Let f:A->B be a bijection

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  • #1
aaaa202
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This is something I have been wondering about. Let f:A->B
be a bijection. If B is a subset of X. Can there still exist a bijection from A to X?
 

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  • #2
UltrafastPED
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But A is already fully mapped onto B, so which elements of A are left to map onto the X - B?
 
  • #3
aaaa202
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Well the cardinality of the rationals, natural numbers and even natural numbers are the same.
So there exists a bijection between each of these sets even though, the even natural numbers are a subset of the natural numbers. So I think you're actually wrong?
 
  • #4
UltrafastPED
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I asked a question; I did not provide an answer.

However it seems that you now know the answer: I think you are saying "for finite sets the answer is NO; for infinite sets the answer is YES". Next you should go ahead and construct an example using the set of natural numbers and subsets of odds and evens.
 
  • #5
HallsofIvy
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You might want to re-write the question to specify that B is a proper subset of X.
 
  • #6
jbunniii
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I asked a question; I did not provide an answer.

However it seems that you now know the answer: I think you are saying "for finite sets the answer is NO; for infinite sets the answer is YES". Next you should go ahead and construct an example using the set of natural numbers and subsets of odds and evens.
It should be mentioned that it isn't always true for infinite sets. E.g., take ##A = B = \mathbb{R}## and ##X = \mathbb{Z}##. In general it will be possible only if ##X## has the same cardinality as ##B## (and hence the same cardinality as ##A##).
 

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