset \ell^\infty([0,1])$, $i \in I$, i.e. each $B_i$ denotes a class of bounded functions mapping from $[0,1]$ to $\mathbb{R}$. Let $d(f,g)=\sup_{u \in [0,1]}|f(u)-g(u)|$, $f,g \in \ell^\infty([0,1])$ denote the sup-norm and let $d_i$ denote its restriction to $B_i$, $i \in I$. 
Define the metric spaces $\mathcal{B}_i=(B_i,d_i)$, $i\in I$. Which is the difference between the topology of the disjoint union topological space $\coprod_{i \in I}\mathcal{B}_i$ and the metric topology induced by the restriction of $d$ on $\cup_{i\in I}B_i$? 
Is one of the two topologies contained in the other or shall we consider those as two different topologies on two different spaces? Is there a meaningful way to relate them?

REPLY [0 votes]: Let's say you have two metric spaces X and Y whose metrics are "compatible" on their intersection. You could interpret $X \cup Y$ as "X glued to Y along their intersection". And you can construct $X \sqcup Y$ as "replace X and Y with two spaces X' and Y' which have empty intersection, then define $X \sqcup Y = X' \cup Y'$. So $\mathbb R \cup \mathbb R = \mathbb R$ but $\mathbb R \sqcup \mathbb R = \mathbb R \times \{0,1\}$, the space consisting of two real lines that are disjoint.
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