I considered deleting the post, but this seems more cowardly than just admitting I was wrong. But TIL something!
this post was submitted on 06 Jan 2024
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It's a similar problem. Both are infinity but one is a bigger infinity than the other.
The core reason why the infinities are different sized is different. The ways you prove it are different. It's kinda the first thing you learn when they start teaching you about different types of infinities.
All real numbers 0-1 is infinite, but all real numbers is equal to that infinity times the infinite set of real integers.
Logically this makes some sense, but this is fundamentally not how the math around this concept is built. Both of those infinities are the same size because a simple linear scaling operation lets you convert from one to the other, one-to-one.
The ∞ set between 0 and 1 never reaches 1 or 2 therefore the set of real numbers is valued more. You're limiting the value of the set because you're never exceeding a certain number in the count. But all real numbers will (eventually in the infinite) get past 1. Therefore it is higher value.
The example they're trying to say is there are more real numbers between 0 and 1 than there are integers counting 1,2,3... In that case the set between 0 and 1 is larger but since it never reaches 1 it has less value.
Infinity is a concept so you can't treat it like a direct value.
There is a function which, for each real number, gives you a unique number between 0 and 1. For example,
1/(1+e^x). This shows that there are no more numbers between 0 and 1 than there are real numbers. The formalisation of this fact is contained in the Cantor-Schröder-Bernstein theorem.There is a function which, for each real number, gives you a unique number between 0 and 1. For example,
1/(1+e^x). This shows that there are no more numbers between 0 and 1 than there are real numbers. The formalisation of this fact is contained in the Cantor-Schröder-Bernstein theorem.ah, but don't forget to prove that the cardinality of [0,1] is that same as that of (0,1) on the way!
This is pretty trivial if you know that the cardinality of (0, 1) is the same as that of R ;)
Isn't cardinality of [0, 1] = cardinality of {0, 1} + cardinality of (0, 1)? One part of the sum is finite thus doesn't contribute to the result
technically yes, but the proof would usually show that this works by constructing the bijection of [0,1] and (0,1) and then you'd say the cardinalities are the same by the Schröder-Berstein theorem, because the proof of the latter is likely not something you want to demonstrate every day