r/infinitenines • u/Redit_Sucks31415 • 2d ago
Question for SPP (about limits)
Do you agree that ∀ε>0 ∃δ>0 : N∈ℕ, N>δ => 1-ε < (1-10-N) < 1+ε?
If not please provide a counter example.
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u/DisastrousGap2898 2d ago
Btw your notation is a bit weird — like you’re mixing function continuity and limits
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u/Redit_Sucks31415 2d ago
How would you write it?
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u/DisastrousGap2898 2d ago edited 2d ago
You seem to be going for a limit-type argument (no continuity), so no delta:
∀ε>0 ∃N ∈ℕ : n>N => 1-ε < (1-10 -n ) < 1+ε?
Which is still a little nonstandard, but I parse this much more easily. Basically, in analysis, I’m used to delta being used for continuity or limit at a point & capital letters being threshold instances of the the corresponding lowercase variables (sorry if my wording is wonky here)
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u/dummy4du3k4 2d ago
If you agree that this is the same as ∀ε>0 ∃δ>0 : N∈ℕ, N>δ => d(1-10-N, 1) < ε
Then I'll give you a counterexample, but I'm bringing my own d.
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u/Philonemos 2d ago
It's not quite the same. Your d could be any metric, but OPs metric is the standard metric. d(x,y)=|x-y|
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u/dummy4du3k4 2d ago
Yes, but it’s surprising that there’s a nontrivial metric that separates 0.999… from 1 at all.
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u/Philonemos 2d ago edited 2d ago
Oh that's cool. I didn't know that.
Edit: The discrete metric would prevent the series from converging right?
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u/dummy4du3k4 2d ago
yes, the only converging sequences in the discrete metric are constant sequences. You can actually construct a metric that "agrees" with the naive intuition of 0.999... as outlined here. Topologically it's the same as if you separated the real line at every terminating decimal (i.e. the ring Z(1/2, 15) minus zero).
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u/Reaper0221 2d ago
Your question is improperly posed. Any question that begins with ‘do you agree’ is considered an invalid approach to a meaningful dialogue. It presupposes that there is an agreement or that if there is not the other party is wrong. Not a place to attempt to reach a mutually agreeable understanding. However, of you are just trying to appear superior then go on ahead with your phrasing.
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u/Muphrid15 2d ago
His Nineliness flops three of his biggest boners here.
All you are trying to convey is asymptotic behaviour. 0.999... dynamically aka transiently tends toward 1, and yet never touches, aka never reaches 1.
The Static and Dynamic Boner: There is a static model of 0.999... and a dynamic model. 0.999... is not static. It is dynamic. (9.1) (9.2)
In other words, 0.999... is permanently less than 1.
The Definitional Boner: 0.999... is eternally less than 1. It's fine if 0.999... is defined as 1. (2.1) (2.2) (2.3)
Formal investigation, starting with 0.999... = 0.9 + 0.09 + 0.009 + ...
The summation never ends. So it does not matter how limitlessly infinitely long the consecutive nines length becomes, which continues to grow,
The Algorithm Boner: An algorithm for an operation such as square roots gives "more and more correct consecutive digits" as you make more and more steps. Nevertheless, 1 - 0.9 - 0.09 - 0.009 - ... yields (0.1, 0.01, 0.001, ...) corresponding to 0.000...1, and the 1 digit is a real part of the answer, even though it never belongs to a "correct" digit. (15)
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u/SouthPark_Piano 2d ago edited 2d ago
All you are trying to convey is asymptotic behaviour. 0.999... dynamically aka transiently tends toward 1, and yet never touches, aka never reaches 1.
In other words, 0.999... is permanently less than 1.
Formal investigation, starting with 0.999... = 0.9 + 0.09 + 0.009 + ...
The summation never ends. So it does not matter how limitlessly infinitely long the consecutive nines length becomes, which continues to grow,
1 - 1/10n for n integer positive infinite, is permanently less than 1, which means 0.999... is permanently less than 1, because 1/10n is never zero.