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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.

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    $\begingroup$ 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. $\endgroup$ – fleablood Oct 16 '19 at 2:17
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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'.

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  • $\begingroup$ 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? $\endgroup$ – Jr194 Oct 16 '19 at 2:20
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    $\begingroup$ 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. :) $\endgroup$ – Cornman Oct 16 '19 at 2:23
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    $\begingroup$ @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. $\endgroup$ – Cornman Oct 16 '19 at 2:25
  • $\begingroup$ 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. $\endgroup$ – Jr194 Oct 16 '19 at 2:28
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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*}

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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 $x\in B$ $$x\in B \text { and } x\in C \implies x\in B\cap C$$

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

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

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