The property US ("Unique Sequential limits") is a classic example of property implied by $T_2$ and implying $T_1$. In fact, it's the weakest assumption out of a chain of several distinct properties studied in the literature:

During a recent meeting of the Carolinas topology seminar, some comments of Alan Dow got me to thinking about another intermediate property. We say that a long sequence is a continuous function $f:\kappa\to X$ for an infinite cardinal $\kappa$, and it has a limit $x\in X$ provided for every neighborhood of $x$, the neighborhood contains $f[\kappa\setminus\alpha]$ for some $\alpha<\kappa$.

Then let's call a space SUS ("strongly US") if whenever a long sequence has a limit, this limit is unique. It's immediate that all SUS spaces are US. Furthermore, every $k_2H$ space $X$ is SUS: let $x,y$ be limits of a sequence $f:\kappa\to X$. Then consider the compact Hausdorff space $K=(\kappa+1)\times\{0,1\}$ and the continuous function $g:K\to X$ defined by $g(\alpha,i)=f(\alpha)$, $g(\kappa,0)=x$, and $g(\kappa,1)=y$. Note that there do not exist open neighborhoods $U,V$ of $(\kappa,0),(\kappa,1)$ with $f[U],f[V]$ disjoint, so it follows that $f(\kappa,0)=f(\kappa,1)$ and $x=y$.

The standard example of a US-not-$k_2H$ space is $\omega_1+1$ with the endpoint doubled: it's US as only trivial $\omega$-length sequences converge to $\omega_1$, but it fails to be SUS as the identity on $\omega_1$ is a long sequence with two distinct limits at the doubled endpoint.

So, is it possible to construct a SUS space that's not $k_2H$? Let's call such a space an imposter.

EMERGENCY MEETING: This space seems pretty SUS: $X=[0,1]\cup\{\infty\}$ where points of $[0,1]$ have their usual neighborhoods, and neighborhoods of $\infty$ must contain an open dense subset of $[0,1]$.

Of course its open subspace $[0,1]$ is definitely SUS: it's metrizable and thus Hausdorff. So it's sufficient to show that if $x\in[0,1]$ is a limit of a long sequence, $\infty$ is not a limit of that sequence.

Let $x\in[0,1]$ be the limit of a long sequence given by $f:\kappa\to X$. Then for each $n<\omega$, there exists $\alpha_n<\kappa$ such that $B(x,1/2^n)$ contains $f[\kappa\setminus\alpha_n]$.

Suppose $\sup_{n<\omega}\alpha_n=\alpha<\kappa$. Then $f[\kappa\setminus\alpha]=\{x\}$, and $X\setminus\{x\}$ is a neighborhood of $\infty$ missing $f[\kappa\setminus\alpha]$.

Suppose $\sup_{n<\omega}\alpha_n=\kappa$. Then $f\upharpoonright\{\alpha_n:n<\omega\}$ is a countable sequence converging to $x$ in $[0,1]$. Then $\{\alpha_n:n<\omega\}$ is a nowhere dense subset of $[0,1]$, so $X\setminus\{\alpha_n:n<\omega\}$ is a neighborhood of $\infty$ that fails to contain a final subsequence of $f$.

However, while I can tell that this space is not weakly Hausdorff (it contains a non-closed copy of the compact Hausdorff space $[0,1]$), it's not clear to me whether or not it is $k_2H$ (and therefore whether or not it is an imposter).

Update: This space is not an imposter. Whoops. Here's why it's $k_2H$.

Let $K$ be compact Hausdorff, let $f:K\to X$ be continuous, and let $f(l)\not=f(k)$. To check the $k_2H$ criterion, we need only consider the case where $f(l)=\infty$ as pairs of points in $[0,1]$ are separated by open sets in $X$.

Let $V$ be an open neighborhood of $l$ whose closure is contained in $f^\leftarrow[X\setminus\{f(k)\}]$; in particular, so $f(k)\not\in f[cl(V)]$. It follows that $cl(V)$ is compact and thus $f[cl(V)]$ is compact. We claim it is closed in $X$, so let $x$ be a limit point. If $x=\infty$, $x=f(l)$ and we're done. Otherwise, pick $x_n\in B(x,1/2^n)\cap f[cl(V)]$ for each $n<\omega$. Then the open collection $\{X\setminus\{x_n:N\leq n<\omega\}:n<\omega\}$ covers $X\setminus\{x\}$ but has no finite subcover for $\{x_n:n<\omega\}$. It follows it cannot be a cover of $f[cl(V)]$, showing $x\in f[cl(V)]$.

