Does it make any sense to prove $0.999\ldots=1$? I have read this post which contains many proofs of $0.999\ldots=1$. 

Background

The main motivation of the question was philosophical and not mathematical. If you read the next section of the post then you will see that I have asked for a "meaning" of the symbol $0.999\ldots$ other than defining it to be $1$. Now here is a epistemological problem and this is mainly the problem from which the question arose. Suppose you know that $1$ is a real number. Now I give you a symbol, say $0.999\ldots$ which from now on I will denote as $x$. Now I ask you whether $x$ is a real number. To answer this, if you define $x=1$ then you are already attributing the properties of $1$ to $x$ among which one is it being a real number without proving whether to $x$ we can indeed attribute the properties of $1$. 

A common response to this question has been to define the symbol $x$ as the limit of the sequence $\left(\displaystyle\sum_{i=1}^n \dfrac{9}{10^i}\right)_{n\in\mathbb{N}}$ and then prove that the limit of this sum is indeed $1$. But again the problem is that you are defining the symbol $x$ to be a real number and hence are assuming a priori that the symbol $x$ denotes a real number.

As per the discussion that has been conducted with Simply Beautiful Art let me state my position again in brief,


  Also let me say that I do not disallow $0.999…$ to be a real number. My impression that if you assume $0.999…$ to be a real number then there is no sense in proving that $0.999…$ is indeed equal to $1$ because either you define it to be $1$ or you prove the equality as a theorem. But if you are going to use the limit definition of $0.999…$ then what you are a priori assuming it to be a real number which is an assumption that I don't allow. What can be allowed is that $0.999…$ is a number (but not necessarily a real number). 


Question

My question is, 


  Does it make any sense to prove this equality? 


Can one give any "meaning" of the symbol $0.999\ldots$ other than defining it to be $1$?
We have to agree about what the symbols

$$
0.99999\dots
$$

are supposed to mean. The symbols capture an intuitive idea, but it doesn't have meaning unless we agree on what that meaning is. When you write these symbols down, everyone will agree that what you mean is the following:

$$
.9 + .09 + .009 + \dots = \frac{9}{10} + \frac{9}{100} + \frac{9}{100} + \dots = \sum_{i = 1}^\infty \frac{9}{10^i}
$$

There is no proof of that -- it is an agreement.

If you wrote this, perhaps similar-looking, string of symbols down

$$
\begin{aligned}
0.00000\dots1\\
\end{aligned}
$$

then there is no agreement on what you mean. You would seem to be talking about a real number which is smaller than all other real numbers -- an object that doesn't exist.

So what string of symbols means what, rigorously, is a matter of agreement. Usually we elide that fact when it seems intuitive what we mean, but we do not all share the same intuition (or the same knowledge of how to express that intuition), so we must drag it out on occasions like now.

Anyway, the content of a proof of $0.9999\dots = 1$ is not that we must agree to define $0.9999\dots$ as $1$. The content is to define $0.9999\dots$ as the sum above, then by deduction show this sum is equal to $1$.
We don't simply define $0.9999.... $ to be equal to 1.

We do a lot of background analysis first which involves constructing the real numbers out of the rational numbers.

In very short summary:

Consider this basic fact about rational numbers:  no matter how close together two rational numbers are we can find a rational number between them (which also means for any value greater than zero, no matter how small, you can find a smaller one between it and zero). 

This is nice.  It means we can get as close as we like to any value using rational numbers.

Now consider this monkey wrench:  There are many values we can't express (for example: pi and the square root of 2).  [This is weird because we can get as close as we like to to these values by the statement above.]

And consider this horrible result:  Between any two rationals no matter how close we get, there is always one of the inexpressible holes between them!

So how can we think of these ... irrational values...

Well, a lot of subtle analysis and picayune debating we notice that in the rational numbers we can have an infinite set of rationals, where all the rationals are within a range, but the set needn't have a biggest value.  For example: {all the rational numbers less than pi}; all these values are less than 3 1/3 so it's bounded, but there is no precise rational upper bound to this set.

(Another such set is all the numbers $.9, .99, .999, .9999$ etc.  It's infinite but each term is rational.  It has no biggest value.  And all values are less than 1.)

