Redundant finite character family definition

Question

Definition: Let $F$ be a family of sets. $F$ is called of finite character if for each set $A$, we have that:

$A\in F\iff$ Each finite subset of $A$, is also in $F$.

I can't see the point of this definition, I find it redundant.

It seems trivial, if $A$ is in $F$ then every subset (finite or infinite) of $A$ will be always in $F$.

My thinking is that every family $F$ is of finite character, according to the definition.

Answer 1

If you are familiar with Zorn's lemma, try to think about the "typical use" of Zorn's lemma. Let's investigate it a little bit.

Zorn's Lemma. Suppose that $(P,\leq)$ is a partially ordered set in which every chain has an upper bound, then $P$ has a maximal element.

1. Every partial order has a maximal [anti-]chain. Fix $(A,\leq)$ to be some partially ordered set. Let $P$ be the set of [anti-]chains in $A$, and consider $(P,\subseteq)$ as our partial order. We claim it satisfies the conditions of Zorn's lemma:

   Let $D\subseteq P$ be a chain. Namely, every element of $D$ is a[n anti-]chain of $A$, and $\subseteq$ linearly orders $D$. We claim that $X=\bigcup D$ is an upper bound of $D$. Clearly it contains all the members of $D$ by the definition of a union; but we need to verify that $X\in P$. Namely, we need to check that $X$ is a[n anti-]chain.

   Suppose that $a,b\in X$. Then there are $X_a,X_b\in D$ such that $a\in X_a$ and $b\in X_b$. Since $D$ is linearly ordered by inclusion, either $X_a\subseteq X_b$ or vice versa. Therefore without loss of generality, $a,b\in X_b$. Since $X_b$ is a[n anti-]chain, it follows that $a$ and $b$ are [in]comparable. So $X$ is also a[n anti-]chain, as wanted.

   Therefore, by Zorn's lemma, $P$ has a maximal element which is a maximal [anti-]chain in $A$. $\square$

2. Every vector space has a basis. Let $V$ be a vector space over a fixed field, and let $P$ be the set of all linearly independent subsets of $P$. We claim that $(P,\subseteq)$ satisfies the conditions of Zorn's lemma, and must therefore have a maximal element which is of course a basis.

   Suppose that $D\subseteq P$ is a chain. We claim that $X=\bigcup D$ is an upper bound of $D$. Again, we only need to check that $X\in P$, i.e. that $X$ is a linearly independent set. And indeed, suppose that $v_1,\ldots,v_n\in X$, then there is some $X_1,\ldots,X_n\in D$ such that $v_i\in V_i$. But since $D$ is a chain, there is some $k\leq n$ such that $X_i\subseteq X_k$ for all $i\leq n$. Therefore $v_1,\ldots,v_n\in X_k$, and by the assumption, $X_k$ is linearly independent, so if $\sum_{i=1}^n\alpha_i v_i=0$, it has to be that $\alpha_i=0$ for all $i$. In other words, $X$ is also linearly independent. $\square$

3. Every two sets have an injection from one into the other. Suppose that $A$ and $B$ are two sets, and without loss of generality, there is no injection from $B$ into $A$. Let us show that there is an injection from $A$ into $B$. Define $P$ to be the set of all partial injections from $A$ into $B$. We claim that $(P,\subseteq)$ satisfies Zorn's lemma conditions and therefore has a maximal element, which corresponds to an injection from $A$ into $B$. $\DeclareMathOperator{dom}{dom}\DeclareMathOperator{rng}{rng}$

   Suppose that $D\subseteq P$ is a chain, then $f=\bigcup D$ is an injective function, as a the union of a chain of function: if $\langle a,x\rangle,\langle a,y\rangle\in f$, then there are $f_x,f_y\in D$ such that $f_x(a)=x$ and $f_y(a)=y$. But since $D$ is a chain, either $f_x\subseteq f_y$ or $f_y\subseteq f_x$. By the assumption that both $f_x$ and $f_y$ are functions, we immediately obtain that $x=y$. So $f$ is indeed a function.

   To see that $f$ is injective, suppose that $a,b\in A$ are in the domain of $f$, then there are $f_a,f_b\in D$ such that $a\in\dom f_a$ and $b\in\dom f_b$. By the fact $D$ is a chain, we can assume that $f_a\subseteq f_b$, so $a,b\in\dom f_b$, without loss of generality $f_b\subseteq f_a$. But since $f_b$ is an injective function, $f(a)=f_a(a)=f_a(b)=f_a(b)$ if and only if $a=b$.

   So by Zorn's lemma, there is a maximal element $f$. If $\dom f\neq A$, then there is some $a\in A\setminus\dom f$. By maximality, it means that $\rng f=B$, but then $f^{-1}$ is an injection from $B$ into $A$, which we assume does not exist, so there is some $b\in B\setminus\rng f$. Therefore $f\cup\{\langle a,b\rangle\}$ is also an injective function from a subset of $A$ into $B$, strictly extending $f$ which is a contradiction to maximality. Therefore $\dom f=A$, and $f\colon A\to B$ is injective as wanted. $\square$

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What do these things have in common?

Well, in order to check that the upper bound of the chain was an upper bound we had to check some property holds for the union, and we used the fact that the property is given by finite subsets (even just sets of size $2$). Being an injective function, or just being a function; being linearly independent; being a chain or an antichain. These are properties which have certain compactness to them: you're a chain if and only if your finite subsets of chains. Or rather, there is a counterexample if and only if there is a finite counterexample.

So this leads us to an understanding of the typical use of Zorn's lemma. We want to apply it to properties which have this type of finitary characteristics to them, because that way when we look at the union of a chain, any finitely many elements had to come from one of the members of the chain (this need not be true for infinitely many elements, without additional restrictions anyway).

And finite character comes to characterize exactly this idea; and the Teichmüller–Tukey lemma comes to give us the result of Zorn's lemma directly, without passing through Zorn's lemma each time.

Answer 2

As Somos pointed out in the comments, this is not "redundant". Let's consider the vector space $\mathbb{R}^{\alpha}$ for some dimension (possible infinite) $\alpha \ge 2$ and let $F_1$ be the collection of all linearly independent subsets $A \subseteq \mathbb R^{\alpha}$. $F_1$ is of finite character since a subset $A \subseteq \mathbb R^{\alpha}$ is linearly independent if and only if every finite subset $A^* \subseteq A$ is linearly independent over $\mathbb R^{\alpha}$.

Now consider $F_2$ - the collection of all maximal linearly independent subsets $A \subseteq \mathbb R^{\alpha}$. $F_2$ is not of finite character since for any maximal linearly independent $A \in F_2$ and any $a \in A$ we may consider the finite subset $\{a\} \subseteq A$. Since $\alpha \ge 2$ the set $\{a \}$ is no maximal linearly independent over $\mathbb R^{\alpha}$ and hence $\{a\} \not \in F_2$.

Answer 3

A family of sets $\mathcal{A}$ need not be closed under (finite) subsets.

Sometimes they are, like the family of bounded subsets of a metric space, e.g. So in that case we can conclude from $A \in \mathcal{A}$ and $B \subseteq A$ that $B \in \mathcal{A}$. This also holds for all measure $0$ subsets of a measure space or all countable subsets of a set.

Being of a finite character is a way to go back: if you have a set $B \subseteq X$ and we only know that all finite subsets if $B$ are in $\mathcal{A}$, then we know $ B$ is too.

Note that all of the above examples show that finite character is special: all finite subsets of a metric space are bounded, but this doesn’t mean the whole space is bounded; ditto for measure $0$ and countable sets.

Typical examples come from algebra: a subset is linearly independent in a vector space iff all finite subsets are (as we only look at finite linear combinations).

Note that if a family is of finite character, it will be closed under all subsets of members (simply because a finite subset of the subset is also one of the superset).