# Topologies and Borel $\sigma$-fields on disjoint unions

Consider a set of functions $$\mathcal{F}$$ on $$E$$ where $$E \subset\mathbb{R}^k$$ - e.g. the class of $$L_1$$ functions on $$[0,1]$$ - and endow it with a suitable metric $$d$$ that makes it Polish.

1. Consider a partition $$\{\mathcal{F}_i, i\in I\}$$ (say at most countable) of $$\mathcal{F}$$ and denote by $$d_i$$ the restriction of $$d$$ to $$\mathcal{F}_i$$. Is it true that the topological space $$(\mathcal{F},\tau_d)$$ - where $$\tau_d$$ is the $$d$$-metric topology - coincides with the disjoint union topological space $$\coprod_{i \in I} (\mathcal{F}_i,\tau_{d_i})$$ - where $$\tau_{d_i}$$ is the $$d_i$$-metric topology on $$F_i$$ - only if $$\mathcal{F}_i \in \tau_d$$, for all $$i \in I$$?

2. If yes, can we also conclude that the Borel $$\sigma$$-algebras induced by open sets under the two topological structures coincide only if $$\mathcal{F}_i \in \tau_d$$, for all $$i \in I$$? Or would it be sufficient to have $$\mathcal{F}_i \in \sigma_B(\tau_d)$$, for all $$i \in I$$, where $$\sigma_B(\tau_d)$$ is the Borel $$\sigma$$-algebra induced by $$\tau_d$$?

I guess that there's probably a link with an old question:

My questions rise from asking myself whether one could construct a $$\sigma_B(\tau_d)$$ \ $$\sigma_B(G)$$-measurable map from $$\mathcal{F}$$ to some metric space $$G$$, with Borel $$\sigma$$-algebra $$\sigma_B(G)$$, by starting from some continuous maps $$\phi_i:\mathcal{F}_i \mapsto G$$, $$i \in I$$, and then appeal to the universal property of coproducts.

• In addition to Gerald Edgar's comments, I will add that it is just true in general that if you have a measurable space $(X,\Sigma)$ and a countable partition $(F_i)_{i \in I}$ of $X$ into $\Sigma$-measurable sets, and a corresponding family of measurable functions $f_i : F_i \rightarrow G$, where $G$ is some other measurable space, then the function $f : X \rightarrow G$ defined by patching the $f_i$ together is measurable. This follows almost immediately from the fact that a set $S \subseteq X$ is in $\Sigma$ iff for all $i \in I$, $F_i \cap S \in \Sigma$. Dec 4 '19 at 13:56
• So you do not even need to construct the coproducts in the category of measurable spaces to solve the question in your last paragraph. Dec 4 '19 at 13:57
• But what if we now want $f$ to be also continuous w.r.t. the original topology $\tau_d$ (so, not only measurable)? It seems to me that in order to have this, not only we need $F_i$ to be measurbale but also $F_i\in \tau_d$, isn't it? Otherwise we can only build maps $f$ (patching together the continuous maps $f_i$) which are continuous w.r.t. a finer topology. Dec 9 '19 at 21:03

Remarks and hints, not a solution
Question 1 A disjoint union $$\mathcal F = \bigcup_{i \in I} \mathcal F_i$$ in a metric space has the disjoint union topology if and only if all sets $$\mathcal F_i$$ are open in $$\mathcal F$$. Is that question 1? The answer is yes. Why not try to prove it?
In particular, whether $$\mathcal F$$ is a space of functions, or whether the metric is Polish, or indeed whether $$\mathcal F$$ it is metrizable at all: these do not come into it.

Question 2 No, it could happen that some $$\mathcal F_i$$ is not open, but the two Borel sigma-algabras coincide anyway. Perhaps you can find an example where there are just two sets $$\mathcal F_i$$.

• My doubt concerning Q1 was the following. Assume $\mathcal{F}_i$ are not open sets. We can anyway define the relative topologies $\tau_{i}=\{S\cap \mathcal{F}_i: \, S \in \tau_d\}$, and those would correspond to restricted metric topologies $\tau_{d_i}$, defined above. Now: A) assume $S \in \tau_d$: then $S\cap \mathcal{F}_i \in \tau_{d_i}$, thus $S$ is open in the disjoint union topology. That is: the latter contains $\tau_d$; Dec 4 '19 at 14:07
• B) next, assume $S$ is open in the disjoint union topology: then $S \cap \mathcal{F}_i \in \tau_{d_i}$, by definition, and since the latter is also the relative topology, it must be that $S \in \tau_d$. Consequently, we also have the reverse inclusion: i.e. the disjoint union topology is included in $\tau_d$. Dec 4 '19 at 14:12
• The above reasoning, however, does not use the fact that $\mathcal{F}_i$ are open in $\tau_d$, so where's the mistake? Dec 4 '19 at 14:13
• @JackLondon In your second comment "it must be that $S \in \tau_d$" does not follow. Consider $[0,1]$ partitioned into $\{0\}$ and $(0,1]$. Then $\{0\}$ is open in the disjoint union topology but not in $[0,1]$. Dec 6 '19 at 0:36
• Ok, so the "problematic sets" are e.g. $\mathcal{F}_i$, as of course $\mathcal{F}_i \in \tau_{d_i}$ and $\mathcal{F}_i \cap \mathcal{F}_j=\emptyset \in \tau_{d_j}$, so $\mathcal{F}_i$ is in the disjoint union topology but, by assumption, $\mathcal{F}_i\notin \tau_d$. Thus it was trivial, thanks. Dec 6 '19 at 10:56