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Fri, 01 Jan 2021

I made a big mistake in a Math Stack Exchange answer this week. It turned out that I believed something that was completely wrong.

Here's the question, are terminating decimals dense in the reals?. It asks if the terminating decimals (that is, the rational numbers of the form !!\frac m{10^n}!!) are dense in the reals. “Dense in the reals” means that if an adversary names a real number !!r!! and a small distance !!d!!, and challenges us to find a terminating decimal !!t!! that is closer than !!d!! to point !!r!!, we can always do it. For example, is there a terminating decimal !!t!! that is within !!0.0000001!! of !!\sqrt 2!!? There is: !!\frac{14142135}{10^7} = 1.4142135!! is closer than that; the difference is less than !!0.00000007!!.

The answer to the question is ‘yes’ and the example shows why: every real number has a decimal expansion, and if you truncate that expansion far enough out, you get a terminating decimal that is as close as you like to the original number. This is the obvious and straightforward way to prove it, and it's just what the top-scoring answer did.

I thought I'd go another way, though. I said that it's enough to show that for any two terminating decimals, !!a!! and !!b!!, there is another one that lies between them. I remember my grandfather telling me long ago that this was a sufficient condition for a set to be dense in the reals, and I believed him. But it isn't sufficient, as Noah Schweber kindly pointed out.

(It is, of course, necessary, since if !!S!! is a subset of !!\Bbb R!!, and !!a,b\in S!! but no element of !!S!! between these, then there is no element of !!S!! that is less than distance !!\frac{b-a}2!! of !!\frac{a+b}2!!. Both !!a!! and !!b!! are at that distance, and no other point of !!S!! is closer.)

The counterexample that M. Schweber pointed out can be explained quickly if you know what the Cantor middle-thirds set is: construct the Cantor set, and consider the set of midpoints of the deleted intervals; this set of midpoints has the property that between any two there is another, but it is not dense in the reals. I was going to do a whole thing with diagrams for people who don't know the Cantor set, but I think what follows will be simpler.

Consider the set of real numbers between 0 and 1. These can of course be represented as decimals, some terminating and some not. Our counterexample will consist of all the terminating decimals that end with !!5!!, and before that final !!5!! have nothing but zeroes and nines. So, for example, !!0.5!!. To the left and right of !!0.5!!, respectively, are !!0.05!! and !!0.95!!.

In between (and around) these three are: $$\begin{array}{l} \color{darkblue}{ 0.005 }\\ 0.05 \\ \color{darkblue}{ 0.095 }\\ 0.5 \\ \color{darkblue}{ 0.905 }\\ 0.95 \\ \color{darkblue}{ 0.995 }\\ \end{array}$$

(Dark blue are the new ones we added.)

And in between and around these are:

$$\begin{array}{l} \color{darkblue}{ 0.0005 }\\ 0.005 \\ \color{darkblue}{ 0.0095 }\\ 0.05 \\ \color{darkblue}{ 0.0905 }\\ 0.095 \\ \color{darkblue}{ 0.0995 }\\ 0.5 \\ \color{darkblue}{ 0.9005 }\\ 0.905 \\ \color{darkblue}{ 0.9095 }\\ 0.95 \\ \color{darkblue}{ 0.9905 }\\ 0.995 \\ \color{darkblue}{ 0.9995 }\\ \end{array}$$

Clearly, between any two of these there is another one, because around !!0.????5!! we've added !!0.????05!! before and !!0.????95!! after, which will lie between !!0.????5!! and any decimal with fewer !!?!! digits before it terminates. So this set does have the between-each-two-is-another property that I was depending on.

But it should also be clear that this set is not dense in the reals, because, for example, there is obviously no number of this type that is near !!0.7!!.

(This isn't the midpoints of the middle-thirds set, it's the midpoints of the middle-four-fifths set, but the idea is exactly the same.)

Happy New Year, everyone!