Topological spaces among US

[StevenClontz]

This is easily my most popular Math.SE question, though not due to the mathematical content I'm afraid...

Butterfly meme - Reaching for "inserting among us puns into my research" - Caption: "is this math outreach"

The "SUS" property came out of Alan Dow's comments the other day after my talk, though the memes are completely original. I raised it as a thing during today's Pitt seminar presentation I gave, and Paul Gartside suggested a more natural property between $k_2H$ and $US$ might consider one-point compactifications more generally, rather than the limit of a "long sequence". I need to work out details, but I think this would live between $k_2H$ and SUS. EDIT: (SUS will herefore be defined as USR, for "Unique Strongly Radial limits")

I decided to start a thread because I've been really enjoying doing my math in the open on Math.SE and MO, but this is a space where I can do so without quite as much concern for moderation. So I'm going to try working out some details on this property (and a few others Gartside suggested) in this thread; others are welcome to participate as they like (or mute the thread).

[StevenClontz]

Definition

A space $X$ is UK ("Unique Kompactifying limits") if for every non-compact locally compact Hausdorff $L$ and continuous function $f:L\to X$, there exists at most one compactifying limit $x\in X$ such that for each neighborhood $N$ of $x$, there exists co-compact $C\subseteq L$ with $f[C]\subseteq N$.

Proposition

If $f:L\to X$ is continuous and $x\in X$ is a compactifying limit, then $f[L]\cup{x}$ is compact.

Proof: Let $K=L\cup{k}$ Let $L$ be open with its usual neighborhoods, and let neighborhoods of $k$ be co-compact subsets of $L$. It follows that $K$ is compact Hausdroff.

Now let $f':K\to X$ be defined by $f'(l)=f(l)$ for $l\in L$, and $f'(k)=x$. We claim $f'$ is continuous and thus $f[K]=f[L]\cup{x}$ is compact. To see this, let $U$ be open in $X$. If $x\not\in U$, then by continuity of $f$ we have the inverse image of $U$ is open. If $x\in U$, choose co-compact $C\subseteq L$ with $f[C]\subseteq U$. Then ${k}\cup C\subseteq K$ is co-compact and thus a neighborhood of $k$ with $f'[{k}\cup C]\subseteq U$, and we're done.

Theorem

All $k_2H$ spaces are UK.

Proof: Consider $f:L\to X$ continuous such that there exist points $x,y\in X$ such that for each pair of neighborhoods $N_x,N_y$ of $x,y$, there exist co-compact sets $C_x,C_y\subseteq L$ with $f[C_x]\subseteq N_x,f[C_y]\subseteq N_y$. We define $g:K\times\{0,1\}\to X$ by $g(l,i)=f(l)$, $g(\infty,0)=x$, and $g(\infty,1)=y$ and note $g$ is continuous. Pick open neighborhoods of $(\infty,0),(\infty,1)$, and note their images must intersect. Thus by $k_2H$ we have $x=y$.

[StevenClontz]

Definition

A transfinite sequence is a map from an infinite cardinal to the space.

Definition

A space is USR ("Unique Strongly Radial limits") provided the limit of every countinous transfinite sequence is unique.

Proposition

Every UK space is USR, and every USR space is US.

Proof: Every limit ordinal is a non compact, locally compact Hausdorff space, and a standard sequence is a continuous transfinite sequence.

[StevenClontz]

EDIT: none of this was good (not deleting in case the edit history is helpful though)

[prabau]

Interesting stuff. The definition of gwH may not be completely clear, at least to me. The idea seems to be that you want to consider one-point compactifications $L\cup\{\infty\}$ and consider $x$ as a limit as points of $L$ tend to $\infty$. So $L$ should also be specified as non-compact, right? Also if we want to implicitly talk about $x=\lim_{z\to\infty}f(z)$ with $z\in L$, shouldn't it be that each nbhd ot $x$ should contain $f[L\setminus K]$ for some compact $K\subseteq L$? That does not seem to be the same as containing a co-compact subset of $f[L]$?

[StevenClontz]

Yeah, those were the issues that came to me last night when I decided to put a pin in things and get ready for bed. :-) I'll chew on this and make an edit above.

