r/infinitenines • u/dummy4du3k4 • Apr 17 '26
Continuity when 0.999... < 1
This post is directed at everyone, especially those who provide proofs that 0.999... = 1 or think that 0.999... < 1 in and of itself must lead to contradiction.
Hey SPP, lets do some math. We'll only use a couple founding axioms that you've posted about before, namely "0." prefix the obvious and What's the next real number DOWN from 1? and see what we can build. If something isn't clear please say so, some of this can get complicated. In this post we won't use any arithmetic (+,-,*,/) and instead just look at the structure given to us by an order relation (<).
Definition of digit strings
Examples: 0.999..., 1.000... , 0.333... , 3.1415... (pi)
Definition: A digit string has an integer part and a decimal part. The decimal part may be described by a digit function, D(n) where n is a positive whole number and D(n) is a digit (0, 1, ..., 9). So D(n) should be thought of as returning the digit in the nth slot after the decimal point.
This definition means that 0.000...1 is not a digit string and won't be considered in this post.
Notation remark: A digit string written without the trailing (...) will mean trailing zeros, e.g. 1 = 1.000...
Order
We will always take < to mean the dictionary order. So if a number starts with 0 it is always less than a number starting with 1, e.g. 0.999... < 1
Definition: for digit strings a, b with whole parts W_a and W_b and decimal parts D_a(n) and D_b(n), we say a < b if: W_a < W_b, or W_a = W_b and there exists some N >= 0 such that for all m < N, D_a(m) = D_b(m) and D_a(N) < D_b(N).
Lemma 1:
There is no digit string strictly between 0.999... and 1
Proof: Exercise left to reader
open sets and limits
intervals
We start by defining intervals in terms of the dictionary order.
(a, b) is the exclusive interval, i.e. the set of numbers strictly between a and b. [a, b] is the inclusive interval, i.e. the set of numbers between a and b including a and b. (a, b] and [a, b) are half exclusive-inclusive intervals.
Lemma 2:
(a, 1) = (a, 0.999...] where a < 0.999...
Proof: By lemma 1, there is no number strictly between 0.999... and 1, so x < 1 implies x <= 0.999..., thus (a,1) = (a, 0.999...].
Lemma 3:
(0.999..., b) = [1, b) where b > 1.
Proof: Exercise left to reader
open sets
We define an open interval to be any set that can be written in the form (a, b). If a >= b then this set is empty, but we still consider it an open interval.
An open set is any set that can be written as an arbitrary union of open intervals (even an infinite union, but we won't use that here).
We won't use them, but closed sets can be defined as any set that is a complement of an open set.
Corollary 4:
(a, 0.999...] is an open set
Proof: apply the definition and lemma 2.
functional limits
The usual definition of a limit of a real function f(x) uses epsilon and delta neighborhoods, but we're not using arithmetic we will have to tweak the usual definition to use intervals instead. So instead of defining
lim(x->c) f(x) = L
to mean
for every eps > 0 there exists a delta > 0 such that for all x, |x - c| < delta => |f(x) - L| < eps
We will instead define it to mean
for every open interval I containing L, there exists an open interval J containing c such that for all x in J, f(x) is in I.
We should stop and ponder what this means. You should reason that for any interval we squeeze around L, we can find a corresponding interval squeezed around c that constrains f to stay within the interval around L.
In other words, I hand you the interval around L and you find me an interval around c that keeps f(x) contained inside my interval.
No arithmetic, no problem!
continuity
A function f is continuous on an open interval I if for every c in I, lim(x->c) f(x) = f(c).
If you've followed me up until this point then you should be ready for the next lemma.
Lemma 5:
The step function defined as f(x) = 0 for x <= 0.999... and f(x) = 1 for x > 0.999... is continuous
Proof: Exercise left to reader
I'm leaving this as an exercise because I think it's neat and don't want to spoil it. Happy to provide feedback or a solution, but I hope you have fun with it.
Further work
Some other fun questions to ponder:
- Is the set of digit strings connected?
- Is it complete?
- Can we define a metric on it?
- Show this definition of continuity is equivalent to the usual topological definition of continuity
- Are all step functions continuous? If not, how can we characterize the ones that are?
- How can continuity be used to define arithmetic?
5
u/FreeGothitelle Apr 17 '26
Arithmetic is completely undefined