Specific question about this set theory proof

Factual Questions
1

I actually follow this proof, but have a question about the endgame.

Why can’t mission accomplished be declared at this line:

x ∈ A ∩ (B ∩ C) (third line above equation 1)

At that point, x has been shown to be in both (A ∩ B) ∩ C and A ∩ (B ∩ C), so I would have thought one could write therefore (A ∩ B) ∩ C = A ∩ (B ∩ C). QED.

Instead, the proof goes on to show that (A ∩ B) ∩ C and A ∩ (B ∩ C) are subsets of each other, and therefore are equal. (Yes, I understand this, so no need to explain why this is so.) I just don’t see why it is necessary.

2

Not quite. x has shown to be in (A ∩ B) ∩ C and A ∩ (B ∩ C) on the assumption that x is in (A ∩ B) ∩ C. (Thus the next step is saying that if x is in (A ∩ B) ∩ C, then it is in A ∩ (B ∩ C).)

3

Remember equality is read left to right and right to left. The first part shows that if x starts out in (A ∩ B) ∩ C then it is also in A ∩ (B ∩ C) but that does nothing to show us that if it is in A ∩ (B ∩ C) that it is also in (A ∩ B) ∩ C.

In other words, if (A ∩ B) ∩ C is a proper subset of A ∩ (B ∩ C), the first part of the proof works but the two sets are not equal.

4

Gotcha ya. Thank you. When I did my relearning of high-school math, in the simple proofs that were covered, it was enough simply to get some manifestation of the LHS to equal some manifestation of the RHS without working backwards.

5

But step 1 did not show they were equal; it showed only one inclusion. It would be like showing that x <= y. You would then show that y <= x and conclude x = y.

6

Why don’t they teach set theory like algebra, starting with some basic axioms and working up from there? Set theory has almost the same basic axioms as conventional algebra:

  1. Intersection is commutative: A ∩ B = B ∩ A

  2. Intersection is associative: (A ∩ B) ∩ C = A ∩ (B ∩ C)

  3. Union is commutative: A ∪ B = B ∪ A

  4. Union is associative: (A ∪ B) ∪ C = A ∪ (B ∪ C)

Ah, but which is distributive over the other? This gets interesting. They are BOTH distributive over the other!

  1. Intersection is distributive over union: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
  2. Union is distributive over intersection: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

These rules exactly parallel the logical operations of conjunction and disjunction (I can never remember which is which), if you just replace ∩ with ∧ and ∪ with ∨.