# Does Intersection Distribute Over Subset in Set Theory?

• marovan
In summary: So if you can't prove it, then it's still possible to have A⊂B.In summary, if A is always a subset of B, and so is A intersection C a subset of C, then A⊂B.
marovan

## Homework Statement

$$A \subset B \Rightarrow A \cap C \subset B \cap C$$

2. Homework Equations [/B]
$$A \subset B \Leftrightarrow A \cup B \subset B$$
$$A \cap C \Leftrightarrow A \cap C \subset A \wedge A \cap C \subset C$$

## 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

What's the problem here?

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

marovan said:

## Homework Statement

$$A \subset B \Rightarrow A \cap C \subset B \cap C$$

2. Homework Equations [/B]
$$A \subset B \Leftrightarrow A \cup B \subset B$$
$$A \cap C \Leftrightarrow A \cap C \subset A \wedge A \cap C \subset C$$

## The Attempt at a Solution

For sets A and C
$$A \cap C \Rightarrow A \cap C \subset A$$

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

Likewise for sets B and C
$$B \cap C \Rightarrow B \cap C \subset B$$

Same here.

Therefore
$$A \subset B \Rightarrow A \cap C \subset B \cap C$$

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

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?

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:
marovan said:
All right

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

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

marovan said:
As for A∩C⊂A

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

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##

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.

fireflies said:
yes, it is okay. The proof is correct.
It's not.

fireflies said:
And, if A is always a subset of A, and so does A intersection C a subset of C.
This sentence doesn't make sense.

marovan said:
$$If x \in C then x \in A \cap C \Leftrightarrow x \in A$$

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

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:
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.

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.

fireflies said:
Why won't?
Why won't what?
fireflies said:
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.
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.
fireflies said:
Whatever be the case, (A∩C) becomes
a subset of C.
The technical term for this kind of argument is "arm waving."

Fredrik said:
This sentence doesn't make sense
Mark44 said:
Why won't what?
"
Why won't it make sense?
Mark44 said:
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."

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.

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

Am I any wrong here?

fireflies said:
Like, let x∈A

And A⊂C

Then, x may or may not belong to C.
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.

fireflies said:
If x belongs to C, then x∈(A∩C)

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

Here Φ= empty set.
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?

fireflies said:
Whatever be the case, (A∩C) becomes
a subset of 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##.

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

fireflies said:
And A⊂C

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

Fredrik said:
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##.

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

ma [QUOTE="Fredrik said:
It's not.

Yes, it's actually not.

marovan said:
Therefore
$$A \subset B \Rightarrow A \cap C \subset B \cap C$$

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

It doesn't come.

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}.

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

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

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.

## 1. What is a basic set theory proof attempt?

A basic set theory proof attempt is a mathematical process used to prove the validity of a statement or theorem in the field of set theory. It involves using logical reasoning and mathematical principles to demonstrate that a given statement is true or false.

## 2. How do I approach a basic set theory proof?

There are several steps involved in approaching a basic set theory proof. First, carefully read and understand the statement or theorem you are trying to prove. Then, identify any relevant definitions and axioms that can be used in your proof. Next, use logical reasoning to construct a clear and concise argument that demonstrates the validity of the statement. Finally, check your proof for any errors or inconsistencies.

## 3. What are some common techniques used in basic set theory proofs?

Some common techniques used in basic set theory proofs include proof by contradiction, direct proof, and proof by induction. These techniques involve using logical reasoning and mathematical principles to demonstrate the validity of a statement.

## 4. How do I know if my basic set theory proof is correct?

It is important to carefully check your proof for any errors or gaps in reasoning. You can also ask a colleague or mentor to review your proof and provide feedback. Additionally, you can compare your proof to other existing proofs of the same statement to ensure that your approach is valid.

## 5. Are there any common mistakes to avoid in basic set theory proofs?

Some common mistakes to avoid in basic set theory proofs include using incorrect definitions or axioms, making faulty assumptions, and using circular reasoning. It is also important to clearly and logically present your proof, avoiding any vague or ambiguous language. Additionally, double-checking your work and seeking feedback from others can help catch any mistakes before submitting your proof.

• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
1K
• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
4K
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
1K