Since $f(k)\not\in f[cl(V)]$, let $U=f^\leftarrow[X\setminus f[cl(V)]]$. We have now obtained open neighborhoods $U,V$ for $k,l$ such that $f[U]\cap f[V]\subseteq (X\setminus f[cl(V)])\cap f[cl(V)]=\emptyset$.

Sidenote: after presenting related work to the Pitt topology seminar earlier today, Paul Gartside pointed out that perhaps a more natural definition for a "strong US" property would involve something like continuous functions from arbitrary Hausdorff spaces, rather than ordinals, and then consider unique limits (where a limit must have neighborhoods with co-compact intersection with the image). If we called this SUS', then we'd have $k_2H$ implies SUS' implies SUS. I think it's more likely for SUS' to be equivalent to $k_2H$ than the ordinal-focused SUS discussed here, but there's something else for folks to consider thinking about.

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    $\begingroup$ I really admire the devotion to the meme here. $\endgroup$
    – Alex Ortiz
    Sep 30, 2023 at 20:05
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    $\begingroup$ extremely real ඞ $\endgroup$
    – Max0815
    Oct 1, 2023 at 0:12
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    $\begingroup$ Terminology-wise, using transfinite sequences versus ordinary sequences is the difference between sequential spaces and (pseudo)radial spaces, and between sequentially closed sets and radially closed sets. Similarly, one has "unique sequential limits" (US) versus "unique radial limits" (UR ?). $\endgroup$
    – PatrickR
    Oct 4, 2023 at 3:52
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    $\begingroup$ It's not nearly as fun as SUS, but I'm forced to admit it's a better term. Thanks Patrick. $\endgroup$ Oct 4, 2023 at 13:22
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    $\begingroup$ Up until now I had not realized the importance of one crucial word: the transfinite sequences from $\kappa$ to $X$ are meant to be continuous when $\kappa$ is given its usual order topology. $\endgroup$
    – PatrickR
    Dec 22, 2023 at 5:00

1 Answer 1


The side note gives a hint at how to proceed with the main question, i.e., if we find a space that's $SUS$ but not $SUS'$ then we're done.

Consider a non-isolated point $p\in\beta\omega$, where $\beta\omega$ is the Stone-Čech compactification of $\omega$. Double the point $p$, i.e., let $X=\beta\omega\cup\{p'\}$, where each $x\in\beta\omega$ has the usual topology from $\beta\omega$, and a base at $p'$ is given by sets of the form $\{p'\} \cup U\backslash\{p\}$ for $U\subseteq \beta\omega$ open.

Since $\beta\omega$ is sequentially discrete (convergent sequences are eventually constant), it follows that $X$ is as well. Moreover, in a sequentially discrete space, convergent long sequences must also be eventually constant (i.e., if $f\colon \kappa \to X$ is a long sequence converging to $x$, then for some $\lambda<\kappa$, $f([\lambda,\kappa))=\{x\}$). To see this note that otherwise, by $T_1$, $f$ would have infinite range, giving some sequence $\alpha_n<\kappa$ for which the values $f(\alpha_n)$ were distinct. But by continuity $f(\alpha_n)\to f(\sup\alpha_n)$ (unless $\sup\alpha_n=\kappa$, in which case we have $f(\alpha_n)\to x$), so $f(\alpha_n)$ is eventually constant, contradicting distinctness.

Hence $X$ is $SUS$.

On the other hand, if $f\colon \beta\omega\to X$ is the inclusion and $g\colon \beta\omega\to X$ is given by $$ g(x)= \begin{cases} x & x\neq p\\ p' & x=p, \end{cases} $$ then $f$ and $g$ are continuous maps to $X$ from a compact Hausdorff space that agree on a non-closed subset, hence $X$ is not $k_2H$ (since $f\times g$ witnesses that the diagonal of $X\times X$ is not $k_2$-closed).

Chuck that imposter out the airlock!


We could unify understanding of these properties a bit as follows. If $\mathcal C$ is a class of pairs $(A,Z)$, where $A\subset Z$ is a proper subset of a topological space $Z$, then say that $X$ has the $\mathcal C$-unique extension property ("$UE_{\mathcal C}$") if given any $(A,Z)$ in $\mathcal C$ and continuous function $f\colon A\to X$, there is at most one continuous extension $\tilde{f}\colon Z\to X$.

