Is my proof that $ B- \bigcup_{i=i}A_i \subseteq \bigcap_{i=I}(B - A_i)$ correct?

Question

Suppose $x$ is arbitrary . Let $$ x \in B- \bigcup_{i=i}A_i$$

Then $$ x \in B \land x \notin \bigcup_{i=i}A_i$$

So $$ x \in B \land \forall i \in I \rightarrow x \notin A_i $$

Thus,

$$ x \in \bigcap_{i=I}(B - A_i)$$

Since $x$ is arbitrary,

$$B- \bigcup_{i=i}A_i \subseteq \bigcap_{i=I}(B - A_i)$$

I'm not really sure about my conclusion since $ x \in \bigcap_{i=I}(B - A_i)$ seems to be equivalent to $ \forall i \in I \rightarrow x \notin A_i \land x \in B $ and not the third line of my proof. Or are they equivalent?

Answer 1

Your proof is perfectly fine. Good job! Concerning your question, maybe you should think about why $\forall i \in I \rightarrow x \notin A_i \land x \in B$ is equivalent to the third line of your proof. That isn't very hard to see, since $x \in B$ doesn't depend on the $i$'s.