# Homework Help: Basic set theory proof attempt

1. Aug 15, 2015

### marovan

1. The problem statement, all variables and given/known data
$$A \subset B \Rightarrow A \cap C \subset B \cap C$$

2. Relevant equations

$$A \subset B \Leftrightarrow A \cup B \subset B$$
$$A \cap C \Leftrightarrow A \cap C \subset A \wedge A \cap C \subset C$$

3. The attempt at a solution
For sets A and C
$$A \cap C \Rightarrow A \cap C \subset A$$
Likewise for sets B and C
$$B \cap C \Rightarrow B \cap C \subset B$$
Therefore
$$A \subset B \Rightarrow A \cap C \subset B \cap C$$

I don't know if I'm completely off the map

2. Aug 15, 2015

### fireflies

What's the problem here?

3. Aug 15, 2015

### marovan

I'm sorry..

I have to proof the first statement.

I have attempted at a proof, but I don't know if it's the right way to do so. I'm asking for a little guidance

4. Aug 15, 2015

### PeroK

This makes no sense. I think what you mean is simply:
$$A \cap C \subset A$$
As an aside, can you prove this?

Same here.

The conclusion certainly doesn't follow.

To help you prove this properly, the first question is: can you state (formally) the definition for one set to be a subset of another? In this case $$A \subset B$$

Can you state formally what that means?

5. Aug 15, 2015

### marovan

All right

$$A \subset B \Leftrightarrow (x \in A \Rightarrow x \in B)$$

As for A∩C⊂A

$$A \cap C \Leftrightarrow (x \in A and x \in C)$$

$$If x \in C then x \in A \cap C \Leftrightarrow x \in A$$

Therefore, since A ∩ C ⊂ A and likewise for B ∩ C then A ∩ C ⊂ B ∩ C

Last edited: Aug 15, 2015
6. Aug 15, 2015

### PeroK

That's right. You should use this to try to prove the original proposition.

That doesn't make sense. I think you mean:

$$x \in A \cap C \Leftrightarrow (x \in A \ and \ x \in C)$$
You must be more careful how you use these symbols.

One final point. To prove something it's often best to start the proof with what you have, so I would start with:

Let $A \subset B$

7. Aug 16, 2015

### fireflies

yes, it is okay. The proof is correct.

And, if A is always a subset of A, and so does A intersection C a subset of C.

8. Aug 16, 2015

### Fredrik

Staff Emeritus
It's not.

This sentence doesn't make sense.

I can't follow your reasoning here. You want to prove that if A⊂B, then A ∩ C ⊂ B ∩ C. So you need to start by stating that you're assuming that A⊂B. Then, since what you want to prove now is that every element of A ∩ C is an element of B ∩ C, the straightforward way to continue is with something like "let x be an arbitrary element of A ∩ C". (It's OK to just say "Let $x\in A\cap C$). Then you explain how you can be sure that x is an element of B ∩ C.

9. Aug 16, 2015

### Zondrina

You should use a direct proof to show $A \subseteq B \Rightarrow A \cap C \subseteq B \cap C$.

Start by saying "If $A \subseteq B$, then $\forall x, \space x \in A \Rightarrow x \in B$".

Now choose an arbitrary $y \in A \cap C$. Then $y \in A$ and $y \in C$.

You know if $y \in A$, then $y \in B$ because of the starting statement.

Now we know $y \in B$ and $y \in C$ because of the two previous statements, so you only have one conclusion to make.

Last edited: Aug 17, 2015
10. Aug 16, 2015

### fireflies

Why won't?

Like, let x∈A

And A⊂C

Then, x may or may not belong to C.

If x belongs to C, then x∈(A∩C)

If x does not belong to C, then Φ∈(A∩C)

Here Φ= empty set.

Whatever be the case, (A∩C) becomes
a subset of C.

11. Aug 16, 2015

### fireflies

Well, it's not how you prove it. It's if you can prove it. In class ten, I did this proof the same way.

12. Aug 16, 2015

### Staff: Mentor

Why won't what?
The empty set is a subset of every set, whether or not x ∈ C. It's silly to conclude a statement that is always true.
The technical term for this kind of argument is "arm waving."

13. Aug 17, 2015

### fireflies

Why won't it make sense?

I know, but since they saying that it's
unreasonable that A∩C is a subset of C (which is also always true), so, I described it elaborately.

14. Aug 17, 2015

### fireflies

I don't understand what you meant by arm-waving.

Am I any wrong here?

15. Aug 17, 2015

### Fredrik

Staff Emeritus
x certainly belongs to C, since you assumed both that x is an element of A, and that every element of A is an element of C.

Here's an empty set symbol that you can copy and paste: ∅
You can also find it by clicking the ∑ symbol above the field where you type your post.

Why would ∅ be an element of A∩C?

If you want to prove that A∩C is a subset of C, this is how you do it: Let x be an arbitrary element of A∩C. Since x is an element of A∩C, it's an element of A and an element of C. In particular, it's an element of C.

Alternatively you just write down these implications along with a comment that says that they hold for all x:

$x\in A\cap C\Rightarrow \left(x\in A\text{ and }x\in C\right)\Rightarrow x\in C$.

16. Aug 17, 2015

### fireflies

Well, the statement is still true if you don't consider A is a subset of C.

17. Aug 17, 2015

### fireflies

I think I would rather exclude this part. It is a mistake, other part is okay.

All right. But the same thing comes that this statement is true. That makes the proof correct too.

18. Aug 17, 2015

### fireflies

Yes, it's actually not.

It doesn't come.

19. Aug 17, 2015

### Fredrik

Staff Emeritus
Marovan's argument in posts #1 and #5 appears to be that since $A\cap C\subseteq A\subseteq B$ and $B\cap C\subseteq B$, we have $A\cap C\subseteq B\cap C$. To see that this line of reasoning doesn't work, try replacing $A\cap C$ and $B\cap C$ with two arbitrary sets. Does the following implication hold for all E,F?

If $E\subseteq A\subseteq B$ and $F\subseteq B$, then $E\subseteq F$​

It does not. Consider e.g. E={1,2}, A=B={1,2,3}, F={2,3}.

20. Aug 17, 2015

### Zondrina

Reading anything past post #9 has fried my logic processor.

I don't see how anything fireflies has said can hold any merit.

21. Aug 17, 2015

### fireflies

I think I should say sorry for that.

I put more emphasis on the subset statements of the OP than the final conclusion. The final conclusion is, anyways, not correct.

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