[StevenClontz]

Looking into spaces that aren't radial to find a USR-not-k2H example. Unfortunately this one turns out to be k1H, but it's still interesting I think.

Let $X=(\omega_1\times\omega)\cup\{\infty_0,\infty_1\}$ have $\omega_1\times\omega$ with the product of their order topologies, and neighborhoods of $\infty_i$ are of the form $\{\infty_i\}\cup(\omega_1\setminus\alpha)\times(\omega\setminus n)$.

$X$ is $k_1H$ because compacts are subsets of some Hausdroff subspace $(\alpha\times n)\cup\{\infty_0,\infty_1\}$. But I was hoping it would end up being more interesting, because long sequences cannot converge to either $\infty_i$.

[StevenClontz]

M W came to the rescue and pointed out a couple things.

Definition

A space is UR ("Unique Radial limits") provided any limit of a transfinite sequence is unique.

Proposition

$T_2\Rightarrow UR\Rightarrow USR$

Proof: standard arguments.

Example

The Fortissimo space on $\omega_1$ with a doubled extra point is USR but not UR. This is the space $\omega_1\cup{\infty_1,\infty_2}$ where neighborhoods of either $\infty$ are co-countable and $\omega_1$ is discrete. Then the identity of $\omega_1$ is a transfinite sequence with two limits, but the space is anticompact and $T_1$, and thus $k_1H$ and $USR$.

[StevenClontz]

Prior art suggests the use of "strong" rather than "continuous" for radial/pseduoradial proprerties: https://arxiv.org/abs/1904.04416

[prabau]

Good find about "strong pseudoradial". The 2015 article by Brazas & Fabel: https://arxiv.org/pdf/1301.4624.pdf

[StevenClontz]

The authors there use unique strong pseudoradial convergence. But is there a reason to use "pseudo"? I.e. what would *unique strong radial convergence be?

[prabau]

I agree there is no reason to use "pseudo" in their terminology. Presumably they were only interested in pseudoradial and not radial spaces in that paper, so they just carried over the "pseudo" prefix without thinking it through.

[StevenClontz]

I've updated most posts here to use USR for "Unique Strongly Radial limits" accordingly.

[StevenClontz]

Cleaning up definition of UK...

Definition

A space is $X$ is UK if for every locally compact, non-compact Hausdorff space $L$ with continuous $f:L\to X$, there is at most one point $x\in X$ such that $f$ extended to $L$'s one-point compactification $K=L\cup{\infty}$ by $f(\infty)=x$ is continuous.

Proposition

$k_2H \Longrightarrow UK \Longrightarrow USR$

Proof: For the first, take $L$, $f:L\to X$ continuous, and points $x_0,x_1$ such that extending $f$ to $K=L\cup{\infty}$ by either $f(\infty)=x_i$ is continuous. Define $g:K\times{0,1}\to X$ by $g(l,i)=f(l)$ and $g(\infty,i)=x_i$. Then $g$ is continuous, and there are no neighborhoods of $(\infty,0),(\infty,1)$ whose images are disjoint, so $g(\infty,0)=g(\infty,1)$, that is, $x_0=x_1$.

The second follows as every infinite cardinal is a locally compact, non-compact Hausdorff space.

[marswill]

We actually do indeed have $UR\implies UK$.

To see this, we first note that every non-isolated point in a compact Hausdorff space is radially accessible (i.e., it is the limit of some transfinite sequence that is not eventually constant). To prove this, let $X$ be compact Hausdorff, let $x\in X$ be non-isolated, and let $\mathcal C$ be a maximal chain of compact subsets of $X$ that contain $x$ as a non-isolated point, ordered by reverse inclusion. Then $\bigcap \mathcal C$ is compact, and must contain $x$ as an isolated point (otherwise $\bigcap \mathcal C$ would be a maximal member of $\mathcal C$, but then we could intersect with a closed neighborhood of $x$ that omits some other member of $\bigcap \mathcal C$ to obtain a strictly smaller compact set still containing $x$ as a limit point.)

