2
$\begingroup$

Consider the topological spaces $X$ with the following property:

For every compact $K\subseteq X$ there is a compact set $L$ such that $K\subseteq L\subseteq X$ and $L$ is a retract of $X$.

Let us call this property RBC (Retract to a Bigger Compact). This property seems natural and useful, but I have been unable to find anything directly on it.

It is not hard to see that the following is true:

Proposition 1: Every closed convex topological subspace $C$ of a locally convex metrizable topological vector space $V$ has the RBC property.

Details on Proposition 1: Take any nonempty compact $K\subseteq C$. Then, by known results (note, in particular, Ref. [7] there), the closed convex hull $L:=\overline{\text{conv}\,K}$ of $K$ in $V$ is a retract of $V$. Since $L\subseteq C\subseteq V$, it follows that $L$ is a retract of $C$. Since $K$ is compact, $L$ is also compact, and of course $K\subseteq L\subseteq C$. Thus, $C$ has the RBC property. $\Box$

It is also clear that the RBC property is topological and thus invariant with respect to homeomorphisms. Trvially, any compact topological space has the RBC property.

Question: Can one characterize the RBC property?

That is, can one give a usable necessary and sufficient condition for it? Or a usable sufficient condition somewhat close to necessity? By "usable", I mean without the quantifier "there is" in the definition of the RBC property.

I do not know the answer even to this question: Is there an example of a topological space without the RBC property?

Thinking about the latter question, I have in mind the "non-retract" example of the $(n-1)$-sphere, which is compact but not a retract of the corresponding closed ball, whereas the ball is compact as well and of course is a retract of itself and of the corresponding $n$-space.

Update: The latter question has been answered in comments by erz, Anonymous, and Taras Banakh, who provided examples of topological spaces without the RBC property.

At this point, I would like to make the first question, to characterize the RBC property, more specific:

Specific question: Is it true that all Polish spaces have the RBC property?

My motivation for all these questions comes from probability. Indeed, any retraction $r$ from $X$ to a compact subset $K$ of $X$ naturally induces the truncation map $\xi\mapsto r\circ\xi$ of random elements $\xi$ of $X$, so that the truncated version $r\circ\xi$ of $\xi$ is a random element of the compact set $K$. Moreover, this truncation map is continuous: if $\xi_t\to\xi$ in distribution, then $r\circ\xi_t\to r\circ\xi$ in distribution.

Update 2: Following the latest comment by Anonymous, I have found a recent paper by Lipham, where on the first page one finds an example of a connected completely metrizable space which has no compact connected subset of cardinality greater than one, and therefore does not have the RBC property. By Mazurkiewicz's theorem, this space is also separable and hence Polish. Thus, it is not true that all Polish spaces have the RBC property.

$\endgroup$
  • $\begingroup$ Please ignore my previous comments, I was thinking of deformation retracts. $\endgroup$ – Wojowu Apr 30 at 19:44
  • $\begingroup$ @Wojowu : I did not see your previous comments, but I think for me, with my total inexperience in this field, almost any comments could be useful. $\endgroup$ – Iosif Pinelis Apr 30 at 19:49
  • $\begingroup$ My comments were saying that the infinite discrete space and the long line do not satisfy your stated property for deformation retracts, though they have pretty obvious retracts onto large compact sets. I'm afraid I don't really see a way to adapt these to the question at hand. $\endgroup$ – Wojowu Apr 30 at 19:52
  • $\begingroup$ @IosifPinelis Apologies I've misread the question. $\endgroup$ – Denis Nardin Apr 30 at 21:14
  • 2
    $\begingroup$ For the second question, let X be a connected space such that no compact subset having at least two elements is connected. Then X obviously fails to have the RBC property. $\endgroup$ – Anonymous May 1 at 11:49

Your Answer

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

Browse other questions tagged or ask your own question.