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Terence Tao's Analysis I 4th edition defines set intersection as follows:

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.

Question

My question is - why does he define it in this complicated way?

A simpler definition might be:

The intersection of a family of sets $A_\alpha$ is the set whose elements exist in every set $A_\alpha$. That is,

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

Clearly I am missing something important.

Thoughts

I recall Tao discussing earlier in the book how subsets of sets are sets, but sets defined by logical statements can cause paradoxes. Is his definition an attempt to start with a set and then use axioms he introduced to guarantee the intersection is also a set?

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    $\begingroup$ It's not that complicated if you've read the specification axiom. That's a nice definition of $\bigcap_{\alpha \in I} A_{\alpha}$ $\endgroup$ Mar 16 at 19:51
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    $\begingroup$ The problem with your proposed definition is formally that you don't use the set builder notation correctly (you can only use it in this way, with a condition on the right, to form a subset of a given set); Tao does this properly. Your problem comes to light if you try to apply it to an empty family of sets: the condition becomes vacuous, and your notation wants to build the universe as a set, which is wrong. Just saying you forbid the empty family does not make your notation correct; Tao uses that assumption to make the notation correct. But he should prove independence of the choice. $\endgroup$ Mar 17 at 5:23
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    $\begingroup$ One way that can be defined, and this is how I usually teach this, is that $\bigcap_{i\in I}A_i=\{x\mid\forall i\in I: x\in A_i\}$ is the intersection class, and then a good exercise is to show that if $I$ is not empty, then this is a set by applying Separation to any of the $A_i$. But this still requires proof. $\endgroup$
    – Asaf Karagila
    Mar 17 at 18:59
  • $\begingroup$ @Asaf Karagila. This approach seems the most natural. Is it possible to read it somewhere in written form? $\endgroup$
    – zkutch
    Mar 17 at 23:06
  • $\begingroup$ @zkutch: My set theory lecture notes on my website. $\endgroup$
    – Asaf Karagila
    Mar 18 at 7:16

3 Answers 3

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The thought in your last paragraph is the answer. To avoid the set theoretic paradoxes Tao does not allow you to create sets with an expression like $$ A = \{ x \mid \text{some property of } x \}. $$ You can use it only to find a subset of a known set $$ A = \{ x \in \text{some set } \mid \text{some property of } x \}. $$

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    $\begingroup$ "A set made of elements of known sets" is not an allowable construction in general. It is OK if you happen to know that those known sets are all subsets of some set $U$. So your definition of intersection would work if your $A_a$ were all subsets of some set $U$, say, the set of real numbers. Since Tao does not assume the existence of such a $U$ he needs the clever workaround that you ask about. $\endgroup$ Mar 16 at 20:57
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    $\begingroup$ Ethan thank you - I have learned something today. $\endgroup$
    – Penelope
    Mar 16 at 21:15
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    $\begingroup$ @Penelope You're welcome. I learned Tao's definition from you today. That makes it a good day for me too. $\endgroup$ Mar 16 at 21:23
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    $\begingroup$ @Penelope Note that your definition would allow $\bigcap_{i \in \emptyset} A_i$ which is the proper class (not a set) $\{x\ :\ {\sf true}\}$, i.e. the class of all sets. Allowing this as a set is problematic in many set theories because it triggers Russell's paradox. By contrast, $\bigcup_{i \in \emptyset} A_i$ is $\{x\ :\ {\sf false}\}$ which is $\emptyset$, so it is not harmful. Tao avoids this issue by sticking to the set builder notation $\{x\in A \ :\ \ldots\}$ where $A$ is a known set. $\endgroup$
    – chi
    Mar 17 at 15:18
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    $\begingroup$ @Penelope Tao is rightfully careful to get this right from the start, but later I suspect you will be dealing with sets of subsets of the real numbers or sets of sets of functions and the important issue you raise in this question will be moot. In those contexts mathematicians regularly use the unrestricted set builder notation and don't encounter the paradoxes it allows. $\endgroup$ Mar 17 at 20:48
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In axiomatic set theory, it is not true that everything we can write down using the symbols { and } and the language of mathematical logic is a set. You have to prove something is a set by using the axioms. The most common way to do this is to prove it's a subset of some set you already know about, using Specification. That's what Tao is doing. Note that your definition isn't obviously a subset of anything.

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I must admit that I might take a slightly different pedagogical approach than Tao. I would define a set according to your proposed definition, and then prove that it exists. But both approaches have their complications.

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