# A reasonable topology on the group of minimal usco maps

An usco map is an abreviation for an upper semicontinuous multi-valued map with non-empty compact values. An usco map $f:X\multimap \mathbb R$ is called minimal is it coincides with each usco map $g:X\multimap \mathbb R$ with $g\subset f$. Here we identify multi-valued maps with their graphs.

It is well-known that each minimal usco map $f:X\multimap \mathbb R$ on a Polish space $X$ is single-valued on a dense $G_\delta$-subset $\mathsf{single(f)}$ of $X$. This allows us to define an addition operation $\oplus$ on the set $\mathsf{musco}(X)$ of all minimal usco maps letting $f\oplus g$ be the closure of the (graph of the) function $f+g|(\mathsf{single}(f)\cap\mathsf{single}(g))$. It can be shown that this operation turns $\mathsf{musco}(X)$ into a group containing the vector space $C(X)$ of continuous real-valued functions as a subgroup.

Problem. Is any reasonable (natural) topology turning $\mathsf{musco}(X)$ into a (semi)topological group?

• Cannot you similarly define $\lambda f$ as the closure of $\lambda f|_{\mathsf{single}(f)}$? It would then presumably give a vector space... – მამუკა ჯიბლაძე Jul 12 '18 at 12:02
• What's a "multivalued map"? Is it just a relation? – Qfwfq Jul 12 '18 at 16:35
• @მამუკაჯიბლაძე Yes, we can also define a multiplication by a number and make $\mathsf{musco}(X)$ a vector space. – Taras Banakh Jul 13 '18 at 4:09
• @Qfwfq You are right: a multi-valued map is just a relation (i.e., a subset of $X\times\mathbb R$). – Taras Banakh Jul 13 '18 at 4:10

## 2 Answers

I have just realized that the group $\mathsf{musco}(X)$ carries a natural supremum metric $$\rho(f,g)=\sup_{x\in X}\{|y|:\exists x\in X\mbox{ with }y\in (f-g)(x)\}=\sup\{|f(x)-g(x)|:x\in\mathsf{single}(f)\cap\mathsf{single}(g)\}.$$

Warning: The approach described below does not work. The problem is the pointwise addition on the continuous partial functions is not the same operation as pointwise addition on the musco maps; the domain of the results in the former case can be a strict subset of the set of single-valued values of the latter. As a consequence, we don't get the group structure, as pointed out by Tarakh in the comments.

Side remark: I am reasoning about this natively in the setting of represented spaces, not of topological spaces. There is a chance that to translate properly, some sequentializations need to be added in some places.

I would start by looking at the space $\mathcal{P}(X,\mathbb{R})$ of continuous partial functions from $X$ to $\mathbb{R}$. These naturally have $G_\delta$-domains (not necessarily dense though), and pointwise addition is a continuous operation.

The topology on this space is found as follows: Fix a dense sequence $(a_n)_{n \in \mathbb{N}}$ in $X$ and a universal Turing function $\Phi$. We say that $x \in 2^\omega$ codes $f : \subseteq \mathbf{X} \to \mathbb{R}$ iff whenever $p \in \omega^\omega$ is such that $d(a_{p(i)},x) < 2^{-i}$ for some $x \in X$, then if $x \in dom(f)$, $\Phi^x(p)$ is total, and $\Phi^x(p)(n)$ codes a rational $q$ with $d(q,f(x)) < 2^{-n}$, and if $x \notin dom(f)$, then $\Phi^x(p)$ is partial. Take the subspace of $2^\omega$ of those $x$ that do code a function, and quotient by coding the same function.

Assuming that we understand the space of upper semicontinuous maps with compact images to be $\mathcal{C}(\mathbf{X},\mathcal{K}(\mathbb{R}))$ (where $\mathcal{K}(\mathbb{R})$ essentially carries the topology induced by $\{\{K \mid K \subseteq U\} \mid U \in \mathcal{O}(\mathbb{R}))$, then we have that:

Proposition: $\operatorname{single} : \mathcal{C}(\mathbf{X},\mathcal{K} (\mathbb{R})) \to \mathcal{P}(\mathbf{X},\mathcal{R})$ is continuous.

The proof is essentially noting that we can continuously extract a point from a compact singleton.

We could then just look at the subspace of $\mathcal{P}(X,\mathbb{R})$ arising as the image of the musco's under that function to obtain a candidate for the desired topology. This has the feature that the total continuous functions $\mathcal{C}(\mathbf{X},\mathbb{R})$ show up as a subspace with the usual topology.

This topology is a quite weak one, though, and the natural reversal operation of taking the closure of the graph of a partial function is not continuous with any desirable topology on the right hand side.

• Thank you for the answer. The pointwise addition does not turn $\mathcal P(X,\mathbb R)$ into a group, but merely into an inverse semigroup. So, the map $C(X,\mathcal K(\mathbb R))\to \mathcal P(X,\mathbb R)$ is not a homomorphism. On the other hand, I have realized that the group $\mathbb{musco}(X)$ carries a natural supremum metric, in which is a topological group, maybe even complete! – Taras Banakh Jul 13 '18 at 4:17