But now, we can assume with no loss of generality that in fact $\bigcap \mathcal C=\{x\}$, since otherwise we could take a closed neighborhood $D$ of $x$ with $D\cap \bigcap \mathcal C=\{x\}$, and replace $\mathcal C$ with the family $\{C\cap D\mid C\in\mathcal C\}$.

Therefore, we may choose a well-ordered co-final chain $\{C_\alpha\}\subseteq \mathcal C$, and for each $\alpha$ choose $x_\alpha\in C_\alpha\backslash \{x\}$. We know every open neighborhood $U$ of $x$ must contain some $C_\alpha$, since otherwise the family $\{C_\alpha\backslash U\}$ is a family of closed sets with the finite intersection property but empty intersection, contradicting compactness. Thus $U$ contains $x_\beta$ for each $\beta>\alpha$, since $C_\beta\subset C_\alpha$. As this holds for all neighborhoods, the sequence converges to $x$.

Finally, since every non-isolated point in a compact Hausdorff space is radially accessible, if $X$ is $UR$ and $f\colon Z\to X$ is a map from a locally compact noncompact Hausdorff space $Z$ with compactifying limits $x_1,x_2\in X$, then let $z_\alpha\in Z$ be a transfinite sequence converging to $\infty$ in $Z^+$. Then $f(z_\alpha)$ converges to $x_1$ and $x_2$, so by $UR$, $x_1=x_2$.

[marswill]

Hmm, haven't been able to do serious research on this yet but I gather just from googling and inferring from how things are phrased that it is at least consistent that there are compact Hausdorff spaces that fail to be pseudoradial. See e.g. this link http://at.yorku.ca/p/a/a/g/11.htm. I'm sure with a bit more digging we can figure out whether this is always true in $ZFC$, or whether additional axioms are necessary.

[Edit: Never mind the striken comment above, I forgot about $\beta\omega$ not being pseudoradial from this question., so the proof below (taking $Z=\beta\omega$ and $A=\omega$) actually shows that from $\beta\omega$ one can construct a $UR$ -not- $k_2H$ space.]

---

In any case, we have $UR\implies k_2H$ if and only if every compact Hausdorff space is pseudoradial.

Proof

If every compact Hausdorff space is pseudoradial [edit this is never true in ZFC, but I'll leave the argument for posterity in case the logic is useful elsewhere], then suppose $X$ is $UR$, $Z$ is compact Hausdorff, and $f_1,f_2\colon Z\to X$ are continuous, and let $A\subseteq Z$ be the set on which $f_1=f_2$. Then for any transfinite sequence $a_\alpha\in A$ converging to $z\in Z$, $f_1(z)=\lim f_1(a_\alpha)=\lim f_2(a_\alpha)=f_2(z)$, so $z\in A$. Thus $A$ is radially closed, so by pseudoradiallity of $Z$, $A$ is closed. Since this holds for arbitrary compact Hausdorff $Z$, we see that $X$ is $k_2H$.

In the other direction, if $Z$ is compact Hausdorff and not pseudoradial, let $A\subset Z$ be radially closed but not closed. Let $B=Z\backslash A$, and let $X=A\cup B_1\cup B_2$ where $B_1$ and $B_2$ are disjoint homeomorphic copies of $B$, via homeomorphisms $\pi_i\colon B_i\to B$. Give $X$ the topology where $U\subseteq X$ is open if and only if $(U\cap A)\cup \pi_i(U\cap B_i)$ is open in $Z$, and let $\pi\colon X\to Z$ be such that $\pi$ is the identity on $A$ and agrees with $\pi_i$ on $B_i$. Note that $\pi$ is continuous.

We also have continuous maps $f_i\colon Z\to X$, where $f_i$ restricts to the identity on $A$, and to $\pi_i^{-1}$ on $B$.

