2
$\begingroup$

Let $\omega_1$ have the discrete topology. Its Fortissimo space is $X=\omega_1\cup\{\infty\}$ where neighborhoods of $\infty$ are co-countable.

A space is radial provided for every subset $A$ and point in its closure, there exists a transfinite sequence within $A$ converging to it (see d-4 Pseudoradial spaces).

Then $\infty\in cl(A)$ for any uncountable $A$, and given the "transfinite sequence" $a_\alpha=\min\{a\in A:\beta<\alpha\Rightarrow a>a_\beta\}$ for $\alpha<\omega_1$, we have that every neighborhood of $\infty$ contains a final subsequence. Since $\infty$ is the only possible point added by the closure operator, this shows $X$ is radial.

The problem is: is $a_\alpha$ actually a transfinite sequence? This mapping from the ordinal space $\omega_1$ to $X$ is not continuous as its image is discrete. In fact, every convergent continuous image of an infinite limit ordinal in $X$ is eventually constant. Under this interpretation, the space is not radial.

After more careful review of my reference and JDH's comment, I think it's clear that the first interpretation is correct, and the space is in fact radial.

However, does the more restrictive concept appear in the literature anywhere?

Improve this question
$\endgroup$
7
  • 4
    $\begingroup$ I would find it odd to have an unstated continuity requirement when referring to transfinite sequences in a topological space, and when continuity was required, I would rather want to refer to them as continuous transfinite sequences. So I think you do have an $\omega_1$-sequence here. But I know little about radial spaces or what the standard terminology is in connection with them. $\endgroup$
    – Joel David Hamkins
    Dec 18 at 19:57
  • $\begingroup$ It's not in front of me at the moment, but the reference I give for the term uses the ambiguous word "map", which some authors use to mean continuous. $\endgroup$
    – Steven Clontz
    Dec 18 at 20:36
  • 1
    $\begingroup$ Back at the laptop - in an earlier chapter, "map" is clearly defined to be an arbitrary function, with "continuous map" being more restrictive. So I would say the space is radial; the question remains if there is an existing notion for the more restrictive interpretation. $\endgroup$
    – Steven Clontz
    Dec 18 at 21:29
  • 2
    $\begingroup$ Every time I've seen radial defined all that is required is that every neighbourhood of the limit contains a tail of the sequence. The only places where I've seen continuity be an issue is when the question was: "is there a closed copy of the ordinal space in the space under consideration?" $\endgroup$
    – KP Hart
    Dec 19 at 14:19
  • 2
    $\begingroup$ @StevenClontz Rummage around in Alan's work; he has a fair number of publications involving radial and pseudoradial spaces. There was a change of terminology in the '80s/'90s radial spaces were called chain-net spaces for a while. The LOTS theorem is an easy exercise: if $x\in\bar A$ then it is in the closure of $A\cap(\gets,x)$ or of $A\cap(x,\to)$. Take a well-ordered cofinal sequence in the former or a well-ordered initial sequence in the latter. $\endgroup$
    – KP Hart
    Dec 20 at 17:37

1 Answer 1

Reset to default
4
$\begingroup$

Summarizing comments as an answer.

The standard interpretation of radial does not require that $\alpha\mapsto a_\alpha$ be continuous. Therefore, the Fortissimo space is radial, despite the lack of non-trivial continuous maps from limit ordinals into the space. This also aligns with the standard result that every LOTS is radial: when taking the necessary well-ordered initial/cofinal transfinite sequence in $A\cap(\leftarrow,x)$ or $A\cap(x,\rightarrow)$ in KP's proof in the comments, there's no guarantee that this sequence will be closed in the space.

There does not seem to be existing terminology for the strengthening of radial that requires the transfinite sequences be continuous maps from limit ordinals. However, "strongly pseudoradial" appears in https://arxiv.org/abs/1904.04416, suggesting "strongly radial" as a candidate.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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