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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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.
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.
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 :)
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.
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)
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)
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.
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