Various choices of $\mathcal C$ equate to various properties, e.g.,

  1. Taking $\mathcal C=\{(\omega,\omega+1)\}$ gives $US$.
  2. The class of pairs $(\kappa,\kappa+1)$ for $\kappa$ an infinite cardinal gives $SUS$.
  3. The class of pairs $(Z,Z^+)$ of noncompact locally compact Hausdorff spaces $Z$ with their one-point compactifications $Z^+$ give something like $SUS'$ from the side-note. (While not mentioned in that side note, it seems one needs local compactness to get the desired implication $k_2H\implies SUS'$, though I haven't thought of a counterexample otherwise.)
  4. The class of pairs $(A,Z)$ where $A$ is dense in the compact Hausdorff $Z$, gives $k_2H$.
  5. Dropping "Hausdorff" from 4. gives $k_1H$.
  6. If $\mathcal C$ includes some pair $(A,Z)$ with $A$ open in $Z$, then $UE_{\mathcal C}\implies T_1$, since for $X$ not $T_1$ we have some $x\neq y\in X$, $x\in \overline{\{y\}}$, and then we can define $f\colon A\to X$ as $f\equiv y$, and can extend $f$ to either the constant function $f_1\colon Z\to X$, $f_1\equiv y$, or the function $$f_2(z)=\begin{cases}y & z\in A\\ x & z\notin A.\end{cases}$$
  7. If $A$ is dense in $Z$ for each pair $(A,Z)$ in $\mathcal C$, then $T_2\implies UE_{\mathcal C}$, since given two extensions $f_1,f_2$ of $f\colon A\to X$, $(f_1\times f_2)^{-1}(\Delta X)$ is a closed set containing $A$, hence is all of $Z$.

From this perspective the (known) implications $$T_2\implies k_1H\implies k_2H\implies SUS'\implies SUS\implies US\implies T_1$$ are easy to see. As we hinted earlier, the example given above actually shows that the implication $SUS'\implies SUS$ is strict, since $\beta\omega$ is the one-point compactification of $\beta\omega\backslash\{p\}$. It's not clear to me whether, as the side note proposes, we might have $k_2H\iff SUS'$. Of course all the other implications are known to be strict from the answers to this question.

There's one other class that might be worth mentioning in this picture:

  1. The class of pairs $(\kappa,\kappa+1)$ for an infinite cardinal $\kappa$, but now with $\kappa$ discrete and $\kappa \in \kappa+1$ having a base $\{(\lambda,\kappa]\mid \lambda<\kappa\}$, gives what we might call "$UR$", or "unique radial limits."

Note that this is stronger than $SUS$. Also, it is not even implied by $k_1H$, let alone any of the weaker properties, since $\omega_1+1$ (with the aforementioned topology) with a doubled endpoint is anticompact and $T_1$, hence $k_1H$, but not $UR$. Certainly $T_2\implies UR\implies SUS$, but otherwise I haven't thought about this in detail, and don't know what else is known.

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    $\begingroup$ The best part about this question is that SUS can remain the canonical term, since UR seems to be distinct (though I need to work through the details of your final paragraphs a bit more carefully). I greatly appreciate you finding the imposter in any case. :-) $\endgroup$ Dec 18, 2023 at 14:48
  • $\begingroup$ So, it's unclear to me if the definition of radial requires that the long sequence be a continuous mapping. $\endgroup$ Dec 18, 2023 at 16:50
  • $\begingroup$ There's definitely two distinct properties here; whichever matches with the radial property should be UR, and the other can be SUS $\endgroup$ Dec 18, 2023 at 16:53
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    $\begingroup$ @StevenClontz I was going by math.stackexchange.com/a/283331/1210477, which suggested that to be a radial limit a long sequence only need be continuous at the (added) endpoint, not at any other point, but I am new to all of this and don't know what is actually standard. $\endgroup$
    – M W
    Dec 18, 2023 at 19:34
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    $\begingroup$ Thanks again M W. I'm interested in writing this up. If you'd like to collaborate, I have some notes I'm working on at github.com/orgs/pi-base/discussions/450 $\endgroup$ Dec 22, 2023 at 2:43

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