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I was going through the book Ultrafilters Throughout Mathematics and I came across the notion of ultralimits, defined below.

Ultralimit. Let $(X,\tau)$ be a topological space, $(x_i)_{i\in I}$ be a sequence from $X$ and $\mathcal{U}$ is an ultrafilter on $I$. Then, an $\mathcal{U}$-ultralimit of $(x_i)_{i\in I}$ is defined to be a point $x\in X$ such that for all $U\in \tau$, $x\in U$ implies that $\{i\in I: x_i\in U\}\in \mathcal{U}$. A sequence $(x_i)_{i\in I}$ from $X$ is said to have an ultralimit if there exists an $\mathcal{U}$-ultralimit for $(x_i)_{i\in I}$.

I was thinking of a stronger version of this notion, defined below.

Strong Ultralimit. Let $(X,\tau)$ be a topological space, $(x_i)_{i\in I}$ be a sequence from $X$ and $\mathcal{U}$ is an ultrafilter on $I$. Then, a strong $\mathcal{U}$-ultralimit of $(x_i)_{i\in I}$ is defined to be a point $x\in X$ such that for all $U\in \tau$, $x\in U$ iff $\{i\in I: x_i\in U\}\in \mathcal{U}$. A sequence $(x_i)_{i\in I}$ from $X$ is said to have an strong ultralimit if there exists an strong $\mathcal{U}$-ultralimit for $(x_i)_{i\in I}$.

One of the reasons I am interested in this question is the following criteria of compactness via ultralimits.

Theorem. A topological space $(X,\tau)$ is compact iff every sequence $(x_i)_{i\in I}$ from $X$ and has an $\mathcal{U}$-ultralimit for any ultrafilter $\mathcal{U}$ on $I$.

My questions are:

  1. Is it true that a sequence has ultralimits iff it has strong ultralimits? (I suspect a negative answer to this question, but can't find a counterexample.)
  2. Assuming the answer to the above question is negative, what kind of compactness arises iff every sequence $(x_i)_{i\in I}$ from $X$ and has a strong $\mathcal{U}$-ultralimit for any ultrafilter $\mathcal{U}$ on $I$? Is it something already known?

Any help is appreciated.

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    $\begingroup$ Since $U=X\setminus\{x\}$ is an open set, and since $x\notin U$, doesn't that mean that $\{i:x_i\in U\}\notin\mathcal U$? So $\{i:x_i\notin U\}\in\mathcal U$, i.e., $\{i:x_i=x\}\in\mathcal U$? $\endgroup$
    – bof
    Commented Nov 20 at 6:05
  • $\begingroup$ $x\in U\iff\{i:x_i\in U\}\in\mathcal U$, so $x\notin U\iff\{i:x_i\in U\}\notin\mathcal U\iff\{i:x_i\notin U\}\in\mathcal U$ $\endgroup$
    – bof
    Commented Nov 20 at 6:10
  • $\begingroup$ I deleted my first comment because for a moment there I thought I'd made a mistake. $\endgroup$
    – bof
    Commented Nov 20 at 6:13
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    $\begingroup$ I will use $a$ instead of $x$ for the limit - so that we do not have the same letter for the limit and the function $x\colon I\to X$. Just in case it helps to clarify things, this is more-or-less what bof said, but rephrased a bit: Let us assume $X$ to be $T_1$ and let $a\in X$. Then $U=X\setminus\{a\}$ is open. Since $\mathcal U$ is ultrafilter, either $x^{-1}[U]\in\mathcal U$ or $x^{-1}[\{a\}]\in\mathcal U$. If $x^{-1}[U]\in\mathcal U$, then $a$ is not a strong ultralimit. That means that for $a$ to be a strong ultralimit, we need $x^{-1}[\{a\}]\in\mathcal U$. $\endgroup$
    – Martin Sleziak
    Commented Nov 20 at 6:13
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    $\begingroup$ @bof; I see. So, virtually every $T_1$-space constitutes a counterexample. Not sure how I missed that. $\endgroup$
    – Ray
    Commented Nov 20 at 6:23

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The first thing to to with a definition like that is test it against some familiar examples. The sequence $\langle 2^{-n} : n\in\mathbb{N}\rangle$ converges to $0$ in $\mathbb{R}$ and it would seem desirable then to have $0$ be a strong ultralimit as well. That is not going to work: the open sets that contain $0$ give us the cofinite filter in the sequence, hence you do not get any strong ultralimits this way. But the $0$ is an ultralimit (for every free ultrafilter on $\mathbb{N}$) of this sequence. So the answer to the first question is no.

The second question has been answered in the comments: If $a$ is a strong $\mathcal{U}$-limit of $\langle x_i:i\in I\rangle$ in a $T_1$-space then the `sequence' must be constant on a member of $\mathcal{U}$. This implies that your kind of compactness is equivalent to finiteness in the class of $T_1$-spaces. That finite discrete spaces have the compactness property is clear; on the other hand the comments show that no injective sequence can have a strong ultralimit along a free ultrafilter.

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  • $\begingroup$ The definition does not work because of the "if and only if". If you weaken it to "The family of sets of the form $\{i\in I:x_i\in O\}$, where $O$ runs through the neighbourhoods of $x$, generates the ultrafilter $\mathcal{U}$" then the answer to question 1 remains the same but question 2 may get a more interesting answer. $\endgroup$
    – KP Hart
    Commented Nov 25 at 9:47

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