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?
 A: $x \in A$ 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.
A: You are correct to start with let $x\in A\cap B$ so $x\in A\text{ and } B$. Since x is in A we know that $A\cap B\subseteq A$.
Now let $x\in A$. By definition, x is in the union if it is in A or B so $A$ is a subset of the union.
A: To show this, show that any element $x\in A\cap B$ (read as "x in A intersection B") must be in A, and similarly for the union, show that any $x\in A$ must be in $A\cup B$.
