Is there anyone among us who can identify a certain SUS space

compactnessgeneral-topologyseparation-axioms

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.

Best Answer

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!

Remarks.

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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