0.999... cannot be defined as the largest number less than 1.
There is someone on this thread that is trying to peddle the nonsense that 0.999... is defined as the largest number less than 1.
Can we define a number by some arbitrary property? Yes, absolutely.
For example we define the number sqrt(2) as the number x>0 such that x^2=2.
We define the golden ratio ψ=a/b for two numbers a>b>0 if a/b=(a+b)/a. This is also equivalent to the solution to the quadratic equation ψ^2 = ψ+1.
Both of these numbers are roots of quadratic equations. Can we define numbers that aren’t roots of equations? Yes, the transcendentals like pi and e are.
As everybody knows pi is defined as the ratio circumference/diameter.
e is defined in at least two ways: the formula for the annual interest where the rate r is compounded n periods per year is (1+r/n)^n, then e^r= lim_{n→∞}(1+r/n)^n. Alternatively f(x)=e^x can be defined as the function that equals its own derivative f(x)=f’(x). In both cases e=lim_{n→∞} Σ_{k=1}^n 1/k!.
pi solves the integral pi=int_{-1,1}(1-x^2)^(-1/2)dx and can be expressed as a variety of infinite sums.
Both are transcendental numbers that take their value from the way the number is literally defined.
So can we define a number x as the largest number less than 1?
Firstly, this is not a solution to some polynomial equation, nor is it defined in some sensible fashion like pi or e.
But ignoring this insurmountable problem, we run into the issue that there simply is no number that can possible satisfy that definition. There is a trivial proof by contradiction that has been made by many of this sub (but some with learning difficulties refuse to accept the proof and keep peddling disingenuous nonsense). If x purports to be the largest number less than y the then x<y and for all numbers z<y we must have x≥z. Since x<y it is the case that x<(x+y)/2<y directly contradicting the claim that x is weakly greater than all numbers less than y. There is no way to even define a largest number less than another. (Archimedes drops his mic.)
Why? Because the set of numbers z<y is open (or half open depending on the LHS of the interval). There are always numbers between any z and y.
A number u is an upper bound of a set S if u≥s for all s∈S. The least upper bound is the smallest of the upper bounds (if it exists). If x was the largest number less than one it would be the least upper bound for the set of all numbers less than one. However the least upper bound of that set is one itself. One is an upper bound of all numbers less than one. Then the proof we just did for some x<1 as a candidate upper bound proves that one is the least upper bound.
Of course the fact that 0.999...=1 and 1 is the least upper bound of all numbers less than one mean that 0.999 is the LUB. However even so it's not the largest number strictly less than 1 for obvious reasons.
