Show that $A\cap B\subseteq A$ and $A\subseteq A\cup B$

$$A \cap B \subseteq A$$

My first step would be to write it as $(x \in A \land x \in B) \subseteq A$. Then I know by the following implication that is always true $P \land Q \implies P$. But I am not sure how to write it down mathematically correct.

$$A \subseteq A \cup B $$

I would write it as $A \subseteq (x \in A \lor x \in B)$ . Then I know by following implication that is always true $P \implies P \lor Q$. But again I do not know how to write in down mathematically correct.

May you could help a beginner?

Solution 1:

$x \in A$ is a logical expression and so is $\left((x \in A) \land (x \in B)\right)$. The operator $\subseteq$ operates on sets, the result is a logical value. The definition is

$U \subseteq V$ if and only if $(x \in U) \implies (x \in V)$

The other definitions you use ar

$ x \in U \cap V$ if and only if $ (x \in U) \land (x \in V)$
$ x \in U \cup V$ if and only if $ (x \in U) \lor (x \in V)$

So the expression $((x \in A) \land (x \in B)) ⊆ A$ does not make sense because you put a logical operator on the left side of $\subseteq$.

$P$ and $Q$ are logical variables in

$$P \land Q \implies P$$

If you substitute $P$ by $(x \in A)$ and substitute $Q$ by $(x \in B)$ you get

$$(x \in A) \land (x \in B) \implies (x \in A)$$

But the left side of the implication is the definition of $x \in (A \cap B)$, so you get $$x \in (A \cap B) \implies (x \in A)$$ which again is the definition of $\subseteq$.

Your second statement can be proved in a similar way.