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 July 9th, 2019, 01:33 PM #1 Newbie   Joined: Jul 2019 From: Brazil Posts: 1 Thanks: 0 Definition and sets The following sentence, a set is infinite if and only if the set have a bijection with itself, the negation of the affirmation is true? A set is finite if only if there doesn't exist any bijection with itself? The idea is what is a better definition of a bijection in this case. (Sorry, guys, I don't know English very well.) Last edited by skipjack; July 9th, 2019 at 03:29 PM.
 July 10th, 2019, 06:27 PM #2 Global Moderator   Joined: May 2007 Posts: 6,820 Thanks: 722 Any finite set can have a bijection with itself. For example the elements form a finite list. Reverse the order to form another list. The two lists are in one to one correspondence. Thanks from topsquark
July 10th, 2019, 06:37 PM   #3
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Quote:
 Originally Posted by phytonguy The following sentence, a set is infinite if and only if the set have a bijection with itself, the negation of the affirmation is true? A set is finite if only if there doesn't exist any bijection with itself? The idea is what is a better definition of a bijection in this case. (Sorry, guys, I don't know English very well.)

The definition is: A set is infinite if and only if it has a bijection with a proper subset of itself. So for example the set of positive integers is infinite because it can be bijected with the even positive integers via the map $f(n) = 2n$.

Since this is a DEFINITION, it's automatically reversible. A set is finite if (and only if) there is NO bijection with a proper subset of itself.

For example there is certainly no bijection between a set of five elements and any set of 1, 2, 3, or 4 elements. By generalizing that idea, we can prove that each of the usual counting numbers 0, 1, 2, 3, ... are finite.

Quote:
 Originally Posted by phytonguy The idea is what is a better definition of a bijection in this case.
A bijection is a one-to-one correspondence. Likewise it's both injective (one-to-one) and surjective (onto). It means you can pair the elements of one set to the elements of another so that each element in one set corresponds to each element of the other set, and vice versa.

For completeness I'll mention that this definition of infinity, that a set has a bijection to a proper subset of itself, is technically called Dedekind-infinite.

There are other definitions of infinite sets that may or may not be equivalent to Dedekind-infinite sets, depending on which axioms of set theory you choose.

Last edited by Maschke; July 10th, 2019 at 06:40 PM.

July 10th, 2019, 06:41 PM   #4
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 Originally Posted by mathman Any finite set can have a bijection with itself. For example the elements form a finite list. Reverse the order to form another list. The two lists are in one to one correspondence.
Do infinite sets have bijections with themselves? If they did, would that make them finite?

July 11th, 2019, 01:46 PM   #5
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 Originally Posted by Maschke Do infinite sets have bijections with themselves? If they did, would that make them finite?
Yes, no.

Simple example: infinite set: $a_1,a_2,a_3,a_4....$ Rearrange to $a_2,a_1,a_4,a_3,...$ Bijection by matching by place in sequence.

July 11th, 2019, 02:35 PM   #6
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Quote:
 Originally Posted by mathman Yes, no. Simple example: infinite set: $a_1,a_2,a_3,a_4....$ Rearrange to $a_2,a_1,a_4,a_3,...$ Bijection by matching by place in sequence.
So every infinite set is order-isomorphic to the usual order on the natural numbers? How would this work for, say, an amorphous set?

July 12th, 2019, 12:18 PM   #7
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Quote:
 Originally Posted by mathman Yes, no. Simple example: infinite set: $a_1,a_2,a_3,a_4....$ Rearrange to $a_2,a_1,a_4,a_3,...$ Bijection by matching by place in sequence.
Do you think that perhaps the identity function would serve as a better example of a bijection that every set, no matter how weird or exotic, must have with itself?

And what is the relevance of that to the OP's question?

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