The question was inspired by this excellent blog post re: the equation 0.999... = 1.

I see a connection between:

· Zeno’s Dichotomy (1/2 + ¼ + 1/8 + … = 1)

· 0.999… = 1

· The derivative as a limit (i.e., ever approaching a value, never reaching it, after all “at” a point you get 0/0)

Enter the epsilon-delta definition of a limit:

· But this doesn’t resolve any of the above “paradoxes”

· Rather, it specifies what we mean by “limit” and “convergence”

· Absent is an explanation for why we are justified in equating the sum of an infinite series with its limit. For example, you often read that calculus “solves” Zeno’s Dichotomy paradox by defining it away: the truth that 1/2 + ¼ + 1/8 + … = 1 follows from our definition of limit

The blog post made me wonder: does the Archimedean property help provide us with the missing explanation?

· If we have found the limit of the function in question, that means we have won the “epsilon-delta” game (for every epsilon, we can find the appropriate delta and so forth)

· Then the difference between the limit and the function can be shown to be less than any given epsilon > 0

· Borrowing a line from the blog post, another way to say that is the difference between the function and the limit is some positive number that is less than or equal to 1/10^k for every k

· But by Archimedes property, this is impossible – no such number can exist

· The function and the limit must therefore be the same number

What’s your reaction to the above line of reasoning? To me, it seems to provide the reason why we’re able to equate an infinite series with the limit to which it converges.

  • $\begingroup$ ... that, if we go back further, the answer lies entirely in the axiom of completeness of the real numbers. $\endgroup$ – Nameless Apr 8 at 21:58
  • $\begingroup$ Related Is-it-true that $0.9999999\dots = 1$? $\endgroup$ – Brian Apr 8 at 22:17
  • $\begingroup$ @Nameless Interestingly, it depends on the type of completeness you assume. If you have an ordered field with Dedekind completeness (the supremum axiom), then the field is uniquely determined to be the reals. If you assume Cauchy completeness (every Cauchy sequence converges), it's not enough! You need to also assume the Archimedean property to recover the reals, precisely to stop a failure of Zeno's paradox from occurring! $\endgroup$ – Theo Bendit Apr 8 at 22:21
  • $\begingroup$ "Absent is an explanation for why we are justified in equating the sum of an infinite series with its limit." Absent is a meaningful definition of summing an infinitude of numbers. It is, in a sense, "defining the problem away", but given that there was no definition to the problem to start with, properly defining the problem was always going to have to be the first step in solving it. So, we might as well define it to be something useful. $\endgroup$ – Theo Bendit Apr 8 at 22:25
  • $\begingroup$ @TheoBendit yes, it's true. I was referring to the first kind of completeness you cited, which is equivalent to upper/lower bound lemma, to nested interval lemma, to finite covering lemma and to (limit point lemma + Archimedean property). $\endgroup$ – Nameless Apr 9 at 7:22

Im going to restate your last 3 bullet points in the way I like to think about this. Overall I agree with what you're saying.

I will use the Zeno Dichotomy for concreteness.

When we say $\sum_{k=1}^{\infty} \frac{1}{2^k } = 1$, we are saying 2 things mathematically.

(1) The infinite sum never exceeds that value, so $$ \sum_{k=1}^{\infty} \frac{1}{2^k } \leq 1 $$

(2) At the same time, it's never less than that value. No matter how hard we try to separate them, we cannot: $$ \mbox{Given any number } x \mbox{ less than 1, no matter how close to 1, we have } $$ $$ x < \sum_{k=1}^{\infty} \frac{1}{2^k } \leq 1 $$

So no matter how we try to view the infinite sum as a separate number than what it converges to, we can't.

This is the same line of reasoning is the squeeze theorem, which you probably heard of or if not can easily find on Wikipedia

  • $\begingroup$ I like this and not just because you agree with me! The squeeze perspective strikes me as equally valid. Thank you. $\endgroup$ – drzaius7 Apr 8 at 23:31

These ancient “paradoxes” are paradoxes for the same reason as to why there are so many unresolved philosophical disputes: they are open to different interpretations. Modern real analysis puts an end to that.

Sure, our modern definitions of real numbers, limit and continuity etc. does not, strictly speaking, solve those old paradoxes. To say that $\lim_{n\to\infty} a_n=L$, these wise people 2000 years ago certainly didn't mean that for any given $\varepsilon>0,$ they can find $N\in\Bbb N$ such that $$ n\ge N \implies |a_n - L|<\varepsilon. $$ What these definitions does is laying down an agreement on what the symbol $$ L = \frac 12 + \frac 14 +\frac 18+\dots $$ means. Only after that we can begin to contemplate its value as a real number.

That being said, is this really it? Is there any gap between our modern definitions, e.g. the $\varepsilon$-$\delta$ definition of limit, and, says, Zeno paradox that ought to be filled with the Archimedian property? Philosophically speaking, you can think of it that way if it helps forming your intuition of how real numbers work. However, please be reminded that the modern incarnation of the Archimedian property is a result of the Axiom of Completeness of $\Bbb R$ (or the Least Upper Bound Property), so the problem doesn't go away completely but shifted to "What is $\Bbb R$?" instead (which is very different from the ancient view of the real line, whatever that means).


Your question seems to boil down to this observation: if $|x|<\epsilon$ for all $\epsilon>0$, then $x=0$. This is true. However, you say "The function and the limit must therefore be the same number" which is vague - the function where? The limit of what?

Say we have a function $f(x)$, a point $a$, and $\lim_{x \to a}f(x)=L$. This means for all $\epsilon>0$, there exists $\delta>0$ so that if $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$. But this does not mean that $|f(x)-L|<\epsilon$ for all $\epsilon$. The value of $x$ depends on $\delta$, which in turn depends on $\epsilon$. In general, $\delta$ gets smaller as $\epsilon$ gets smaller, so there is no value of $x$ that satisfies $|x-a|<\delta$ for each $\delta$ (corresponding to each $\epsilon$). So it does not make sense to say "the function and the limit must be the same number," since $f(a)$ need not equal $L$, and $f(x)$ need not equal $L$ for any value of $x$.

Similarly, if we have a sequence $\{a_n\}$ and $\lim_{n\to \infty} a_n=L$, this means that for all $\epsilon>0$ there exists $N$ such that for all $n>N$, $|a_n-L|<\epsilon$. But again this does not mean that $|a_n-L|<\epsilon$ for all $\epsilon$, since the values of $n$ that "work" depend on the chosen $\epsilon$, and in general $N$ will get larger as $\epsilon$ gets smaller. Once again, there need not be a value of $n$ that works for all $\epsilon$, and there need not be any $a_n$ that actually equals $L$.

  • $\begingroup$ I'm saying that for every epsilon > 0 we can find the delta such that the difference between the function and its limit is less than the epsilon we chose. That is, the difference can be made smaller than any epsilon > 0. $\endgroup$ – drzaius7 Apr 8 at 23:30
  • $\begingroup$ The "difference between the function and the limit" doesn't make sense unless you are specifying where you are evaluating the function. It's true that the difference between some $f(x)$ and the limit can be made less than any chosen $\epsilon$, but the value of $x$ changes dependent on $\epsilon$. We could write $x_\epsilon$ to emphasize this dependence. So we don't have $|f(x)-L|<\epsilon$ for all $\epsilon$, rather $|f(x_\epsilon)-L|<\epsilon$ for all $\epsilon$. From this inequality we cannot conclude that anything is equal to $L$. $\endgroup$ – kccu Apr 9 at 15:17

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