Well, there are more integers than naturals, yet both share the same cardinality. Also, I thing hilbert’s hotel problem shows that rationals and naturals also share the same cartinality, somehow. You could arrange every rational in a line like the naturals and the integers.
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BarbedDentalFloss@lemmy.dbzer0.com 2 months ago
There are more rational numbers than natural numbers. There are more real numbers than rational numbers.
Checkmate meme.
TheGuyTM3@lemmy.ml 2 months ago
borokov@lemmy.world 2 months ago
That’s not how cardinality works when dealing with infinite. For ex, there are the same number of prime number than number of integer. Yes, there are many non prime inter between 2 prime integer, but as long as you can “count” them, they have the same cardinality, which is called “aleph 0”.
But you cannot “count” real number. There are actually more real between 0 and 1 than there are interger. This value is called “aleph 1”.
Yes, there is also aleph 2, aleph 3,… There is not a single “infinite”, but there are several one that don’t have the same size.
Have a look to Hilbert’s hotel paradox en.wikipedia.org/…/Hilbert's_paradox_of_the_Grand…
procrastitron@lemmy.world 2 months ago
The problem is that rational numbers can be mapped to the integers (e.g. just encode each rational number as an integer), so there are not more rational numbers than integers.
BarbedDentalFloss@lemmy.dbzer0.com 2 months ago
No that’s not true. There are rational numbers in between the integers. Therefore the mapping between integers to rational numbers is injective and thus there are more rational numbers than integers.
berber@feddit.org 2 months ago
“the” mapping? there is no “the” mapping.
you are talking about the canonical inclusion mapping 1 in N to 1 in Z (restriction of the canonical inclusion of rings of integers Z into any other ring, Z is an initial object), which can be seen as a non-generic canonical mapping of semigroups.
but as sets, there is no inherent structure, there are injection, surjections, and of course bijections in both directions.
the only way one can call one set “bigger” is in the very strict sense of sets, N being a true subset of Q. however, this assumes N to be an actual subset of Q, which is a matter of definition and construction. so we say there is some embedding included, which is the same as (re)defining N as that embedded subset, so we are at your canonical inclusion of semigroups again. if you view this as inherent to N and Q, then there are “more” elements in Q as in N, but not in terms of cardinality.
BarbedDentalFloss@lemmy.dbzer0.com 2 months ago
transfinite hocuspocus bullshit is what it is