Proof by induction - $\bigcup_{k=1}^n (X \setminus A_k) = X \setminus \bigcap_{k=1}^n A_k$ I am trying to prove this for any positive integer $n$, and for any subsets $A_1, A_2, \ldots, A_n$.
The below is what I have so far.
Proof.
Let $X$ be a set. 
The proof is my induction on $n$. -------------Should this say anyting about induction on the subsets?
Basis step: For $n = 1$, we have $(X \setminus A_1)$ for the LH side, and $X \setminus (A_1)$ on the RH side, so the statement is true for $n = 1$. 
Induction hypothesis: Assume that the statement is true for some positive integer $w$. That is, $$\bigcup_{k=1}^w (X \setminus A_k) = x \setminus \bigcap_{k=1}^w A_k$$
I need help moving onto the induction step and showing that this is true for $w + 1$. I've been given a hint that tells me I need to prove the two sets are equal without using  the formal set equality template during my induction step. So I believe I am needing to use DeMorgan's Laws.
 A: $\displaystyle\bigcup_{k=1}^{w+1}(X\setminus A_{k})=X\setminus A_{w+1}\cup\bigcup_{k=1}^{w}(X\setminus A_{k})=X\setminus A_{w+1}\cup X\setminus\bigcap_{k=1}^{w} A_{k}=X\setminus\bigcap_{k=1}^{w+1}A_{k}.$
A: I wouldn't do it by induction because that would limit the result to finitely many indices.  One can write
$$
\bigcup_{A\,\in\,\mathcal A} (X \setminus A) = X \setminus \bigcap_{A\,\in\,\mathcal A} A.
$$
It doesn't matter whether $\mathcal A$ is finite nor which infinite cardinality it has if it's infinite.
Suppose $x\in \bigcup_{A\,\in\,\mathcal A} (X \setminus A)$.
Then for some $A\in\mathcal A$, $x \in X \setminus A$.
So for some $A\in\mathcal A$, $x\in X$ and $x\not\in A$.
So $x\in X$ and for some $A\in \mathcal A$, $x\notin A$.
So $x\in X$ and not$\left( \text{for all }A\in\mathcal A,\  x\in A \right)$.
So $x\in X$ and $x\notin\bigcap_{A\,\in\,\mathcal A} A$.
Finally, $x\in X\setminus \bigcap_{A\,\in\,\mathcal A} A$.
This shows $\bigcup_{A\,\in\,\mathcal A} (X \setminus A) \subseteq X\setminus \bigcap_{A\,\in\,\mathcal A} A$.
Then you need to prove $\supseteq$, and that's done similarly.
