So, 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.

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.

So, 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 c 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.

So, 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.