Bijection between two infinite sets

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Offline marcm200

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« on: April 20, 2019, 07:34:51 AM »
If a bijection exists between two sets A and B, then they have the same cardinality.

I'm wondering whether the reverse is also true: If two sets have the same cardinality, then there exists a bijection.

I am trying to construct one in the case of two uncountable intervals of the real numbers: The fully-cosed one [0..1] and the half-open [0..1). I think, if it's possible at all, splitting up the intervals into the true real numbers (R-Q) and the rationals might allow me to use the trick for Hilbert's hotel (moving every guest one room number up, freeing room 1) to put the additional 1 somewhere. But I haven't figured that out yet.

Does the notion of an "almost bijection" exist? Being a bijection except for a finite number of arguments.


Offline marcm200

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« Reply #2 on: April 22, 2019, 10:43:38 AM »
Thanks for the link. That's quite an interesting approach - using an infinite countable sum.

My splitting approach worked. Using a couple of bijections in both directions, Hilbert's hotel and Cantor's diagonalization scheme, I was able to get a bijection.

But it was a lot of effort for just one more number to add. I wonder how difficult it would be to, say, put an uncountable number of objects more into an even bigger set (say, the power set of the real numbers). But I won't try that.

Offline gerrit

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« Reply #3 on: April 24, 2019, 05:42:57 AM »
I wonder how difficult it would be to, say, put an uncountable number of objects more into an even bigger set (say, the power set of the real numbers).
That's easy. Take the set S of all subsets of [0 1] with more than 1 element. Then add the set of all subsets of [0 1] with just 1 element.

Offline marcm200

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« Reply #4 on: April 24, 2019, 09:20:28 AM »
That's a clever way to get the final set. Is it possible to explicitly write down a bijection from your set S to the complete power set of the real numbers? (Both are cardinality aleph2, if I recall correctly).

Offline gerrit

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« Reply #5 on: April 24, 2019, 09:03:51 PM »
That's a clever way to get the final set. Is it possible to explicitly write down a bijection from your set S to the complete power set of the real numbers? (Both are cardinality aleph2, if I recall correctly).
No idea.

Cardinality of the real numbers is \( \mathcal{C}=2^{\aleph_0} \) and of S it's \( 2^{\mathcal{C}} \).
\( \mathcal{C} > \aleph_0 \) but it could be anything like \( \mathcal{C} = \aleph_{316} \) or
Search for "continuum hypothesis" for more on that.


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