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Basic set theory proof attempt

  1. Aug 15, 2015 #1
    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. jcsd
  3. Aug 15, 2015 #2
    What's the problem here?
     
  4. Aug 15, 2015 #3
    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
     
  5. Aug 15, 2015 #4

    PeroK

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    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?
     
  6. Aug 15, 2015 #5
    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
  7. Aug 15, 2015 #6

    PeroK

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    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##
     
  8. Aug 16, 2015 #7
    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.
     
  9. Aug 16, 2015 #8

    Fredrik

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    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.
     
  10. Aug 16, 2015 #9

    Zondrina

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    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
  11. Aug 16, 2015 #10
    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.
     
  12. Aug 16, 2015 #11
    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.
     
  13. Aug 16, 2015 #12

    Mark44

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    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."
     
  14. Aug 17, 2015 #13

    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.
     
  15. Aug 17, 2015 #14
    I don't understand what you meant by arm-waving.

    Am I any wrong here?
     
  16. Aug 17, 2015 #15

    Fredrik

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    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##.
     
  17. Aug 17, 2015 #16
    Well, the statement is still true if you don't consider A is a subset of C.
     
  18. Aug 17, 2015 #17
    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.
     
  19. Aug 17, 2015 #18
    Yes, it's actually not.

    It doesn't come.
     
  20. Aug 17, 2015 #19

    Fredrik

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    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}.
     
  21. Aug 17, 2015 #20

    Zondrina

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

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