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In the given solution of this Problem in Spivak's "Calculus", 3rd ed., there are some details, which I fail to comprehend. I think that in order to be clear I have to include two images.

There is a short preliminary text on pg. 73., the last part of which reads as follows:

There is one ambiguity about infinite decimals which must be eliminated: Every decimal ending in a string of $9$'s is equal to another ending in a string of $0$'s (e.g., $1.23999...=1.24000...$). We will always use the one ending in $9$'s.

The problem reads as follows:

19. Describe as best you can the graphs of the following functions (a complete picture is usually out of the question). (i) $f(x)=$ the 1st number in the decimal expansion of $x$.

The following are the given solution and my own handwritten solution:

The given solution.

My solution.

(The dots mean that these ends of the intervals are "closed" and the arrows mean that these ends of the intervals are "open".)

I agree with the part of Spivak's solution which is to the right of the vertical axis. Note that $f(0.2)=1$ because in the preliminary text it is made clear that $0.2000...=0.1999...$. (To be completely rigorous, shouldn't he replace $1$ on the horizontal axis with $0.999...$?)

However, I don't understand the indicated Intervals to the left of the vertical axis in Spivak's solution. Isn't it rather the case that for example $f(-0.1)=0$ because $-0.1000...=-0.0999...$ like I indicated in my solution? Am I missing something about negative real numbers? Technically $0=0.000...$, so is there a way to express $0$ with another number ending in $9$'s?

Carlevaro99
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The labelling on the horizontal axis merely identifies numbers. There's no need to choose to identify them in a way that matches the particular representation he's using in defining $f$. What if he also defined a second function, $g$, using the ALTERNATIVE representation of finite-decimals, and asked you to draw $f + g$? What labels would you then have him use on the $x$-axis?

For $x = 0$, his rule about $9$s doesn't apply, for there's no decimal ending in a (nonempty) string of $9$s that's equal to zero. So the first digit in the decimal expansion of $0$ is certainly $0$. The indicated solution seems to suggest that the first digit is $10$, which makes no sense at all. (Indeed, in general, it's tough to know what the first digit in a decimal expansion means, unless it's very carefully defined. For instance, is $0.11\ldots$ or $.11\ldots$ the decimal expansion of $1/9$? The first starts with $0$, the second starts with $1$.

If you say "the first nonzero digit," then there's no answer for $0$.

A typical number between $-0.1$ and $0$ is something like $-0.0734$; I guess that one might say that this starts with $0$ (but not 10 ... that's crazy!). So the first dot-line-arrow shape to the left of the $y$-axis in the solutions manual is just plain wrong. What about the second one? A typical number in there is $-0.1302938\ldots$, where we'd have to say that the first digit is either $0$ (probably not what's intended) or $1$, but certainly not $0$.

So ... the solution-manual answer is wrong to the left of the $y$-axis.

John Hughes
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  • Thank you very much! I'm sorry, but it is not clear to me what you are getting at in your first paragraph. Could you give me a hint? – Carlevaro99 Sep 25 '20 at 20:41
  • Suppose that Spivak defined the "height" of a number as its largest decimal digit and asked you to plot height(x) as a function of x [That'd be nearly impossible, but...]. You wouldn't insist that the numbers labelling the x-axis be at different heights, right? Well here, he's defined a function on numbers that happens to relate to how those numbers CAN be written as decimals. Does that require him to use that particular written form for the x-axis labels? What if the function he defined depended on two DIFFERENT ways to write numbers -- would he have to use BOTH ways to label the axis?(cont.) – John Hughes Sep 25 '20 at 22:32
  • Of course not. The labelling of the axis and the definition of the function are completely independent operations. If this still doesn't make sense...ignore it. It's really a meta-point having to do with your "To be completely rigorous..." side-remark, and not germane to the main point. – John Hughes Sep 25 '20 at 22:33