Then $X$ is $UR$. To see this, observe that if $x_1\neq x_2\in X$, then $x_1$ and $x_2$ can always be separated by disjoint open neighborhoods, except in the case $x_i=\pi_i(b)$ for some $b\in B$. In this latter case, suppose $x_\alpha$ is a transfinite sequence converging to $x_1$. Let
$$S_\alpha:=\overline{\bigcup_{\beta>\alpha}{\pi(x_\beta)}}.$$
Since $A$ is radially closed in $Z$, we eventually have $S_\alpha\cap A=\emptyset$, since $S_\alpha$ eventually lies in every closed neighborhood of $b$, so if the intersection were always nonempty we could extract a radially convergent sequence of points in $A$ converging to $b\notin A$.

Therefore we may choose some $\alpha$ with $S_\alpha\cap A=\emptyset$. Since $X\backslash f_2(S_\alpha)$ is by definition a neighborhood of $x_1$, we eventually have $x_\beta\notin f_2(S_\alpha)$, so eventually $x_\beta\in B_1$, and thus $x_\beta\in f_1(S_\alpha)$. But then $x_\beta\notin X\backslash f_1(S_\alpha)$, which is a neighborhood of $x_2$, so the sequence cannot converge to $x_2$. We conclude that $X$ is $UR$.

On the other hand, the maps $f_i$ are continuous maps from the compact Hausdorff $Z$ into $X$, which agree precisely on the non-closed set $A$, and therefore witness that $X$ is not $k_2H$.

[marswill]

Also, since $UR\implies UK$ is true and $UR\implies k_2H$ is false, that also gives us that $UK\implies k_2H$ is false, so that completely determines all implications between the properties, i.e., we have implications

$$T_2\implies UR\implies UK,$$

$$T_2\implies k_1H\implies... k_2H\implies UK,$$

and

$$UK\implies USR \implies US\implies T_1,$$

all strict, and no other implications.

[prabau]

@marswill fyi, for braces "{" in math you need to precede them with double backslashes here.

[StevenClontz]

@marswill I'm going to DM you on the code4math Zulip to coordinate a time to hop on Zoom and move this forward.

[marswill]

We can generalize some of this discussion. I'll outline the general concepts here, with the rigorous proofs TBA in the comments when I get a chance.

---

Let $\mathcal C$ be a class of topological spaces. For a topological space $X$:

• the $\mathcal C$-ification $\mathcal CX$ is the set $X$, together with the final topology from the family of all continuous maps $f\colon Z\to X$ with $Z\in \mathcal C$.
• A subset of $X$ is $\mathcal C$-closed if it is closed in $\mathcal C X$.
• $X$ is $\mathcal C$-generated ($\mathcal C$-Gen)if $\mathcal C X=X$ (equivalently, the closed subsets and $\mathcal C$-closed sets coincide).
• $X$ is $\mathcal C$-Hausdorff if the diagonal is $\mathcal C$-closed (equivalently, the diagonal in $\mathcal C(X\times X)$ is closed).

(@prabau had a math se post more or less outlining this approach for $k_n$-topologies, $n=1,2,3$, though only $n=1,2$ conform to what we're doing here).

---

Then consider the following classes:

• $\mathcal{SIERP}$ consists solely of the Sierpinski space.
• $\mathcal{SEQ}:=\{\omega+1\}$.
• $\mathcal{ORD}$ is the class of ordinals with their order topology.
• $\mathcal{K_H}$ is the class of compact Hausdorff spaces.
• $\mathcal{K}$ is the class of all compact spaces.
• $\mathcal{RAD}$ is the class of successors to limit ordinals, with discrete topology except at their maximum point, which has the same neighborhoods as the order topology. (We could in fact limit it to successors to regular cardinals and nothing changes).
• $\mathcal{HAUS}$ is the class of all Hausdorff spaces.

Then a space $X$ is

• $\text{Alexandrov} \iff \mathcal{SIERP}\text{-Gen}$. *
• $\text{Sequential}\iff \mathcal{SEQ}\text{-Gen}$.
• $CG_2\iff \mathcal{K_H}\text{-Gen}$.
• $CG_1\iff \mathcal{K}\text{-Gen}$.
• $\text{Pseudoradial}\iff \mathcal{RAD}\text{-Gen}$.
• Every topological space is $\mathcal{HAUS}\text{-Gen}$. *

Moreover, $X$ is

• $T_1\iff \mathcal{SIERP}\text{-Hausdorff}$. *
• $US\iff \mathcal{SEQ}\text{-Hausdorff}$.
• $USR\iff \mathcal{ORD}\text{-Hausdorff}$. *
• $k_2H\iff \mathcal{K_H}\text{-Hausdorff}$.
• $k_1H\iff \mathcal{K}\text{-Hausdorff}$.
• $UR\iff \mathcal{RAD}\text{-Hausdorff}$.
• $T_2\iff \mathcal {HAUS}\text{-Hausdorff}$. *

[*These will be elaborated on in comments.]

