The intersection of two sets is non-empty if and only if there exists a point in one set that is in the other set

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Obvious and easy "theorem", created to make the proof of claims in Dense more applicable to other pages and thus worth covering

Statement

Let [ilmath]A[/ilmath] and [ilmath]B[/ilmath] be sets. Then we claim:

• [ilmath](A\cap B\ne\emptyset)\iff(\exists x\in A[x\in B])[/ilmath]. As intersection is commutative, it follows that [ilmath]\iff(\exists y\in B[y\in A][/ilmath] of course, as if we applied the claim to [ilmath]B\cap A[/ilmath] instead.

Discussion

• Discuss relation (but not equivalence) to the implies-subset relation, which looks like:
  • [ilmath]\forall a\in A[a\in B][/ilmath] (which is equivalent to [ilmath]A\subseteq B[/ilmath])
• Extend that to the intersection of sets is a subset of each set

Proof

[ilmath]\implies[/ilmath]

• Suppose [ilmath]A\cap B\ne\emptyset[/ilmath], then:
  • Obviously there exists something in the intersection Caution:Trickier than it sounds
    • Such a thing is in [ilmath]A[/ilmath] and in [ilmath]B[/ilmath]
    • Thus there is a thing in [ilmath]A[/ilmath], thus in [ilmath]B[/ilmath].

[ilmath]\impliedby[/ilmath]

• Suppose [ilmath]\exists x\in A[x\in B][/ilmath]
  • Recall [ilmath][x\in A\cap B]\iff[x\in A\wedge x\in B][/ilmath] by definition of intersection
  • We have an [ilmath]x[/ilmath] such that [ilmath]x\in A[/ilmath] and [ilmath]x\in B[/ilmath]
    • Thus [ilmath]x\in A\cap B[/ilmath], as mentioned above.
      • Thus [ilmath]A\cap B\ne\emptyset[/ilmath] (as if it were equal to the emptyset, there wouldn't exist any [ilmath]x[/ilmath] to be in [ilmath]A\cap B[/ilmath]!)

Grade: C

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