Show that $A \subset B \implies A \cap B = A$ I am trying to show that:
$$A \subset B \implies A \cap B = A$$
So far I got:
$$A \subset B$$
$$A \cap B \subset A$$
$$A \cap B \subset B$$
$$A \cap B \subset A \subset B$$
 A: $A\cap B \subset A$.
On the other hand if $x\in A $ then $x\in B$ (because $A\subset B$). 
Since $x\in A$ and $x\in B$ we have $x\in A\cap B$, so $A\subset A\cap B$
A: Clearly  $A\cap B \subseteq A.$
Conversely $A \subseteq A\cap B$ since:
 Let $x\in A$ then, by assumption, $x\in B$. So $x\in A\cap B$.
A: Doing little diagrams can help to see what happens :

From this, you see that :


*

*$x \in A \cap B$ implies $x \in A$ and $x \in B$, so in particular $x\in A$

*$x \in A$ implies $x \in A \cap B$, since every element of $A$ is also an element of $B$.

A: Suppose that $A\subseteq B$
Let $x\in A\cap B$
Then $x\in A$ and $x\in B$
In particular this implies that $x\in A$ which shows $A\cap B\subseteq A$
Now, suppose instead that $y\in A$
Since $A\subseteq B$ that implies that $y\in B$ also
As a result, $y\in A$ and $y\in B$
This implies $y\in A\cap B$ which shows $A\subseteq A\cap B$
Since $A\cap B\subseteq A$ and $A\subseteq A\cap B$, it must be that $A\cap B=A$

Alternatively, with $A\subseteq B$ you can write $B=A\cup (B\setminus A)$
You have then $A\cap B = A\cap (A\cup (B\setminus A))=(A\cap A)\cup (A\cap B\cap A^c)=A\cup \emptyset = A$
A: Equality of sets is equivalent to the conjunction of two inclusions. You have established the inclusion $A\cap B\subset A$ and you need to establish the other direction of the inequality, namely
$$A\subset A\cap B$$
For this, observe that $X\subset Y\wedge X\subset Z\implies X\subset Y\cap Z.$ Alternatively use the definition of the intersection and prove that any element in $A$ is also in $A\cap B.$
A: A definition of $A\subseteq B$ is $x\in A \implies x\in B$.
A definition of $A\cap B$ is $A\cap B=\{x:x\in A \land x\in B\}$.
Therefore if we have $A\subseteq B$, we have $A\cap B=\{x:x\in A \}=A$.
A: Notice that
$$\begin{aligned}
A\cap B\neq A\quad&\Longleftrightarrow\quad A\cap B\not\subset A\quad\text{or}\quad A\not\subset A\cap B\\
&\Longleftrightarrow\quad A\not\subset A\cap B\\
&\Longleftrightarrow\quad A\not\subset A\quad\text{or}\quad A\not\subset B\\
&\Longleftrightarrow \quad A\not\subset B
\end{aligned}$$
and thus
$$A\subset B\quad\Longleftrightarrow\quad A\cap B=A$$
