# if $A \subseteq B$, then $A \cap C \subseteq B\cap C$

So I tried doing this problem myself, and the answer that I got seems right, yet at the same time I feel like the way I did it is kind of.... wonky? It seems weird basically, and I was hoping someone can help me validate my answer

proof:

Suppose $$x \in A \subseteq B$$.
Then $$x \in A$$ and $$x \in B$$.
Thus, if $$x \in A \cap C$$,
Then $$x\in A$$ and $$x\in C$$ .
Since $$x \in A$$ and $$x \in B$$ and $$x \in C$$,
Then $$x \in B$$ and $$x \in C$$,
which implies that $$x \in B \cap C$$ .
Thus $$A\cap C \subseteq B \cap C$$ .
Therefore, if $$A \subseteq B$$, then $$A\cap C\subseteq B\cap C$$

I'm just not sure if it was alright to assume that $$x \in A \cap C$$, which is what makes me feel like my proof may be wrong and weird.

• The first two lines should use a different variable then the rest because you are talking about any element in $A$ whereas the rest you are talking about an element in $A\cap C$. Or you should leave the first to lines out. – fleablood Oct 16 '19 at 2:17

You did a 'beginners mistake' and started with an assumption. Namely that $$A\subseteq B$$.

You should start with what you have to show. That is $$A\cap C\subseteq B\cap C$$ under the assumption that $$A\subseteq B$$.

Then the proof reads like this:

Let $$x\in A\cap C$$. We have to show that $$x\in B\cap C$$.

Since $$x\in A\cap C$$, we have $$x\in A$$ and $$x\in C$$. Since $$x\in A\subseteq B$$, it is $$x\in B$$. So $$x\in B$$ and $$x\in C$$. Thus $$x\in B\cap C$$.

It is ok to assume that $$x\in A\cap C$$. That is exactly what we have to do, when we want to show some 'subset relationship'.

• Yes, I am actually just starting out with proofs. One question on your answer That I have, is that I learned that I should start with P and assume P is true, then try to arrive at Q (which is a direct proof). So now, I am kind of confused because your answer states that I should start with a part of what I assume is Q in this statement. Can you help clear up this confusion? – Jr194 Oct 16 '19 at 2:20
• What exactly do you mean? We start with $x\in A\cap C$. Because we want to show that $A\cap C\subseteq B\cap C$. For that (by definition) we have to show that for every $x\in A\cap C$, we have that $x\in B\cap C$. So you can (have) to assume that $x\in A\cap C$, to show that we have this subset relation. It is just the definition. :) – Cornman Oct 16 '19 at 2:23
• @Jr194 So what we actually have to show is that $x\in B\cap C$. Where $x\in A\cap C$ is an additional assumption we can make, because of the definition. Note that without the other assumption $A\subseteq B$ this statement would be false. – Cornman Oct 16 '19 at 2:25
• I see. I just thought that the problem was in the format of P implies Q, where P was A $\subseteq$ B, and Q was the second part of the problem. – Jr194 Oct 16 '19 at 2:28

Remember that $$X\cap Y = X$$ iff $$X\subseteq Y$$. According to this result and the additional hypothesis, one has

\begin{align*} (A\cap C)\cap (B\cap C) = (A\cap B\cap C) = A\cap C \Longrightarrow A\cap C \subseteq B\cap C \end{align*}

You need to start with $$x\in A\cap C$$ and show that $$x\in B\cap C$$

Let $$x\in A\cap C$$ then $$x\in A \text { and } x\in C$$

Since $$x\in A$$ and $$A\subseteq B$$ then $$x\in B$$ $$x\in B \text { and } x\in C \implies x\in B\cap C$$

The "$$x$$" you are referring to in lines 1 and 2 are a different "$$x$$" than you have in the rest of the proof. And don't care about the $$x\in A$$ specifically but just that it leads to a general conclusion that we will use for the later $$x$$.

If I were to edit your proof but leave your thought process and pacing completly in tact, but clarify when we are making general from specific cases I'd do:

We are presuming $$A\subseteq B$$.

So for any $$y \in A$$ we'd have $$y$$ is $$A$$ and $$y \in B$$.

Let $$x$$ be an arbitrary element in $$A\cap C$$.

Then x∈A and x∈C .

Since x∈A thus $$x\in A$$ and $$x \in B$$.

So x∈B and x∈C,

which implies that x∈B∩C .

Thus any element of $$x \in A\cap C$$ is in $$B\cap C$$.

Thus A∩C⊆B∩C .

Therefore, if A⊆B, then A∩C⊆B∩C

.....

But you don't need to be quite so stiff a repetitive.

It'd be enough to say.

For any $$x \in A\cap C$$ we have $$x\in A$$ and $$x\in C$$.

Since $$A\subseteq B$$ and $$x \in A$$ we know $$x \in B$$.

So $$x \in B$$ and $$x \in C$$.

So $$x\in B\cap C$$.

Thus $$A\cap C\subseteq B\cap C$$.

We have that $$A,B$$ and $$C$$ are sets.

For all sets $$A, B$$, and $$C$$, if $$A\subseteq B$$ then $$A\cap C\subseteq B\cap C$$.

From the start of your proof

Suppose $$x \in A \subseteq B$$.
Then $$x \in A$$ and $$x \in B$$.
Thus, if $$x \in A \cap C$$,

It is not necessary to assume that $$x \in A \subseteq B$$. Instead, you should assume that $$x\in A\cap C$$ and deduce that $$x\in B\cap C$$. This method is the direct proof technique.

The proof would then be

Let $$A,B,$$ and $$C$$ be sets and assume that $$A\subseteq B$$. We want to show that $$A\cap C\subseteq B\cap C$$. Let $$x\in A\cap C$$. Then, by the definition of intersection, we have $$x\in A$$ and $$x\in C$$. Since $$x\in A$$ and $$A\subseteq B$$, it follows from the definition of subset that $$x\in B$$. Therefore, we have shown that $$x\in B$$ and $$x\in C$$. Again, by the definition of intersection, we can conclude that $$x\in B\cap C$$. Because $$x$$ was arbitrarily chosen, we can now conclude that $$A\cap C\subseteq B\cap C$$.