Note that $UK$ doesn't seem to fit into this picture, though there is a modifications to $UK$ that might be of interest that I'll try to spell out a little later. Bottom line, though, while I don't have an absolute proof one way or another as to whether $UK$ is $\mathcal C$-Hausdorff for some class $\mathcal C$, I'm leaning against believing there is such a class.

---

We can generalize the proof from a previous comment, which said that $UR\implies k_2H$ if and only if every compact Hausdorff space is Pseudoradial (of course both statements are in fact false), as follows:

Lemma. Let $\mathcal C$ and $\mathcal D$ be classes of topological spaces. If every member of $\mathcal C$ is $\mathcal D$-generated, then every $\mathcal D$-Hausdorff space is $\mathcal C$-Hausdorff.

[proof TBA in comments]

Under a Hausdorff assumption we have a converse:

Proposition. Let $\mathcal C$ and $\mathcal D$ be classes of topological spaces, so that every member of $\mathcal C$ is Hausdorff. If there is some member of $\mathcal C$ that is not $\mathcal D$-generated, then there is a $\mathcal D$-Hausdorff space that fails to be $\mathcal C$-Hausdorff.

[Proof TBA in comments]

The previous result relating $UR$ and $k_2H$ is a special case of the preceding results, with $\mathcal C=\mathcal{K_H}$ and $\mathcal D=\mathcal{RAD}$.

---

Update

I should have known this was too good to be true. After trying to write it up the proof of the proposition carefully, it became clear that in my proof above, I was relying much more that I realized on important properties of $\mathcal{RAD}$ that do not generalize.

I suspect we do have a much less ambitious version of the stricken proposition, though having learned my lesson, I will call this a conjecture until after I've carefully written the proof:

Conjecture. Let $\mathcal C$ be a class of topological spaces. If there is some $T_3$ member of $\mathcal C$ that is not $\mathcal{RAD}$-generated(i.e. Pseudoradial), then there is a $\mathcal{RAD}$-Hausdorff (i.e., $UR$) space that fails to be $\mathcal C$-Hausdorff.

I do feel it's pretty likely my argument does work for the aforementioned "conjecture," and will try to write it up soon. I also suspect it might generalize past transfinite sequences to Fort spaces modeled on directed sets in a certain sense.

[StevenClontz]

I like this approach a lot.

[marswill]

First, let's tackle the starred statements.

$\text{Alexandrov} \iff \mathcal{SIERP}\text{-Gen}$. *

Let $S=\{0,1\}$ be the Sierpinski space with the topology where $1\in \overline{\{0\}}$.

$X$ is Alexandrov if and only if closed sets are precisely those for which
$$x\in A\implies \overline{\{x\}}\subseteq A.\tag{1}$$
(I could dig up a reference but I think this is standard and is an exercise regardless).

(1) is equivalent to the statement that $A$ is closed if and only if for all $x,y\in X$, with $y\in\overline{\{x\}}$, we have $x\in A\implies y\in A$. But continuous maps $f\colon S\to X$ are precisely those maps for which $f(1)\in \overline{\{f(0)\}}$, so every pair $x,y$ with $y\in\overline{\{x\}}$ corresponds to a continuous map $f(0)=x$, $f(1)=y$, and vice versa, The statement $x\in A\implies y\in A$ is then equivalent to the statement that $0\in f^{-1}(A)\implies 1\in f^{-1}(A)$, which in turn is precisely equivalent to the claim that $f^{-1}(A)$ is closed.

