Theorem. Let $M, N, L$ be sets. Then the following assertions hold:
(1) $M \cap N \subseteq M$
(2) $M \subseteq M \cup N$
Proof is left as an exercise.
My attempt:
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(1) We have to show that the following assertion
"$\forall x : x \in (M \cap N) \implies x \in M$" is true.
By definition, $M \cap N := \{x \mid x \in M \land x \in N\}$. If "$x \in M$" and "$x \in N$" are true, then $M \cap N$ is true. Resulting in an implication with two true assertions, and hence is true.
We could write that as $\exists x : x \in (M \cap N) \implies x \in M$. (Not quite really sure if this line is correct)
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(2) We have to show that the following assertion "$\forall x : x \in M \implies x \in (M \cup N)$" is true.
By definition, $M \cup N := \{x \mid x \in M \lor x \in N\}$. If either "$x \in M$", "$x \in N$" or both, are true, then $M \cup N$ is true. Resulting in an implication with two true assertions, and hence is true.
We could write that as $\exists x : x \in M \implies x \in (M \cup N)$ (Again, not quite really sure if this line is correct)
Liesen, J., Mehrmann, V. 2015. Linear Algebra. Berlin, Germany.: Springer.
Best Answer
You should replace $\exists$ with $\forall$. A way to make the proof more compact is: $$x\in M \cap N \implies x\in M \land x\in N \implies x\in M \implies M\cap N \subseteq M$$