For any sets $A, B$, and $C$, if $A \subseteq B$ and $A \subseteq C$, then $A \subseteq (B \cap C)$

I made a Venn Diagram so I know that this is true. Now I just need some help on getting the proof right.

What I did first was obviously assume the premises, and then I tried to unpack them. So now I have $x\in A \to x\in B$ because $A \subseteq B$ and $x\in A \to x\in C$ because $A \subseteq C$.

Any pointers on where I should go from here?

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Let $x$ be an arbitrary element of $A$. We want to conclude that $x$ also belongs to $B \cap C$. (This will show $A \subseteq B \cap C$.)

Now, since $A \subseteq B$, we know $x \in B$. Similarly, $x \in C$. Since $x \in B$ and $x \in C$, this means $x \in B \cap C$, as desired. (The set $B \cap C$ is defined to be the collection of all elements belonging to both $B$ and $C$.)

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The first thing you need to do when approaching something like this is to write the definitions down:

• $X\subseteq Y$ if and only if for every $x\in X$ it is true that $x\in Y$.
• $x\in X\cap Y$ if and only if $x\in X$ and $x\in Y$.

Now we assume that $A\subseteq B$ and that $A\subseteq C$, and we wish to show that $A\subseteq (B\cap C)$ as well.

To show that an inclusion holds we take an arbitrary $a\in A$ and we need to show that $a\in B\cap C$. Namely we need to show that $a\in B$ and $a\in C$. Our assumption was that $A\subseteq B$, therefore every element of $A$ is an element of $B$, in particular the $a$ which we took; similarly we assumed $A\subseteq C$ and therefore $a\in C$ as well.

We have therefore proved that if $a\in A$ is any element then $a\in B$ and $a\in C$, and therefore by definition $a\in B\cap C$. Therefore we have shown that the definition of $A\subseteq (B\cap C)$ holds.

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If $x\in A\implies x\in B$ since $A⊆B$ and $x\in C$ also since $A⊆C$
Now, Since $x\in B$ and $C\implies x\in B\cap C$
Thus, whenever $x\in A\implies x\in B\cap C\implies A⊆B\cap C$