• Every topological space is $\mathcal{HAUS}\text{-Gen}$. *

If $A\subseteq X$ is not closed, then let $p\in\overline{A}\backslash A$. There are two possibilities:

If $p\in \overline{\{a\}}$ for some $a\in A$, then define $f\colon [0,1]\to X$ by

$$f(t)= \begin{cases} a & t<1\\ p & t=1. \end{cases} $$

Then $f$ is continuous and $f^{-1}(A)=[0,1)$ is not closed.

If, on the other hand, we have that $p\notin \overline{\{a\}}$ for each $a\in A$, then let $Z=A\cup \{p\}$ have the topology where the neighborhoods of $p$ are precisely those neighborhoods from the subspace topology in $X$, and every other singleton is open. Then since $p\notin \overline{\{a\}}$ for each $a\in A$, $Z$ is $T_1$ with all but one point isolated, hence $Z$ is $T_2$, and the inclusion map from $Z$ into $X$ is continuous, and is our desired map.

$T_1\iff \mathcal{SIERP}\text{-Hausdorff}$. *

If $X$ is $T_1$, then any continuous map $f\colon S\to X$ must satisfy $f(1)\in \overline{\{f(0)\}}=\{f(0)\}$, hence are constant. Thus any two such maps agree precisely on either all of $S$ or $\emptyset$, so the diagonal is $\mathcal{SIERP}$-closed. Conversely, if $X$ fails to be $T_1$, then let $y\neq x$, $y\in \overline{\{x\}}$. Then $f(0)=g(0)=x$, $f(1)=x$, $g(1)=y$ defines two continuous functions which agree on a non-closed set, so the diagonal is not $\mathcal{SIERP}$-closed.

$USR\iff \mathcal{ORD}\text{-Hausdorff}$. *

If $X$ fails to be $USR$, then let $f,g\colon \alpha+1\to X$ be two continuous maps that agree for all $\beta&lt;\alpha$, and disagree on $\alpha$. Then they agree precisely on a non-closed subset, so the diagonal is not $\mathcal{ORD}$-closed.
Conversely, suppose $f,g\colon \alpha+1\to X$ are two maps that agree precisely on a non-closed subset $A\subset\alpha+1$. Then let $\beta$ be the first element of $\overline{A}\backslash A$.

Then $\beta$ is certainly a limit ordinal, and $\overline{A}\cap (\beta+1)$ is a closed subset of $\alpha+1$ with a maximal element (namely $\beta$), hence homeomorphic to the successor of a limit ordinal. Since the restrictions of $f$ and $g$ agree on every element of $\overline{A}\cap (\beta+1)$ except $\beta$, they witness a contradiction to $USR$.

$T_2\iff \mathcal {HAUS}\text{-Hausdorff}$. *

Since $\mathcal{HAUS}\text{-Gen}$ consists of all topological spaces, the $\mathcal{HAUS}$-ification of a space is the space itself, so $\mathcal{HAUS}$-Hausdorff spaces are simply those in which the diagonal is closed.

[marswill]

Proof of Lemma.

Suppose every member of $\mathcal C$ is $\mathcal D$-generated, and let $X$ be $\mathcal D$-Hausdorff. Let $f,g\colon Z\to X$ be continuous, with $Z\in \mathcal C$, and let $A=\{z\in Z\mid f(x)=g(x)\}$. Then for every continuous $h\colon Y\to Z$, with $Y\in \mathcal D$, we have $h^{-1}(A)=\{y\in Y\mid f\circ h(y)=g\circ h(y)\}$. Since $Y\in \mathcal D$ and $X$ is $\mathcal D$-Hausdorff, it follows that $h^{-1}(A)$, as the set of points on which $f\circ h$ and $g\circ h$ agree, is closed.

Since this holds for all such maps $h$, and $Z$ as a member of $\mathcal C$ is $\mathcal D$-generated, we see that $A$ is closed. As this applies to arbitrary pairs of maps $f,g\colon Z\to X$, with $Z\in \mathcal C$, we conclude the diagonal of $X\times X$ is $\mathcal C$-closed.

[prabau]

I have not read this, but you may find it possibly somewhat related?