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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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4 Answers 4

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 $x\in A\implies x\in B\cap C\implies A⊆B\cap C$

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Indeed, if $A \subseteq B$ and $C\subseteq D$ then:
(i) $A \cap C \subseteq B \cap D$
(ii) $A \cup C \subseteq B \cup D $

(i) ($A \subseteq B) \Leftrightarrow x \in A \implies x\in B$ simillar for $C \subseteq D$
now, $x \in A \cap C \Leftrightarrow x\in A$ and $x\in C$ but this is simply implies $x \in B $ and $x \in D$ which implies $x \in B \cap D$
$\therefore $ we have $x \in A \cap C$ $\Rightarrow$ $x \in B \cap D$ $\Leftrightarrow A \cap C \subseteq B \cap D$

(ii) Similar proof we can give for second one. You can use these results to answer your questions and your sum is obvious consequent of (i).

You can also verify my reasoning using truth tables, but I'm not verifying here because it will be lengthy.

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