1
$\begingroup$

Boas's Primer of Real Functions (4 Ed, Carus Monograph 13) is a classic text which has been read by thousands. I don't expect to find an error in it, and I may well be misinterpreting.

In Exercise 3.13 (page 20) Boas asks "Show that there are just as many sequences of real numbers as real numbers". [He is using informal language, but he is asking for proof that the cardinalities of the two sets are equal. He has just proved the Schroeder-Bernstein Theorem and expects the reader to find an injection from each set to the other which will then show that a bijection exists based on the theorem.]

The solution (given at the end of the book) proceeds in a classic and expected way. He limits his consideration to the interval from 0 to 1 based on a bijective mapping shown in another exercise. Then he easily finds an injection from the reals to the sequences of reals - just use the digit sequence of the real number between zero and one - that is a specific sequence of reals.

For his injection from the real sequences to the reals, he writes out the decimals in a "stack" and reads along the diagonals to produce one long real number (decimal expansion).

Here are his exact words:

"On the other hand, if we have a sequence of real numbers between 0 and 1, then we can prefix the digit 5 to each decimal expansion, write out the decimal expansions one below another, and traverse the resulting array along its diagonals (from upper right to lower left) to obtain the decimal expansion of a real number. Different sequences of real numbers generate different real numbers in this way, since the decimal we obtain cannot terminate. "

I am befuddled about the fact that he "prefixes" a 5 to each decimal expansion. Why would one need to do that? I don't think it accomplishes anything and the proof would go through without it.

Perhaps you think it is to avoid terminating decimals (he has earlier in the book said that terminating decimals would be avoided by only expressing them in their form as decimals ending in 99999...). So if the diagonal process produced a terminating decimal ending in 0000000..., it might vitiate the proof by being a duplicate of the other form of terminating decimal produced by a different real sequence, hence the mapping is not an injection.

But I don't believe either type of terminating decimal could possibly come out of this diagonal process. A decimal ending in 9s only or zeroes only, running along the diagonals as he described, would mean that all the numbers from a certain point on in the sequence had only 9s or only zeros in their decimal expansion, which we know is not the case.

Any thoughts or insights would be appreciated as this problem is keeping me up nights.

-Jonathan, about to start a Masters Program in mathematics.

$\endgroup$
4
  • $\begingroup$ It's just a matter of slightly simpler construction vs slightly simpler analysis. Tastes differ. And Even famous books don't necessarily have highly optimized proofs, particularly for exercise solutions. $\endgroup$
    – Dap
    Mar 22, 2018 at 19:19
  • $\begingroup$ Well I'm still not clear why he would prefix 5's on every real number with no reason or benefit. The construction is his, not mine, and I quoted it verbatim. Did Boas write the solutions in the back? Don't know. But whoever did must have had something in mind when they plunked on $\aleph_0$ fives and squandered a veritable ocean of ink. $\endgroup$
    – Jonathan
    Mar 22, 2018 at 22:36
  • $\begingroup$ Without the 5s it's slightly less obvious the sequence doesn't terminate - your explanation is just as long as "prefix the digit 5 to each decimal expansion". By the way, you can see an earlier edition on archive.org/details/primerofrealfunc00boas $\endgroup$
    – Dap
    Mar 23, 2018 at 7:37
  • $\begingroup$ Hmmm... it was a good idea to look at the older edition, Dap. He has the solution there without adding fives to the decimal expansions and just notes that the decimals from this process will not terminate. Just so. Someone must have told him that it would be more correct or more clear if he added 5s to each decimal. Neither in my opinion. Any unnecessary step taken in a mathematical proof will only confuse the engaged reader. No worries. I emailed Harold. Maybe he introduced the fives in his "Updated and Corrected" edition of his father's book. Someone will have to own up. $\endgroup$
    – Jonathan
    Mar 23, 2018 at 15:07

2 Answers 2

1
$\begingroup$

"But I don't believe either type of terminating decimal could possibly come out of this diagonal process. A decimal ending in 9s only or zeroes only, running along the diagonals as he described, would mean that all the numbers from a certain point on in the sequence had only 9s or only zeros in their decimal expansion, which we know is not the case"

Why do we "know" this is not the case?

$\endgroup$
2
  • $\begingroup$ The answer in the book, in particular, only speaks of numbers "between 0 and 1", which is ambiguous about whether it means $[0,1]$ or $(0,1)$. In the latter case, I would worry about $(0.1999..., 0.999..., 0.999..., ...)$ and $(0.200..., 0.000..., 0.000..., ...)$. $\endgroup$ Mar 25, 2018 at 23:17
  • $\begingroup$ It is not so ambiguous. He is explicitly relying on a result from an earlier exercise to bijectively map the reals onto the unit interval before he starts. The mapping that he has used in the earlier exercise (based on tangent or arctangent) is from the reals to the open unit interval. $\endgroup$
    – Jonathan
    Mar 26, 2018 at 13:09
0
$\begingroup$

It is not so ambiguous whether Boas is looking at the open unit interval or the closed one. He is explicitly relying on a result from an earlier exercise to bijectively map the reals onto the unit interval before he starts. The mapping that he has used in the earlier exercise (based on tangent or arctangent) is from the reals to the open unit interval.

Harold Boas was kind enough to just respond to my query. His exact words: @@@@@@@@@@@@@@@@@@@@@@@@@@@

Thanks for writing.

Assuming that "between 0 and 1" means "strictly between" (which is the clear intent), then I agree that prefixing a digit 5 is solving a non-existent problem. Since the decimal 0.000... is not part of the list, you are correct that the construction cannot possibly produce a decimal ending in all zeroes.

@@@@@@@@@@@@@@@@@@@@@@@@@@@

So go buy his excellent book, one of the few real analysis texts which give full, well written solutions to every problem. And use it in good health.

Best - Jonathan

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .