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The definition of an intersection of family($F$) is:

$$\cap F=\bigl\{x\mid\forall A: (A\in F \implies x\in A)\bigr\} $$

If my understanding serves me correctly this notation means that all the $x$ that pass the elementhood test are in the set. So $\forall A: (A\in F \implies x\in A)$ must be either true or false for each different $x$. But in this case isn't $F$ also a free variable, so we can't assign a truth value to $\forall A:(A\in F \implies x\in A)$ as it contains a free variable $F$ ($x$ and $A$ are bounded). Is there some implicit assumption about $F$ ? Also in case of an indexed family $F=\{A_i\mid i\in I\}= \bigl\{x\mid\exists i\in I: x=A_i\bigr\}$, is $A_i$ (which is a function of $i$) bound or free (if free isn't it the same problem as the first question) ?

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To define $\cap F$, $F$ must be given. Once it is given, it acts as a constant and not a variable. There is an implicit assumption about $F$, namely that $F\neq \varnothing$. In $\{x|\exists i(i\in I \land x=A_i)\}$ the variable $i$ is bounded. – Git Gud May 20 '14 at 12:29
@GitGud Is $A_i$ also given ? – Nameless May 20 '14 at 12:33
A function $\colon I\to \{A_i\mid i\in I\}$ is given. Any $A_i$ is given, if and only, an $i\in I$ is given. Does this help? – Git Gud May 20 '14 at 12:35
@JPi Better than \bigl \big ris \left \right, this will adjust the size of the delimiters to whatever is in between them. – Git Gud May 20 '14 at 12:37
@Nameless Yes, because a function from $I$ to $\{A_i\mid i\in I\}$ is given. Without such a function, you can't even make sense of $A_i$ or $\{A_i\mid i\in I\}$. Edit: Perhaps I can improve what I just said. A bijective function $f\colon I\to F$ is given. Then one denotes each $f(i)$ by $A_i$. With this notation it follows that $F=\{A_i\mid i\in I\}$. – Git Gud May 20 '14 at 12:42

1 Answer 1

$F$ is not a free variable. It denotes a particular family of sets which you're taking the intersection of. For instance, if $F = \{\{1, 2\}, \{1, 3\}\}$ then your definition says that $\cap F$ consists of those $x$ which are in every set which is in $F$. In my example, the only such $x$ is $1$. Thus the statement $\forall A(A \in F \implies x \in A)$ is true for $1$, but false for $2$ and $3$.

The indexed family just gives a way of listing an infinite set of sets. Namely, it says that for each element $i$ of the set $I$, we have a set $A_i$ which is in $F$ (as an element, not a subset). If one assumes that $A_i \neq A_j$ whenever $i \neq j$, then this implies that the number of sets in $F$ has the same cardinality as $I$.

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