A basic question about ideals on natural numbers. Here is a basic question about the power set of the natural numbers. It is related to the goal of understanding convergent sequences in a general topological space.
Suppose $S$ is a collection of subsets of $\{1,2,3,...\}$ so that all of the following hold:

*

*If $A \in S$ and $B \subset A$, then $B \in S$ (i.e. $S$  is closed under taking subsets)


*If $A \in S$ and $B \in S$ then $A \cup B \in S$ (i.e. $S$ is closed under taking finite unions)


*If $F \subset \{1,2,3,...\}$ is finite, then $F \in S$ ( i.e. $S$ contains all finite sets of numbers).
The question: Suppose $A \subset \{1,2,3,...\}$ such that for each infinite subset $B \subset A$, there exists an infinite subset $C \subset B$ such that $C \in S$. Must $A \in S$?

Motivational details: The connection to topology arises from the fact that permuting the terms of a sequence in a space $X$ as no effect on whether the sequence converges or not to $x \in X$.
Suppose $X$ is a countably infinite set with $x \in X$, and $S$ is an arbitrary collection of functions from $\{1,2,3,...\}$ to $X$.
We next impose on $X$ the finest topology so that each $f \in S$ continuously extends to a continuous function defined over the one point compactification of $\{1,2,3,...\}$,with $f(\infty)=x$.
The above conditions 1,2,3 translate loosely to the facts that, subsequences of a convergent sequences converge, we can interleave two sequences converging to $x$, and that appending finitely many terms to a convergent sequence has no effect on its convergence.
The above question translates to the fact that, in a space $X$, if every subsequence of a sequence has a subsequence converging to $x$, then the sequence itself converges to $x$.
If we call the latter condition 4), then the sequential analogue of the original set theoretic question is whether the sequential counterparts to conditions 1) 2) and 3) are adequate to ensure condition 4).
 A: Looks like the summable ideal is a counterexample, yielding a no answer.
$S$ is the collection of subsets of natural numbers, so that the sum of the reciprocals converges (ignoring the reciprocal of zero).
The natural numbers has the property that every infinite subset contains an infinite set in $S$, but the natural numbers are not in $S$.
A: Here is a more sophisticated perspective on your question.  You are asking about ideals in $\mathcal{P}(\mathbb{N})$ which contain the ideal of finite sets, or equivalently ideals in the quotient algebra $\mathcal{P}(\mathbb{N})/\mathrm{fin}$.  An ideal in a Boolean algebra can be identified with an open set in its Stone space: each element of the Boolean algebra corresponds to a clopen set and then the ideal corresponds to the open set which is the union of all the clopen sets corresponding to its elements, and conversely an open set corresponds to the ideal of clopen sets that it contains.  Concretely, the Stone space of $\mathcal{P}(\mathbb{N})/\mathrm{fin}$ is the space $\mathbb{N}^*$ of nonprincipal ultrafilters on $\mathbb{N}$, with the topology generated by the sets $U_A=\{F\in\mathbb{N}^*:A\in F\}$ for each $A\subseteq\mathbb{N}$.  An ideal $I\subseteq \mathcal{P}(\mathbb{N})/\mathrm{fin}$ then corresponds to the open set $U=\bigcup_{A\in I}U_A$, i.e. the set of ultrafilters that contain some element of $I$.
Now what does it mean if every infinite subset $B\subseteq A$ has an infinite subset $C\subseteq B$ that is in $I$?  Well, that just means that every nonempty basic open subset of $U_A$ (i.e. set of the form $U_B$ for $B\subseteq A$) has nonempty intersection with $U$ (i.e., it contains one of the nonempty basic open sets $U_C$ that is contained in $U$).  In other words, it just means that $U$ is dense in $U_B$, or equivalently that $U_B$ is contained in the interior of the closure of $U$.  So to demand that such a $B$ is always actually in $I$ (i.e., that such a $U_B$ is always actually contained in $U$) just means you are demanding that $U$ is the interior of its closure, i.e. it is a regular open set.
So, any non-regular open subset of $\mathbb{N}^*$ is a counterexample.  For instance, the complement of a single point is non-regular since $\mathbb{N}^*$ has no isolated points.  In terms of ideals, this would be any nonprincipal maximal ideal on $\mathbb{N}$ (the maximal ideal that is dual to the ultrafilter that is the single point your open set omits).
