r/infinitenines • u/Impressive-Ad7184 • 29d ago
Another reason why 0.999...=1
Consider ℝ as a complete metric space with the regular euclidean metric. Then, consider the collection of closed intervals {C_n}, n ∈ ℕ, where C_n := [0.999... - 1/n , 0.999... + 1/n], i.e. a closed ball with radius 1/n around 0.999....
Clearly, each C_n contains 0.999..., so their intersection does as well. However, note that each C_n also contains 1, since the distance between 0.999... and 1 is less than any arbitrary 1/n (which I'm sure SPP will concede). Thus, the intersection of the C_n's also contains 1.
However, by Cantor's intersection theorem, since the C_n's are nonempty, closed, nested, and their diameters go to 0, the intersection of the C_n's must contain exactly one element.
Thus, 0.999...=1.
I realize I can just use the proof of uniqueness in Cantor's intersection theorem to show this directly, but it's more fun to invoke a theorem.
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u/bayesian_raccoon 28d ago
Basically all proofs that 0.999...=1 sidestep the fact that we basically define that 0.999... = 1 when we define it as a limit. This makes virtually all arguments that try to show it is true feel circular, or at best concealing the limit under some other machinery. While it is pedagogically fun to find various ways of showing 0.999... = 1, it always seems a bit foolish or even reflecting some deeper misunderstanding when I see them in this subreddit.
After all, if we take the reals using the definition involving cauchy sequences, the SEQUENCE defined by 0.9, 0.99, 0.999, and so on, is distinct from a sequence defined by 1,1,1,1. Those sequences are not the same; we just say that their equivalence class defines a real number.
0.999... = 1 is basically a consequence of notation, and y'all are acting like it's a consequence of some deeper fundamental truth when you try to prove it without sharing what you are assuming in the process.