So the issue was can we come up with a bigger system of numbers than the rationals, where every set will have a precise upper bound? Mathematicians decided they could and it was called the Real numbers[1].

Now here is the definition of the real numbers and how the were created (it's very subtle):  Every real number is the least upper bound limit of a bounded set of rational numbers.  And every bounded set of rational numbers has a real number as an upper bound.

That's the definition of the real numbers.

So {.9, .99, .999, .9999,.....} is an bounded infinite set of rational numbers.  By definition it has a real number least upper bound limit.  We call that real number .999999999999......

Okay, we have defined .99999...... and we DIDN'T define it as 1.  So now we can prove that it does equal 1.

The general idea is that if .99999......= c &lt; 1  then can find a finite number in {.9, .99, .999, .9999,.....}  that is bigger than c.  So we were wrong about c being an upper bound of {.9, .99, .999, .9999,.....}  So .9999....... $\ge$ 1.  But 1 is bigger than any of the {.9, .99, .999, .9999,.....}. So .99999.......  $\not &gt;$ 1.  The only consistent option left is .99999.... = 1.

[1] Well, the didn't just wave their hands and declare it.  They had to prove constructing such a number system was possible.  The proof is ... abstract.  And tedious.  And abstract.

===== 2nd answer: Different approach and philosophy====

The OP has a point that we don't really "need" to prove .999.... = 1.  If .999... equals anything at all, showing that it must by 1 is trivially easy.  The idea of proving .999.... = 1 misses the point.  The point really is how do we know that .999.... equals anything at all.

The OP isn't entirely accurate in stating we defined .999... to be 1.  We defined .999.... to be something that can be shown to be 1.  (Slight difference but a difference with ramifications.)  Other comments claim we make an assumption via definitions that .999... is a limit of bounded sequence of rational numbers and another assumption via definition that limits of bounded sequence of rational numbers are real numbers.

These aren't entirely accurate either.  The definitions weren't assumptions.  They were analysis as to what the real numbers are and the discoveries came along the way.

Suppose we know nothing about the real numbers.  ... well,  not nothing but suppose we have no real sense of them.

With or without real numbers we do have to define $ 0.9999....$ as $ \sum_{n = 1}^{\infty} \frac{9}{10^n}$.  But we don't really know anything about it.  It's possible that it's unbounded and "blows up".  (Actually, we can show that can't happen but I don't want to get into that yet.)  It's possible that it adds up to an actual number.  It could be rational or it could be one of those numbers that can't be written as a rational like square root of two or pi.  

But the natural concern is that it might simply never resolve into anything.  

To me, the surprising and subtle thing about analyzing the real numbers is the realization that the very nature of the real numbers means that this "simply never resolving into anything" is impossible.  Everything in the reals that is bounded resolves to something, and every real is a resolution of something.  This really astounded me when I finally wrapped my head ahead.  I mean it really really surprised me!  (Yes, I'm that much a geek.)

Okay, so this leads to the fundamental question:  Real Numbers.  What the Heck are they?

We know that we can measure discrete units via integers, like $n$.  And we know we can chop these discrete units into $m$ pieces as $1/m$ and as $m$ can be arbitrarily large, these $1/m$ can be arbitrarily precise and this collection of ratios, so all the possible $n/m; m \ne 0$ form a system of arbitrarily precise measurements that span every possible range.  So these Rationals can measure anything to arbitrary precision.

That ought to be enough.

But it isn't.  Because we know there are always numbers like $\sqrt{2}$ and $\pi$ that can't be written as any Rational = $m/n$ where $m$ and $n$ are integers.

So w have gone for a thousand years with a number system that we don't really understand and cant describe.  What are these "holes" and what do we know about them?

Enter the Dedekind cut.

An example:  Let's "cut" the rationals into two sets.  Set $A = ${all the rational numbers, $r$, such that $r^2 &lt; 2$} and $B = ${all the rational numbers, $s$, such that $s^2 &lt; 2$}.  Several things we note about these sets:

1) they are completely disjoint and neither are empty; 2) all the elements or 1 are smaller than all the elements of the other; 3) A does not have a largest element and B does not have a smallest element; and 4) They are arbitrarily close together.  (That is, for any number $e &gt; 0$ no matter how small we can always find an $a \in A$ and a $b \in B$ so that $b - a &lt; e$.  Always.)

