Let's call a space $X$ geometric if its components and quasi-components coincide. Let's also define a property called the boundary bumping property:

$X$ has the boundary bumping property ("bbp") if for every proper, closed subset $A$ and a component $C$ of $A$, we have $C \cap \partial(A) \neq \varnothing$.

It is clear that $X$ must be geometric to have the boundary bumping property by the definitions. It is known that if $X$ is a compact, connected Hausdorff space then it has the bbp; if $X$ is a connected and locally connected $T_1$ space then it, too, has the bbp. Both are elementary but a bit long; I think their proofs are floating around on MSE if you want to see them. It's also clear that the space needs to be connected, just by definition of separation.

QUESTION: 1) What might be some sufficient conditions for a connected metric space to have the bbp, besides locally connected or compact? 2) What are necessary conditions? 3) Classify connected, geometric metric spaces with the bbp.

DISCUSSION: Taking a comb with the bottom half of the 'limit tooth' lopped off gives a different sort of counterexample, though not compact. By renormalizing the metric this shows that topological completeness doesn't get you anywhere in the non-compact case.

When $X$ is not locally connected, it may still have the bpp without being compact: Stick a two-sided topologist's sine curve in the middle of $\mathbb{R}$ and we'll get a space that has the bbp but not be compact or locally connected. Removing a point from the critical limit arc will still have the bbp (check it), and this space won't have an exhaustion by continua.

If you take a 'necklace' of pseudo-arcs (it's a continuum that's not locally connected anywhere) attaching one to the next at a single point, then it will have the bbp but be locally connected nowhere; for this one just trust me, if you're not already familiar with the pseudo-arc. Also note that $\mathbb{R}^\omega$ is connected, locally connected and $T_1$ so has the bbp, but is not locally compact.

Summarizing the above examples:

A connected metric space with the bbp might be nowhere-locally connected, it need not be compact, need not be locally compact, need not be topologically complete, and need not have an exhaustion by continua.

Similar questions can be asked for geometric $T_1$ or $T_2$ spaces rather than metric spaces, ofc.


A subset $A$ of $X$ is called a connector if for every component $C$ of $X \setminus A$ we have $\overline{C} \cap \overline{A} \neq \varnothing$. A space is closely connected if every proper, non-empty subset of $X$ is a connector.

As proved here, a connected space $X$ has the bbp if and only if it is closely connected.

This condition is like asking for the bbp on $X \setminus A$ for arbitrary subsets $A$. It shows that having the bbp implies this seemingly-stronger condition. So in a sense, this characterization is a bit of a cheat since one direction implies the other immediately. I'm still interested in finding properties that every bbp space must have, or sufficient conditions other than the ones mentioned above.

So I'm not gonna give this the check mark.


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.