0
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • $\begingroup$ 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 $i \in I$, $F_i \cap S \in \Sigma$. $\endgroup$
    – Robert Furber
    Commented Dec 4, 2019 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. $\endgroup$
    – Robert Furber
    Commented Dec 4, 2019 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. $\endgroup$
    – Jack London
    Commented Dec 9, 2019 at 21:03

1 Answer 1

Reset to default
2
$\begingroup$

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 $\mathcal F_i$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ 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 $\tau_d$; $\endgroup$
    – Jack London
    Commented Dec 4, 2019 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 $\tau_d$. $\endgroup$
    – Jack London
    Commented Dec 4, 2019 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? $\endgroup$
    – Jack London
    Commented Dec 4, 2019 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 $[0,1]$. $\endgroup$
    – Robert Furber
    Commented Dec 6, 2019 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. $\endgroup$
    – Jack London
    Commented Dec 6, 2019 at 10:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .