# Weaker notion of first-countability for filters?

While trying to understand the motivation behind the definition of a filter I've stumbled upon the following notion, let's call it "almost first-countability":

Let $x$ be a point of a set $X$ and $\mathcal{F}$ be a filter on $X$ such that each $F\in\mathcal{F}$ contains $x$. We say $\mathcal{F}$ is almost first-countable if there is a countable collection $\mathcal{C}=\{C_1,C_2,\dots\}$ of nested subsets of $X$ (so $C_1\supset C_2\supset\cdots$) such that $\Sigma(\mathcal{C})=\Sigma(\mathcal{F})$, where for any $\mathcal{X}\in P(P(X))$ the symbol $\Sigma(\mathcal{X})$ denotes the set of sequences $\phi\colon\mathbb{N}\to X$ converging with respect to the collection $\mathcal{X}$, i.e. each element of $\mathcal{X}$ contains a tail of $\phi$.

Clearly if $\mathcal{F}$ is first-countable, i.e. it can be generated by a countable filter base, then it is almost first-countable (just take $\mathcal{C}$ to be $\{C_1,C_1\cap C_2,C_1\cap C_2\cap C_3,\dots\}$ where $\{C_1,C_2,C_3,\dots\}$ is a countable filter base generating $\mathcal{F}$).

Question: Is this notion strictly weaker than first-countability? In other words, are there a set $X$, a non-first-countable filter $\mathcal{F}$ at $x\in X$ and a countable collection $\mathcal{C}=\{C_1,C_2,\dots\}$ of nested subsets of $X$ such that $\Sigma(\mathcal{C})=\Sigma(\mathcal{F})$?

Observation: Call $\mathcal{C}'$ the filter generated by the base $\mathcal{C}$ by adding all supersets of elements of $\mathcal{C}$. Then $\Sigma(\mathcal{C}')=\Sigma(\mathcal{C})$. It follows that the above question boils down to finding two filters at $x$ sharing the same set of convergent sequences and such that one is first-countable while the other is not (or proving they do not exist, of course).

As an exercise for myself let me rewrite user126154's brilliant argument in a more detailed fashion. In order to fix the ideas, instead of letting $\alpha$ be any non-$\omega$-cofinal limit ordinal I decided to choose a concrete example by picking up the smallest limit ordinal which is not $\omega$-cofinal. If I'm not mistaken this should be the first uncountable ordinal $\omega_1$.

Consider the first uncountable ordinal $\omega_1$ and the filter $\mathcal{F}_\infty$ on $\omega_1$ generated by the filter base $\mathcal{B}_\infty:=\{(a,\infty)\mid a\in \omega_1\}$ where $(a,\infty):=\{x\in \omega_1\mid x>a\}$. Observe that $\Sigma(\mathcal{F}_\infty)$ is empty: indeed, assume there is a sequence $\phi\colon\mathbb{N}\to \omega_1$ converging with respect to $\mathcal{F}_\infty$; by induction $\phi$ has a strictly increasing subsequence $\phi'$ also converging w.r.t. $\mathcal{F}_\infty$; the limit of this sequence is just $s:=\sup_{n\in\mathbb{N}}\{\phi'(n)\mid n\in\mathbb{N}\}$, i.e. a countable union of countable ordinals which is still countable, hence $s\in \omega_1$ and the interval $(s+1,\infty)$ does not contain any tail of $\phi'$, which is absurd. Further observe that $\mathcal{F}_\infty$ cannot be generated by a countable filter base, otherwise we could construct a sequence converging w.r.t. the base and hence also w.r.t. $\mathcal{F}$.

Now consider the set $X:=(\sqcup_{n\in\mathbb{N}} (\omega_1)_n)\cup\{\infty\}$ consisting of the disjoint union of countably many copies of $\omega_1$ together with a new point $\infty$, and construct the filter $\mathcal{F}$ on $X$ consisting of all subsets $F\subset X$ satisfying all the three conditions below:

1. $F\cap(\omega_1)_n\in\mathcal{F}_\infty$ for each $n\in\mathbb{N}$,
2. there is $j\in\mathbb{N}$ such that $F\cap(\omega_1)_k=\omega_1$ for each $k\geq j$,
3. $\infty\in F$.

Notice that $\mathcal{F}$ cannot be generated by a countable filter base, otherwise by restricting each element of the base to one copy of $\omega_1$ we would obtain a countable filter base for $\mathcal{F}_\infty$. Now consider the collection $\mathcal{C}=\{C_n\mid n\in\mathbb{N}\}$ of subsets of $X$ where $C_n$ is defined by the next three conditions:

1. $C_n\cap(\omega_1)_j=\emptyset$ for $j<n$,
2. $C_n\cap(\omega_1)_j=\omega_1$ for each $j\geq n$,
3. $\infty\in C_n$ for each $n\in\mathbb{N}$.

Clearly $\mathcal{C}$ satisfies the axioms of a filter base, and the filter that it generates is obviously first-countable.

Now we show that $\Sigma(\mathcal{F})=\Sigma(\mathcal{C})$. Let $x_n$ be a sequence in $X$ converging w.r.t. $\mathcal{C}$. Each $F\in\mathcal{F}$ contains some $C_n$ so $x_n$ converges w.r.t. $\mathcal{F}$ as well. Conversely, assume $x_n$ converges w.r.t. $\mathcal{F}$ and define $\sigma(n)$ to be the index of the copy of $\omega_1$ where $x_n$ lies (put $\sigma(n)=\star$ if $x_n=\infty$). If for each $n$ there is $j_n$ such that for each $k\geq j_n$ $\sigma(k)$ is either greater than $n$ or equal to $\star$ then $x_n$ obviously converges w.r.t. $\mathcal{C}$ and we are done. Otherwise $\sigma$ has a constant subsequence at some natural number, that is there is a subsequence $x_{n_k}$ staying in a single copy of $\omega_1$. Since $x_n$ converges w.r.t. $\mathcal{F}$ by assumption, then $x_{n_k}$ must converge w.r.t. $\mathcal{F}_\infty$ and this is impossible since $\Sigma(\mathcal{F}_\infty)=\emptyset$.

Conclusion: we have found a set $X$, two filters at some point $x\in X$ one of which is first-countable while the other is not (the former being the filter generated by $\mathcal{C}$, the latter being $\mathcal{F}$) sharing the same set of converging sequences in $X$. In other words, $\mathcal{F}$ is "almost first-countable" but not first-countable, therefore the notion spelled out above of "almost first-countability" is really different than first-countability (in particular, it is strictly weaker).

In order to avoid ambiguities I add the definitions used. A filter on $X$ is a nonvoid collection of subsets of $X$ such that:

1. the empty set does not belong to the collection,
2. any finite intersection of its elements is a superset of some element of the collection,
3. any superset of any of its elements belongs to the collection.

A filter base is like a filter but we drop the third assumption. By adding all supersets of all elements of a filter base $\mathcal{F}$ we obtain the filter $\mathcal{F}'$ generated by the base $\mathcal{F}$. It follows that a nonvoid collection $\mathcal{B}\in P(P(X))$ is a base of a given filter $\mathcal{F}$ if and only if $\mathcal{B}\subset\mathcal{F}$ and each $F\in\mathcal{F}$ contains some $B\in\mathcal{B}$.

• what is your definition of first-countable? are we talking about open sets? if so, the $C_i$ are open? – user126154 Mar 10 '14 at 16:28
• @user126154 in this context I say $\mathcal{F}$ is first-countable if it admits a countable filter base; no mention of openness, there is no topology on $X$ yet, just a filter at some point $x\in X$ – johndoe Mar 10 '14 at 17:17
• in this case I think my answer applies, with the change that $\mathcal C$ is claimed to be a filter base at $x$. – user126154 Mar 10 '14 at 17:21
• @user126154 could you please amend your proof accordingly? I can't follow your argument. – johndoe Mar 10 '14 at 17:24
• could you give the precise definition that you are using? I'm a bit confused. – user126154 Mar 10 '14 at 17:41

Let $X$ be almost-first-countable. Let $\mathcal C=\{C_i\}$ be a nested countable sequence of sets containing $x$ so that $\Sigma(\mathcal C)=\Sigma(\mathcal F)$.

Suppose there is $F\in\mathcal F$ so that $x\in F$ and no $C_i$ is contained in $F$.

In particular there is $x_i\in C_i$ such that $x_i\notin F$. Therefore the sequence $\{x_i\}$ converges to $x$ for $\mathcal C$ but does not converge to $x$ for $\mathcal F$. This contradicts the hypothesis of almost first countability.

We have so proved that for any $F\in\mathcal F$ there is $C_i\in \mathcal C$ such that $C_i\subset F$.

If our filter comes from a topology and the $C_i$'s are open sets, then the converse is true.

Let's now see an example to show that the converse is not necessarily true in general.

Let $\omega$ be the typer of order of $\mathbb N$.

For any ordinal $\alpha$ let $\mathcal F_\alpha$ be the filter given by the order relation. (i.e. the filter of neighborhoods of $\infty$.)

Choose a limit ordinal $\alpha$ which is not $\omega$-cofinal. That is to say, there is no sequence $x_i\stackrel{\mathcal F_\alpha}{\to}\infty$ (i.e. $\mathcal F_\alpha$ is not first-countable.)

Let $X$ be the set obtained by countable disjoint union of copies of $\alpha$, plus the point $\infty$.

$$X=(\bigsqcup_{i\in\mathbb N}\alpha_i)\cup\{\infty\}$$

where $\alpha_i$ is a copy of $\alpha$.

Therefore, any subset $A\subset X$ is of the form $\bigsqcup_{i=0}^\infty A_i$ with $A_i\subset \alpha_i$ for $i\in\mathbb N$ and $A_\infty\subset\{\infty\}$.

Let $\mathcal F$ be the filter defined as follows: A set $F\subset X$ is in $\mathcal F$ if and only if

$\exists n_0=n_0(F)$ s.t.

1) $F_i=\alpha_i$ for any $n>n_0$

2) $\forall n\leq n_0$ we have $\emptyset\neq F_i\in\mathcal F_{\alpha_i}$

moreover,

3) $\infty\in F$

CLAIM: $\mathcal F$ is not first-countable.

PROOF: by conditions 1) and 2), any countable filter base for $F$ would give, by restriction to $\alpha_i$, a countable filter base on $\alpha$, which is assumed not $\omega$-cofinal. $\square$

Now, let's define $\mathcal C$. This is easy, let $$C_i=(\bigsqcup_{n\geq i} \alpha_i)\cup\{\infty\}$$

Clearly $\mathcal C$ is a countable filter.

CLAIM: for any sequence $(x_i)_{i\in\mathbb N}$ in $X$ we have $$x_i\stackrel{\mathcal C}{\to}\infty \Leftrightarrow x_i\stackrel{\mathcal F}{\to}\infty$$

PROOF. Let $x_i\stackrel{\mathcal C}{\to}\infty$. By condition $1)$ for all $F\in\mathcal F$ we have $x_i\in F$ for $i>n_0(F)$. Thus $x_i\stackrel{\mathcal F}{\to}\infty$.

On the other hand, suppose $x_i\stackrel{\mathcal F}{\to}\infty$

The point $x_i$ belongs to a set $\alpha_k$. Define $\sigma(i)$ so that $$x_i\in\alpha_{\sigma(i)}$$

If $\sigma(i)$ does not converge to infinity, then we can find a subsequence $i_k$ so that $\sigma(i_k)$ is a constant natural number, say $n$.

Therefore the sequence $x_{i_k}$ is a sequence in $\alpha_n$ which converges to infinity, which is impossible because $\alpha$ is not $\omega$-cofinal.

Therefore $\sigma(i)\to \infty$ and thus $x_i\stackrel{\mathcal C}{\to}\infty$. $\square$

• Your argument shows that $\mathcal{C}$ is finer than $\mathcal{F}$; it still does not prove that $\mathcal{F}$ is generated by a countable filter base (clearly the condition $C_i\in\mathcal{F}$ would suffice but it is not true in general). Thanks for your input by the way, I appreciate it. – johndoe Mar 11 '14 at 1:18
• ok, I'll try to complete with an example. – user126154 Mar 11 '14 at 11:35
• There are several things I fail to understand, please bear with me. First, I don't understand the condition of non-$\omega$-cofinality on $\alpha$; e.g., does $\alpha=\omega+1$ satisfy it? Second, what is $\mathcal{F}_\alpha$? – johndoe Mar 11 '14 at 20:31
• $\mathcal F_\alpha$ is the filter of sets of the type $\{x>a\}$ with $a\in\alpha$, is the filter fo "neigborhood of $\infty$" – user126154 Mar 11 '14 at 20:44
• Given two ordinals $\omega$ and $\alpha$ we say that $\alpha$ is $\omega$-cofinal if there is an order preserving immersion of $\omega$ in $\alpha$ such that for all $a\in\alpha$ there is $b$ in the image of $\omega$ with $a\leq b$. So, you are right, $\omega+1$ is not $\omega$-cofinal. But it suffices to chose $\alpha$ a limit ordinal which is not cofinal in $\omega$. I edit the answer accordingly. – user126154 Mar 11 '14 at 20:50