Tao Analysis I - 3.4.9

Exercise 3.4.9

Show that if $\beta$ and $\beta'$ are two elements of a set $I$, and to each $\alpha \in I$ we assign a set $A_{\alpha}$, then

$$ \{ x \in A_{\beta} : x \in A_{\alpha} \text{ for all } \alpha \in I \} = \{ x \in A_{\beta'} : x \in A_{\alpha} \text{ for all } \alpha \in I \} $$

and so the definition of $\bigcap_{\alpha \in I} A_{\alpha}$ defined in (3.3) does not depend on $\beta$.

Also explain why (3.4) is true.

Let's remind ourselves what the more general definition of intersection (3.3) says:

Given any non-empty set $I$ , and given an assignment of a set $A_{\alpha}$ to each $\alpha \in I$, we can define the intersection $\bigcap_{\alpha \in I}A_{\alpha}$ by first choosing some element $\beta$ of $I$ (which we can do since $I$ is non-empty), and setting

$$ \bigcap_{\alpha \in I} A_{\alpha} := \{ x \in A_{\beta} : x \in A_{\alpha} \text{ for all } \alpha \in I  \} $$

which is a set by the axiom of specification.

Thoughts

Since the intersection contains only elements which are members of all $A_\alpha$, it doesn't matter which $\beta$ is chosen to select one $A_{\alpha=\beta}$. 

Another way to think about this is that, it is impossible to pick a bad $A_\beta$ because if an element is not in $A_\beta$ then it can't be in the intersection.

Another point worth making is that the definition looks over-complicated. Why can't be as simple as the following?

$$ \bigcap_{\alpha \in I} A_{\alpha} := \{ x : x \in A_{\alpha} \text{ for all } \alpha \in I  \} $$

The reason is that defining sets using logical statements can lead to paradoxes, as Tao introduced earlier. However, defining sets as subsets of known sets guarantees they are sets.

Solution

To show the two sets are equivalent, we need to show an element of one is also an element of the other.

If $x \in \{ x \in A_{\beta} : x \in A_{\alpha} \text{ for all } \alpha \in I  \}$ then it must conform to the membership condition $x \in A_{\alpha} \text{ for all } \alpha \in I$.

Since $A_{\beta'}$ is one of the $A_\alpha$, then $x \in A_{\beta'}$. 

So we have $x \in A_{\beta'}$, and also $x \in A_{\alpha} \text{ for all } \alpha \in I$. This is equivalent to  $x \in \{ x \in A_{\beta'} : x \in A_{\alpha} \text{ for all } \alpha \in I  \}$. This is the definition of the second set.

By a symmetric argument, if $x$ is a member of the second set, it is also a member of the first.

Thus we have shown the two sets are equivalent. $\square$.

Now let's consider why (3.4) is true. Let's restate (3.4):

$$ y \in \bigcap_{\alpha \in I} A_\alpha \iff ( y \in A_\alpha \text{ for all } \alpha \in I ) $$

Let's show this in the usual manner, that each statement implies the other.

If  $y \in \bigcap_{\alpha \in I} A_\alpha$, then by definition (3.3) of the intersection of a family of sets, we have $y \in  \{ x \in A_{\beta} : x \in A_{\alpha} \text{ for all } \alpha \in I  \}$. This means $y$ must conform to the membership condition $y \in A_{\alpha} \text{ for all } \alpha \in I $. We have shown

$$y \in \bigcap_{\alpha \in I} A_\alpha \implies y \in A_{\alpha} \text{ for all } \alpha \in I $$

Now the converse, if $y \in A_{\alpha} \text{ for all } \alpha \in I$ is true then the membership condition $\{ y \in A_{\beta} : y \in A_{\alpha} \text{ for all } \alpha \in I  \}$ for the intersection is met, noting that $A_\beta$ is one of the $A_\alpha$. Thus we have shown

$$ y \in A_{\alpha} \text{ for all } \alpha \in I \implies y \in \bigcap_{\alpha \in I} A_\alpha $$

By showing both implications, we have shown that (3.4) is true. $\square$