this post was submitted on 06 Jan 2024
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I considered deleting the post, but this seems more cowardly than just admitting I was wrong. But TIL something!

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[–] fishos@lemmy.world 2 points 10 months ago* (last edited 10 months ago) (4 children)

Not quite because it's easily shown that the set of all real numbers contains the set of all real numbers between 0-1, but the set of all real numbers from 0-1 does not contain the set of all real numbers. It's like taking a piece of an infinite pie: the slice may be infinite as well, but it's a "smaller" infinite than the whole pie.

This is more like two infinite hoses, but one has a higher pressure. Ones flowing faster than the other, but they're both flowing infinitely.

[–] lemmington_steele@lemmy.world 5 points 10 months ago (1 children)

actually you can for each real number you can exhaustively map a uninque number from the interval (0,1) onto it. (there are many such examples, you can find one way by playing around with the function tanx)

this means these two sets are of the same size by the mathematical definition of cardinality :)

[–] PotatoKat@lemmy.world -1 points 10 months ago (2 children)

You mean integers and real numbers between 0 and 1.

All real numbers would start at 0, 0.1, 0.001, 0.0001.... (a 1:1 match with the set between 0 and 1) all the way to 1, 1.1, 1.01.... Etc.

[–] lemmington_steele@lemmy.world 4 points 10 months ago

no, there aren't enough integers to map onto the interval (0,1).

probably the most famous proof for this is Cantor's diagonalisation argument. though as it usually shows how the cardinality of the naturals is small than this interval, you'll also need to prove that the cardinality of the integers is the same as that of the naturals too (which is usually seen when you go about constructing the set of integers to begin with)

[–] sukhmel@programming.dev 3 points 10 months ago

No, ey mean real numbers and real numbers. Any interval of real numbers will have enough numbers to be equivalent to any other (infinite ones included)

[–] Pipoca@lemmy.world 5 points 10 months ago

That's not really how counting infinite sets works.

Suppose you have the set {1,2,3} and another set {2,4,6}. We say that both sets are of equal cardinality because you can map each element in the first set to a unique element in the second set (the mapping is "one to one"/injective), and every element has something mapped to it (the mapping is onto/surjective).

Compare the number of integers to the number of even integers. While it intuitively seems like there should be more integers than even integers, that's not actually the case. If you map 1 to 2, 2 to 4, 3 to 6, 4 to 8, ..., n to 2n, then you'll see both sets actually have the same number of things in them because that mapping is one to one and onto.

There's similarly the same number of real numbers as numbers between 0 and 1.

But there's more numbers between 0 and 1 than there are integers.

[–] WilloftheWest@feddit.uk 1 points 10 months ago* (last edited 10 months ago)

Actually, the commenter is exactly right. The real line does contain the open interval (0,1). The open interval (0,1) has the exact same cardinality as the real numbers.

An easy map that uniquely maps a real number to a number of the interval (0,1) is the function mapping x to arctan(x)/π + 1/2. The existence of a bijection proves that the sets have the same size, despite one wholly containing the other.

The comment, like the meme, plays on the difference between common intuition and mathematical intuition.