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I corrected the sketch of the construction of a non-p-point. We do not want our ultrafilter to concentrate on on any $A_k$.
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Goldstern
Goldstern
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No, only if $\omega$ is a p-point.

If $(A_k:k\in \mathbb N)$ is a partition of the natural numbers into $\omega$-small sets such that there is no $\omega$-large set meeting each $A_k$ in a finite set, then we can choose a sequence $x_n$ by declaring $x_n:= 1/k$ whenever $n\in A_k$. On each set $J\in \omega$ the sequence $x_n$ does not converge to 0.

To find a non-p-point, partition $\mathbb N$ into countably many infinite sets $A_k$, let $I$ be the ideal of sets meeting each $A_k$ only finitely often (except for finitely many $k$), and let $\omega$ be dual to any maximal ideal extending $I$. Then each $A_k$ is in $I$, hence null modulo $\omega$. The partition $(A_k)$ witnesses that $\omega$ is not a p-point.

Note that ZFC does not prove that there are p-points, but existence of p-points follows from CH (or weaker assumptions).

No, only if $\omega$ is a p-point.

If $(A_k:k\in \mathbb N)$ is a partition of the natural numbers into $\omega$-small sets such that there is no $\omega$-large set meeting each $A_k$ in a finite set, then we can choose a sequence $x_n$ by declaring $x_n:= 1/k$ whenever $n\in A_k$. On each set $J\in \omega$ the sequence $x_n$ does not converge to 0.

To find a non-p-point, partition $\mathbb N$ into countably many infinite sets $A_k$, let $I$ be the ideal of sets meeting each $A_k$ only finitely often, and let $\omega$ be dual to any maximal ideal extending $I$.

Note that ZFC does not prove that there are p-points, but existence of p-points follows from CH (or weaker assumptions).

No, only if $\omega$ is a p-point.

If $(A_k:k\in \mathbb N)$ is a partition of the natural numbers into $\omega$-small sets such that there is no $\omega$-large set meeting each $A_k$ in a finite set, then we can choose a sequence $x_n$ by declaring $x_n:= 1/k$ whenever $n\in A_k$. On each set $J\in \omega$ the sequence $x_n$ does not converge to 0.

To find a non-p-point, partition $\mathbb N$ into countably many infinite sets $A_k$, let $I$ be the ideal of sets meeting each $A_k$ only finitely often (except for finitely many $k$), and let $\omega$ be dual to any maximal ideal extending $I$. Then each $A_k$ is in $I$, hence null modulo $\omega$. The partition $(A_k)$ witnesses that $\omega$ is not a p-point.

Note that ZFC does not prove that there are p-points, but existence of p-points follows from CH (or weaker assumptions).

mathbb
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Goldstern
Goldstern
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No, only if $\omega$ is a p-point.

If $(A_k:k\in N)$$(A_k:k\in \mathbb N)$ is a partition of the natural numbers into $\omega$-small sets such that there is no $\omega$-large set meeting each $A_n$$A_k$ in a finite set, then we can choose a sequence $x_n$ by declaring $x_n:= 1/k$ whenever $n\in A_k$. On each set $J\in \omega$ the sequence $x_n$ does not converge to 0.

To find a non-p-point, partition $\mathbb N$ into countably many infinite sets $A_k$, let $I$ be the ideal of sets meeting each $A_k$ only finitely often, and let $\omega$ be dual to any maximal ideal extending $I$.

Note that ZFC does not prove that there are p-points, but existence of p-points follows from CH (or weaker assumptions).

No, only if $\omega$ is a p-point.

If $(A_k:k\in N)$ is a partition of the natural numbers into $\omega$-small sets such that there is no $\omega$-large set meeting each $A_n$ in a finite set, then we can choose a sequence $x_n$ by declaring $x_n:= 1/k$ whenever $n\in A_k$. On each set $J\in \omega$ the sequence $x_n$ does not converge to 0.

No, only if $\omega$ is a p-point.

If $(A_k:k\in \mathbb N)$ is a partition of the natural numbers into $\omega$-small sets such that there is no $\omega$-large set meeting each $A_k$ in a finite set, then we can choose a sequence $x_n$ by declaring $x_n:= 1/k$ whenever $n\in A_k$. On each set $J\in \omega$ the sequence $x_n$ does not converge to 0.

To find a non-p-point, partition $\mathbb N$ into countably many infinite sets $A_k$, let $I$ be the ideal of sets meeting each $A_k$ only finitely often, and let $\omega$ be dual to any maximal ideal extending $I$.

Note that ZFC does not prove that there are p-points, but existence of p-points follows from CH (or weaker assumptions).

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Goldstern
Goldstern
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No, only if $\omega$ is a p-point.

If $(A_k:k\in N)$ is a partition of the natural numbers into $\omega$-small sets such that there is no $\omega$-large set meeting each $A_n$ in a finite set, then we can choose a sequence $x_n$ by declaring $x_n:= 1/k$ whenever $n\in A_k$. On each set $J\in \omega$ the sequence $x_n$ does not converge to 0.