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Math Help - A few set theory questions

  1. #1
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    A few set theory questions

    I have the following questions which i've anwered but want to check the correctness of before moving on with other similar questions.

    1. Let A = {0,1} and let B = {2,3}
    i. how many ordered pairs are in A x B
    ii. Draw the arrow diagrams for all the relations from A to B that contain exactly 2 ordered pairs.
    iii. How many of these are functions?

    My answers:
    i) 4
    ii) A few set theory questions-op.png

    iii) My understanding is a function requires a 1 to 1 relationship for all elements in the domain so the first two would be functions.
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  2. #2
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    Re: A few set theory questions

    another one that I'm not 100% on:
    Let A = {0,1,2}. Let R be the relation from A to 2^A (power set) defined as follows: if x is an element of A and X is a subset of A then (x, X) is in R if and only if x is not a subset of X. Draw the arrow diagram of R.

    So I'm thinking the arrow diagram would have the A set on the left, the power set of A on the right and arrows going from the elements in A on the left to the sets on the right which don't contain that A. If this is correct every element in A would also point to the empty set? So every A has 4 arrows?
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  3. #3
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    Re: A few set theory questions

    Quote Originally Posted by anonymouse View Post
    My answers:
    i) 4
    Correct.
    Quote Originally Posted by anonymouse View Post
    ii) Click image for larger version. 

Name:	op.png 
Views:	3 
Size:	5.9 KB 
ID:	24435
    You missed {(0,2), (0,3)} and {(0,2), (1,2)}.

    Quote Originally Posted by anonymouse View Post
    iii) My understanding is a function requires a 1 to 1 relationship for all elements in the domain so the first two would be functions.
    "1 to 1 relationship" is not a technical term. There are "one-to-one correspondence" and "one-to-one function," which are not the same thing. See the second paragraph here. Neither of those properties is required for a relation to be a function. Your last example is not a function because it maps one element of the domain to two different elements of the codomain.
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  4. #4
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    Re: A few set theory questions

    Quote Originally Posted by anonymouse View Post
    Let A = {0,1,2}. Let R be the relation from A to 2^A (power set) defined as follows: if x is an element of A and X is a subset of A then (x, X) is in R if and only if x is not a subset of X.
    You probably mean "an element of X."

    Quote Originally Posted by anonymouse View Post
    So I'm thinking the arrow diagram would have the A set on the left, the power set of A on the right and arrows going from the elements in A on the left to the sets on the right which don't contain that A. If this is correct every element in A would also point to the empty set? So every A has 4 arrows?
    Yes, this is correct, though it should say, "Every element of A has 4 outgoing arrows."
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  5. #5
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    Re: A few set theory questions

    Thanks for the corrections and looking over my work emakarov!
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  6. #6
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    Re: A few set theory questions

    Quote Originally Posted by anonymouse View Post
    I have the following questions which i've anwered but want to check the correctness of before moving on with other similar questions.

    1. Let A = {0,1} and let B = {2,3}
    i. how many ordered pairs are in A x B
    ii. Draw the arrow diagrams for all the relations from A to B that contain exactly 2 ordered pairs.
    iii. How many of these are functions?

    My answers:
    i) 4
    ii) Click image for larger version. 

Name:	op.png 
Views:	3 
Size:	5.9 KB 
ID:	24435

    iii) My understanding is a function requires a 1 to 1 relationship for all elements in the domain so the first two would be functions.
    Yes, those are all correct.
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  7. #7
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    Re: A few set theory questions

    Quote Originally Posted by HallsofIvy View Post
    Yes, those are all correct.
    What about remarks in post #3?
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  8. #8
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    Re: A few set theory questions

    Quote Originally Posted by HallsofIvy View Post
    Yes, those are all correct.
    See reply #3. There are four pairs in A\times B so there are \binom{4}{2}=6 relations with exactly two pairs.
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