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I can make sense of why these two equations are not equivalent intuitively but I cannot prove them on paper.

For $x\in\cap_{i \in I}(A_i\cup B_i)$ I end up with:

$\forall(i \in I \rightarrow (x \in A_i \lor x \in B_i)$)

which to me is equivalent to what I get for the other equation $x \epsilon (\cap_{i \in I}A_i)\cup(\cap_{i \in I}B_i)$:

$\forall i \in Ix \in A_i \lor \forall i \in Ix \in B_i$

because if I were to simplify $\forall i \in Ix \in A_i \lor \forall i \in Ix \in B_i$, I would essentially get $\forall (i \in I \rightarrow (x \in A_i \lor x \in B_i))$.

Thank you.

Bavneet
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2 Answers2

3

$$\forall i, x\in A_i \text{ or } x\in B_i$$ and $$(\forall i, x\in A_i ) \text{ or } (\forall i, x\in B_i)$$ is different.

The first expression is larger. For example, $I=\mathbb{N}$, $A_i=\{i\}, B_i=\mathbb{N}-A_i$, then LHS=$\mathbb{N}$, but RHS=$\emptyset$.

Ma Ming
  • 7,482
2

Let $A_1=B_2=\{a\}$, and $A_2=B_1=\{b\}$. Then $a\in (A_1\cup B_1)\cap(A_2 \cup B_2)$, but $a\notin A_1\cap B_1$ and $a\notin A_2\cap B_2$.

vadim123
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