We wish to verify the generalized law of DeMorgan $(\bigcup_{i \in \mathcal{I}} A_i)^c = \bigcap_{i \in \mathcal{I}} A_i^c$.
Let $ x \in (\bigcup_{i \in \mathcal{I}} A_i)^c$. Then $x \notin \bigcup_{i \in \mathcal{I}} A_i$ and $x \notin A_i$ for $i \in \mathcal{I}$, and so $x \in A_i^c$ for all $i$. Hence $x \in \bigcap_{i \in \mathcal{I}} A_i^c $. We have shown that $(\bigcup_{i \in \mathcal{I}} A_i)^c \subset \bigcap_{i \in \mathcal{I}} A_i^c$.
We must now show that $\bigcap_{i \in \mathcal{I}} A_i^c \subset (\bigcup_{i \in \mathcal{I}} A_i)^c$. Now let $x \in \bigcap_{i \in \mathcal{I}} A_i^c.$ Then $x \in A_i^c$ for all $i \in \mathcal{I} $ and so $x \notin A_i$ for all $i \in \mathcal{I}$. Hence $x \in A_i^c$ for all $i \in \mathcal{I}$ and so $x \in (\bigcup_{i \in \mathcal{I}} A_i)^c $.
Then $\bigcap_{i \in \mathcal{I}} A_i^c \subset (\bigcup_{i \in \mathcal{I}} A_i)^c$, and since $(\bigcup_{i \in \mathcal{I}} A_i)^c \subset \bigcap_{i \in \mathcal{I}} A_i^c$ we have that $(\bigcup_{i \in \mathcal{I}} A_i)^c = \bigcap_{i \in \mathcal{I}} A_i^c$, which is what we set out to show.
I just want to make sure that my proof makes sense and I was hoping for some constructive criticism regarding proof style/format — would really appreciate any feedback whatsoever.
Best Answer
You should not write
since you do not know that $\mathcal{I}$ is countable (or even consists of integers - perhaps $\mathcal{I}=\mathbb{Q}$ or $\mathcal{I}=\mathbb{R}^2$ or some set that doesn't even contain numbers).
In the reverse direction, you jump from $y \notin A_i$ for all $i \in \mathcal{I}$ to $y \in (\bigcup_{i \in \mathcal{I}} A_i)^c$. That jump is probably fine in most proofs, though you may want to stick closer to the definitions like you did in the forward direction if this is for an intro to proofs course. I would recommend adding an intermediate step there.