Any "procedure" that can cut the Rational numbers into two sets with those 4 properties is called a "cut" and we can, for sake of vocabulary, refer to such cut as $\overline x$ and the two sets as $A\overline x$ and $B\overline x$. 

And we can consider the collection of all such cuttings. One thing is about these cuts is that we don't necessarily need to know how to describe them for them to exist.

One such cut could be $\overline \omega$ such that $A\overline \omega = $ {all rationals less than some .99...9} and $B\overline \omega = $ {all rationals larger than all .99...9}.  [That's not actually true and I pulled a fast one on you.  A brownie point to any one who can figure out my deception.]

Now the astute reader might have noticed that although these cuts cut the rational in two, the two sets don't always contain all the rationals.  Ex: A = {all rationals &lt; 27 1/2} and B = {all rationals > 27 1/2} is a cut.  But the rational number 27 1/2 is not in either A nor B.  This is okay.  The sets don't have to contain all the rationals.  But notice, 27 1/2 is the only rational that is not in either set. 

We can say a cut like that one cuts the rationals "on" a rational number and others cut the rationals "between" rational numbers.  There is a correspondence between the rational numbers and the cuts that cut on the rational numbers and we can refer to such a cut as $\overline{1/2}$ as the cut that cuts on 1/2.  

The other cuts that cut between rational numbers are extra cuts.

So... back to the collection of all possible cuts...

They form a number system.

For two cuts $\overline x$ and $\overline y$ we say $\overline x &lt; \overline y$ if $A\overline x \subset A\overline y$.  We note that if we define $\alpha$ = $A\overline x + A\overline y = $ {all rationals that are sums of two rationals one in $A\overline x$ and the other in $A\overline y$} and $\beta$ = {all the rationals that are bigger than all the rationals in $\alpha$}, then $\alpha$ and $\beta$ perform a cut.

[Again, I'm pulling a fast one.  Another brownie point to whoever finds it.  It's the same as the fast one I pulled above.]

We define $\overline x + \overline y$ as finding that type of cut.

Similarly we define subtraction, multiplication, and division.  (Divisions kind of a pain to define but we do it.)

So the collection of all cuts form a number system.  What's more, the cuts that cut "on" rational numbers behave equivalently as the rational numbers do.

These "cuts" are the real numbers.  Every real number is a point at which we "cut" the rational numbers into two sets.  If we cut "on" a rational number that point is the rational number, if we cut "between" rational numbers that point is some irrational number.

That's it.  That's what the heck Real Numbers are.

... deep sigh and coffee break....

Okay, WHAT THE #@&amp;! DOES THAT HAVE TO DO WITH .9999....?

Notice that we now have discovered (NOT "defined"; NOT "assumed") that every real number cuts the rationals in two.  So every bounded sequence of rationals can be fit into the $A\overline x$ set of some cut $\overline x$ and we can find precisely the lowest cut that will do that.  And that cut point is a precise and unique real number.

Thus the FUNDAMENTAL PROPERTY OF REAL NUMBERS!!!!  Every real number is a limit of a sequence of rational numbers and every bounded sequence of rational numbers has a real number limit.  And this real number is the point that cuts the rational numbers at precisely the "edge" of the sequence.

So let's look at $.9999...... = \sum_{n = 1}^{\infty} \frac{9}{10^n}$ and let's look at this bounded sequence of rational numbers: {.9; .99; .999; etc.} and let's look at the number $1$ which is larger than all the elements of the sequence.

So we know:

1)There is a cut $\overline {\omega}$ where $A\overline{\omega}$ = {all the rationals that are less than or equal to sum .999...9}

2)This cut occurs at a real number, $\omega$.

3)$\omega = \lim $ of the sequence {.9;.99;.999; etc.}

4).999.... "lingers" between the elements of $A\overline {\omega}$ and the elements of $B\overline{\omega}$.

Therefore we can conclude $.999.... = \omega$.

And that's it:

$.9999..... = \sum 9/10^n = \lim \{.9'.99;.999;...\}$ is a real number.

Now it's trivial to prove $.99999..... = 1$.

And there was not a single assumption or circular definition.