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:

Is there a "disjoint union" sigma algebra?

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.

  • $\begingroup$ 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 $i \in I$, $F_i \cap S \in \Sigma$. $\endgroup$ Dec 4 '19 at 13:56
  • 1
    $\begingroup$ So you do not even need to construct the coproducts in the category of measurable spaces to solve the question in your last paragraph. $\endgroup$ Dec 4 '19 at 13:57
  • $\begingroup$ 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. $\endgroup$ 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 $\mathcal F_i$.

  • $\begingroup$ 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 $\tau_d$; $\endgroup$ Dec 4 '19 at 14:07
  • $\begingroup$ 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 $\tau_d$. $\endgroup$ Dec 4 '19 at 14:12
  • $\begingroup$ The above reasoning, however, does not use the fact that $\mathcal{F}_i$ are open in $\tau_d$, so where's the mistake? $\endgroup$ Dec 4 '19 at 14:13
  • 2
    $\begingroup$ @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 $[0,1]$. $\endgroup$ Dec 6 '19 at 0:36
  • $\begingroup$ 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. $\endgroup$ Dec 6 '19 at 10:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.