# Does the “no gaps”-property with transcendental numbers mean that there is only “one number”?

I admit, the question is a little provocative, but is asked in all earnestness.

It is inspired by this one: Does .99999... = 1?

Many of the arguments here point to the fact that 0.999... and 1 are arbitrarily close together. Put differently this means that "there is no gap in-between the two" - which concludes that these numbers are just the same one number: 1.

In another field, concerning the transcendental numbers, it is proved that they are complete. So again put differently there are "no gaps in the number-line".

Interestingly enough this is not seen as a proof that all transcendental numbers are only just one number.

At the moment I can't find a good explanation why "no gaps" here means "is equal" and there it means "they are all different" (=complete). Could anybody please enlighten me?

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The word "gap" is being used in two different informal senses. That's all. The formal property you should be trying to understand is the least upper bound property. – Qiaochu Yuan Nov 8 '10 at 14:08
@Qiaochu: Can you please provide me with some links to understandable material on that? Thanks – vonjd Nov 8 '10 at 14:10
I'm sick of "provocative" titles, why not try for clarity instead. – anon Nov 8 '10 at 14:12
@muad: So you downvoted the question for the title, or what? Strange... – vonjd Nov 8 '10 at 14:18
@vonjd, more that the question just doesn't make sense - you talk about "gaps" a lot without seeming to care that it's not a well defined notion. – anon Nov 8 '10 at 14:24

The property of the real numbers which formalizes these informal statements about gaps is the least upper bound property. It states that any set of real numbers that has an upper bound has a least upper bound, i.e. a supremum.

The set $\{ .9, .99, .999, ... \}$ has an upper bound, so it has a least upper bound. The least upper bound cannot be less than $1$, so it must be exactly $1$. On the other hand, for any reasonable definition of $.999...$, this number is also an upper bound of the same set, and it is less than or equal to $1$. So the two are equal (because least upper bounds are unique). The "gap" which does not exist here is the nonexistent gap between two least upper bounds, and its nonexistence is a purely formal property given the existence of least upper bounds, so the nonexistence of this gap has nothing to do with completeness or the least upper bound property; it's the existence of the least upper bound that is important.

The least upper bound property implies that Cauchy sequences of real numbers converge, which is usually what people mean when they say that there are no "gaps" in the number line. Another way to formalize this statement is that the real line is connected, which implies, among other things, the intermediate value theorem. (Compare with the case of $\mathbb{Q}$, which is totally disconnected.)

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1/ To say that a topological set is "complete" has a mathematical meaning. For example, one definition of completeness is that all Cauchy-sequences converge.

A number is not complete (or at least I have never heard of such a definition for numbers).

We say that the real numbers are complete, because they satisfy the definition.

2/ When we have a sequence, we can (sometimes) see if this sequence converge or not. Again, the notion of convergence has a mathematical meaning, and suppose that the sequence obey a certain definition. When a sequence converges, we name an element "limit", this is an element that obey a certain definition too. It is straightforward from the definitions of convergence and of limit, that if a limit exists, it is unique.

For example we can study the sequence given by 0, 0.9, 0.99, 0.999, ... This sequence converges and has limit 1.

Or we can study another sequence $\frac{p_n}{q_n}$ given by the conditions $q_n = 10^n$ and $\left(\frac{p_n}{q_n}\right)^2 \leq 2 \leq \left(\frac{p_n+1}{q_n}\right)^2$. This sequence will also converge, and the limit will be $\sqrt{2}$.

Now, if observe that this last example is a sequence composed only by elements in $\mathbb{Q}$, we might be surprised to see that the limit is actually not in $\mathbb{Q}$. Mathematicians don't like the fact of not having the limit within the working topological set. Therefore they build $\mathbb{R}$ which is the set all (classes of equivalence of) Cauchy-sequences of elements in $\mathbb{Q}$. Hence $\mathbb{R}$ satisfy the definition of completeness.

Again, saying that $\mathbb{R}$ is complete is just a matter of definition. It has nothing to do with the word "gap". As far as I know, "gap" has no mathematical meaning.

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"Cauchy sequence" is not well-defined for an arbitrary topological space, even one that is metrizable. – Qiaochu Yuan Nov 8 '10 at 15:57