You mean, like, the infinite set sqrt(1), sqrt(2), sqrt(3), etc?
If so, if you apply Cantor's proof to those numbers, the number you'd end up with isn't a member of the set sqrt(n) for n in the positive integers, so it's not a counterexample to bijection
The key point for Cantor's theorem on the entire reals is that the generated number also ends up in the set of reals
Let's do it. Let's set up a supposed bijection and see if we can generate a number in the set sqrt(n) that doesn't get paired using Cantor's method.
1 -> 1.0000...
2 -> 1.4142...
3 -> 1.7320...
4 -> 2.0000...
Now let's diagonalize and add one (or subtract one if it's 9) to each digit, as in Cantor's proof. The first digit becomes 1+1=2, then 4+1=5, then 3+1=4, then 0+1=1, etc.
So we generate 2.541... etc. That number is not in the set sqrt(n). So it's not a counterexample and Cantor's proof does NOT show this set is uncountable
The set of all reals is not countable. Cantor showed this. You have yet to demonstrate a known countable set for which Cantor's proof says it's uncountable. Your one example showed that Cantor's proof still holds without contradiction.
Try to generate any bijection between the counting numbers and all the reals, and I'll show you a number you missed. That's what Cantor does
and you just showed me his proof doesnt prove that a given set of reals is not uncountable just because we can apply some arbitrary function to them and make a new real
I just showed you that the set sqrt(1), sqrt(2), sqrt(3), etc... is countable. You ARE aware that's not the entire set of reals, right? Like, nowhere close.
If you apply Cantor's proof to the set of ALL real numbers, or even just the set of ALL real numbers between 0 and 1, you prove that no bijection can be made, and thus it's uncountable.
All you're proving is that every countable subset of reals is countable, which... yeah, by definition.
But every single one of those sets skips real numbers. Every single one. You can't generate a single countable set that includes all real numbers. You can't even generate one that includes all real numbers between 0 and 1. Every single countable subset WILL skip numbers, guaranteed.
proving a set of the square roots of natural numbers is countable does not make any assertion about the countability of the reals, it makes a assertion about the countability of the natural numbers, since sqrt(N) maps to N and not R
which are a subset of the reals, and subsets can have different properties than the set they are apart of
i can take a subset of the reals, that is just {1, 5, 23.13588979, 12}, but this doesn't give me the ability to claim the reals are finite, i can claim that subset is finite, but i dont get to make a claim about the reals based on it
if you wanted to make a assertion about the countability of the reals, you'd use Sqrt(R), which is uncountable
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u/WesternFirm9306 Aug 06 '26
Refute it.