Prove or disprove that $X\cap Y = X$ implies that $X\cup Y = Y$. I am asked to prove or disprove the following proposition:
$$(X \cap Y = X) \to (X \cup Y = Y)$$
I feel this is true, because this means that X is a subset of Y, and as a result, when we do union with these two, it should be the entire set.
However, I have a hard time constructing a formal proof to prove this. Not sure where to start from.
 A: The proposition is true. This is because $X\cap Y = X$ iff $X\subseteq Y$ iff $X\cup Y = Y$.
Proposition
If $X\cap Y = X$, then $X\subseteq Y$.
Proof
If $x\in X = X\cap Y$, then $x\in X$ and $x\in Y$, which implies that $x\in Y$. Consequently, $X\subseteq Y$.
Proposition
If $X\subseteq Y$, then $X\cup Y = Y$.
Proof
The inclusion $Y\subseteq X\cup Y$ does always hold. Thus we have to prove that $X\cup Y\subseteq Y$.
Indeed, if $x\in X\cup Y$, then $x\in X$ or $x\in Y$. If $x\in X$, then $x\in Y$ because $X\subseteq Y$.
If $x\in Y$, then we are done.
Solution
Based on the previous results, we conclude that
\begin{align*}
X\cap Y = X \Rightarrow X\subseteq Y \Rightarrow X\cup Y = Y
\end{align*}
Hopefully this helps.
A: You can prove it by fairly straightforward element-chasing, showing that each of $X\cup Y$ and $Y$ is a subset of the other.
Suppose that $X\cap Y=X$, and let $x\in X\cup Y$; we want to show that $x\in Y$. Since $x\in X\cup Y$, either $x\in Y$, in which case we’re done, or $x\in X$. But $X=X\cap Y$ by hypothesis, so $x\in X\cap Y$, and therefore $x\in Y$. Thus, in both cases $x\in Y$, and since $x$ was an arbitrary element of $X\cup Y$, we’ve shown that $X\cup Y\subseteq Y$. And it’s clear that $Y\subseteq X\cup Y$, so in fact $X\cup Y=Y$.
