# Thread: Small Set Theory Proof

1. ## Small Set Theory Proof

Let I be a nonempty set and for each $i\epsilon I$ let $X_{i}$ be a set.

Prove that if each $X_{i}$ is a subset of a given set S, then $e( \bigcup_{i\epsilon I} X_{i}) = \bigcap_{i\epsilon I} e X_{i}$

Please note that e represents the complement of a set.

Thank you as always!

Edit: Also note, of course, that we must show inclusion both ways.

2. Hello !

If $x \in e \left( \bigcup_{i\in I} X_i \right)$ we can say that $x \not \in \bigcup_{i \in I} X_i$.
This means that $\forall i \in I ~,~ x \not \in X_i$ (by the definition of the union)
This is equivalent to saying that $\forall i \in I~,~ x \in e\left(X_i\right)$
Therefore $x \in \bigcap_{i \in I} e \left(X_i\right)$ (by the definition of the intersection)

Conclusion : $x \in e \left( \bigcup_{i\in I} X_i \right) \Longleftrightarrow x \in \bigcap_{i \in I} e \left(X_i\right)$
Thus $e \left( \bigcup_{i\in I} X_i \right) = \bigcap_{i \in I} e \left(X_i\right)$

You have to check it, but I'm quite sure the reasoning works with equivalence and not implication, and hence the conclusion that doesn't go through the steps $A \subseteq B$ and $B \subseteq A$
If you want to do the reverse order, it's the same reasoning.

3. More rigorous to prove it by induction, of course.

This is a standard result and it pays to study it. This is known as (one of) "De Morgan's Laws" - the other being with the union and intersection the other way round. I recommend a google session on it.