A little help with elementary set intuition, specifically on intersection Simple example, $A\cup 0=A$ is true because $x\in A$ or $x\in 0$ and that $0\subseteq A$. This I can understand (Actually, I hope I understood it)
But what I am having difficulty with is something like $A\cup(B\cap C)$. I know this implies $x \in A$ or $x\in (B\cap C)$. If $x\in A$, then $x\in(A\cup B)$ and $x\in (A\cup C)$ or alternitavely, $(A\cup B) \cap (A\cup C)$. What I am confused by is how can it be both? Isn't it that the OR in $x \in A$ or $x\in (B\cap C)$ implies that  $x$ is either in $A$ OR $x$ is in $(B\cap C)$ and not both? Obviously not, but I am having a bit of difficulty wrapping my head on why that is exactly.
 A: Sketch of the proof. We want to prove that two sets are equal. Recall the definition of equality between sets.

Definition. Let $A$ and $B$ be any sets. We say that the sets $A$ and $B$ are equal, and we write $A = B,$ if $A \subseteq B$ and $B \subseteq A.$

Hence, we must prove two inclusions, by proving that for an arbitrary object $c$
\begin{align*}
c \in A \cup (B \cap C) \implies c \in (A \cup B) \cap (A \cup C) && \text{and}\\
c \in (A \cup B) \cap (A \cup C) \implies c \in A \cup (B \cap C).
\end{align*}
Recall also the distributive property of the logical operation $\vee.$

Theorem. Let $p,q$ and $r$ be any propositions. Then
$$p \vee (q \wedge r) \iff (p \vee q) \wedge (p \vee r).$$

(The proof of this theorem is quite easy, you just have to construct the truth table for the proposition above stated and conclude that it is a tautology.)

Proof. Let $x$ be an arbitrary object. Then,
\begin{align*}
x \in A \cup (B \cap C) & \iff x \in A \vee x \in B \cap C & \text{(definition of set union)}\\
& \iff x \in A \vee (x \in B \wedge x \in C) & \text{(definition of set intersection)}\\
& \iff (x \in A \vee x \in B) \wedge (x \in A \vee x \in C) & \text{(distributive property of $\vee$)}\\
& \iff x \in A \cup B \wedge x \in A \cup C & \text{(definition of set union)}\\
& \iff x \in (A \cup B) \cap (A \cup C) & \text{(definition of set intersection)}
\end{align*}
Hence, we have prove that each of the sets $A \cup (B \cap C)$ and $(A \cup B) \cap (A \cup C)$ is a subset of the other. Then, by definition of equality between sets we conclude that $A \cup (B \cap C) = (A \cup B) \cap (A \cup C).$ $\square$

Remark. Note that in the above proof we only deal with equivalences and not just with implications, that’s because we are relying on the definitions to justify all of our steps plus the distributive property of $\vee$ which is itself an equivalence.
In fact note that those two sets are the same. Always keep in mind that, in mathematics, we always use an inclusive or which means that saying $x \in A \cup B$ means $x \in A, x \in B$ or $x$ in both $A$ and $B,$ i.e., $x$ is in at least of of those sets $A$ or $B.$
Also, and I leave this as something for you to think about, what would happen if they are not the same? (I showed that they are the same, but answering this question is a really good way to test if you understood).
A: Given a family of sets $A_i$, where $i$ ranges over some index set $I$, an element $x$ lies in the set $A=\bigcup_{i\in I}A_i$ if and only if there is some $i\in I$ such that $x\in A_i$. There is no condition that $x$ has to lie in only one $A_i$; it can lie in arbitrarily many.
On the other hand, given two sets $A$ and $B$, there is a construction $A\triangle B$, called the symmetric difference of the two, which is defined as $(A\setminus B)\cup (B\setminus A)$. So this consists of elements that lie in $A$ or in $B$, but not in both. It seems you are confusing the union construction with the symmetric difference construction; hopefully it helps.
