An exercise from Halmos’s set theory book

I found the following problem in Paul Halmos’s Naive Set Theory. It seemed interesting so I thought I would post it here.

Exercise. A necessary and sufficient condition that is that . Observe that the condition has nothing to do with the set .

The trick here I think is to realize that we can use some past results from set theory to aid us. In particular, I think the following will be helpful.

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One must also keep in mind the distributive laws for sets.

Moving on with the exercise, let us assume that . We first see that the left side becomes . Since under our assumption that , we see that this becomes .

For the right side, we get , which is just , since . Now, we distribute again, which gives us . So we have shown that is a sufficient condition for . We now go on to show that it is also a necessary condition, completing our proof.

Suppose . Our task is to show that . It is probably a good idea to use the let trick. So let . Our task now is to show that . If we let be an arbitrary set, then since , we also know that . This tells us that . Since by assumption, we also have . If , we know that and . So since , we can conclude that . This completes our proof.

Now about the “Observe that the condition has nothing to do with the set ” part. Well, we see that in our proof, the set was arbitrary and wasn’t important at all. So that’s that.