Very tricky problem from Spivak's Calculus, Ch. 4 & 5: Graphs/Limits

[zenterix]

Homework Statement

I have here a pretty tricky problem from Spivak's Calculus. There is a function that appears for the first time in Chapter 4 on Graphs in problem 19 item v, and then reappears in Chapter 5 on Limits in problem 5.

The function is

$f(x)$ = the number obtained by replacing all digits in the decimal
expansion of $x$ which come after the first $7$ (if any) by $0$.

In Chapter 3 we are asked to "describe as best we can the graph" of this function.

In Chapter 5, we are asked for which numbers $a$ the limit $\lim\limits_{x\to a}f(x)$ exists.

Problem 5 doesn't ask us to prove that the limits exist, however. I would like to know how if someone can provide at least a partial proof that the limit above exists at certain values in the domain.

I have quite a few questions about this interesting function.

Relevant Equations

Below you can see a graph of the function as presented in the solution manual

I believe the x-axis is vertical here.

The graph is composed of

i) an infinite number of intervals that start on the $y=x$ line and finish at some $x$ with a decimal expansion ending in $7\bar{9}$. E.g., from $0.67$ to $0.67\bar{9}$ which is considered $0.68$.

Other examples of intervals would be

$$0.7\text{ to }0.7\bar{9}=0.8$$
$$0.6987\text{ to }0.6987\bar{9}=0.6988$$
$$0.6997\text{ to }0.6997\bar{9}=0.6998$$

ii) an infinite number of points on the $y=x$ line; these are points at which $x$ has either no 7's in its decimal expansion or one 7 followed by zeros. $f$ doesn't modify these numbers, as there are no decimal numbers to be removed.

Note: For some reason, the solution manual graph has some open intervals, e.g. $(0.67,0.67\bar{9}]$, $(0.7,0.7\bar{9}]$. I do not know why this is. For example, take $x=0.6\bar{9}=0.7$. $f(0.6\bar{9})=0.6\bar{9}$, so I understand that the point $(0.6\bar{9},0.6\bar{9})$ should be on the graph. This would be my first question about the graph.

Next, I would like to prove that $\lim\limits_{x\to a}f(x)$ exists for any number $a$ that is not on the right (closed) end of one of the aforementioned intervals. These are numbers with a decimal expansion that ends in $7\bar{9}$.

I am developing my ability to do these proofs, and this one seems quite complicated, I will show my initial work. Please let me know if this route is correct, messy as it may be, and if you see a better way.

Proposition: $\lim\limits_{x\to a}f(x)$ exists for any number $a$ that is not on the right (closed) end of one of the aforementioned intervals. These are numbers with a decimal expansion that ends in $7\bar{9}$

Proof:

My approach is to consider three cases:

1) $a$'s decimal expansion contains one digit 7, followed by zeros; ie $a$ is a point on the $y=x$ line but is also part of an interval containing numbers that all have the same value $f(x)=f(a)$
2) $a$'s decimal expansion contains at least one digit 7, followed by at least one non-zero digit
3) $a$'s decimal expansion contains no digit 7, ie $a$ is a single point on the $y=x$ line

Let me try to prove case 1).

$\forall \epsilon>0$, consider the interval $A=(a,a+\delta_1)$, $\delta_1>0$.
Let $p_1$ be the position in which the first digit 7 is located in the decimal expansion of $a$ (tenths is 1, hundredths is 2, etc). Let $p_2=p_1+1$.

If we choose $\delta_1=\frac{1}{10^{p_2}}$ then $|x-a|<min(\delta_1, \epsilon) \implies f(x)=a \implies |f(x)-a|=0<\epsilon$

Now consider the interval $B=(a-\delta_1, a)$.

By assumption, $a$ has one digit 7 followed by zeros.

There does not exist a number $n<a$, such that there is no number between $n$ and $a$ with no digit 7 in its decimal expansion. In other words, for any number $n<a$, there is always a number $n<n_1<a$ such that $n_1$ has no digit 7 in its decimal expansion. (I guess I would need to prove this).

Call $a-n_1=\delta_2<\delta_1$

What can we now say about $|f(x)-a|$ when $x \in (a-\delta_2,a)$?

If $x$ has no digit 7 in its decimal expansion, then $f(x)=x \implies |f(x)-a|=|x-a|<\delta_2$

If $x$ has one or more digits 7 in its decimal expansion, then

$$a-\delta_2<f(x)\leq x<a \implies -\delta_2<f(x)-a<0<\delta_2\tag{1}$$

$$\implies |f(x)-a|<\delta_2<\delta_1<\epsilon$$

Therefore,

$$0<a-x<\delta_2 \implies |f(x)-a|<\epsilon$$

Also

$$0<x-a<\delta_2 \implies |f(x)-a| <\epsilon$$

$$\implies |x-a|<\delta_2 \implies |f(x)-a| < \epsilon$$

Ie, for $a$ values that satisfy case 1), we have that $\lim\limits{x \to a} f(x) = a$

Note: in $(1)$ I claimed that $a-\delta_2<f(x)\leq x$. This is only valid in the very particular case being considered, namely that $a-\delta_2$ has no digit 7's in its decimal expansion, and therefore $f(a-\delta_2)=a-\delta_2$. For any $x \in (a-\delta_2,a)$, x either has no digit 7's in its decimal expansion, such that $f(x)=x>a-\delta_2$ or it does have one or more digit 7's in which case $f(x)$ is the value x with zeros after the first digit 7 in the decimal expansion. This must be larger than $a-\delta_2$, which I think I can prove by contradiction, but I won't do so because this post is too long as it is, maybe separately I can show this.

 

[anuttarasammyak]

The function is
$f(x)$ = the number obtained by replacing all digits in the decimal
expansion of $x$ which come after the first $7$ (if any) by $0$.

Let me just confirm that
$$f(0.73)=0.7<0.73$$
or
$$f(0.73)=0.8>0.73$$?

 

[zenterix]

Let me just confirm that
$$f(0.73)=0.7<0.73$$
or
$$f(0.73)=0.8>0.73$$?

$$f(0.73)=0.7<0.73$$

 

[anuttarasammyak]

Thanks. Here I write my understanding that
For any x $0.7<x<0.8$ $f(x)=0.7$,
and more in general ;
For any x $0.abc...def7<x<0.abc...def8$, $f(x)=abc...def7 < x$ where all the digits $a,b,c,..d,e,f$ are not seven.
Otherwise $f(x)=x$

Obviously $f(x) \leq x$

 

[zenterix]

For any x 0.7<x<0.8 f(x)=0.7,

Isn't it rather $\forall x: 0.7 \leq x \leq 0.7\bar{9}, f(x)=0.7$? (Note that I made the interval closed not open). The solution manual graph that I posted actually shows $\forall x: 0.7 < x \leq 0.7\bar{9}, f(x)=0.7$. This is actually one question I have: why is the interval open on the left side, ie at $x=0.7=0.6\bar{9}$?

Can you prove that $\lim\limits_{x \to a}f(x)$ exists for some $a$?

 

[PeroK]

What does it say about the convention for choosing $0.7\bar 0$ or $0.6\bar 9$ as the decimal expansion for $0.7$?

 

[anuttarasammyak]

Can you prove that limx→af(x) exists for some a?

Yes, for example a=0.73.

Do you think the limit exists for a=0.8 ?

 

[zenterix]

Yes, for example a=0.73.

Do you think the limit exists for a=0.8 ?

I mean actually formally prove using an epsilon-delta proof.

From simply looking at the graph and the function definition, I would say the limit exists for every $x$ except at places such as $0.8$. The reason is that the limit from below exists and is equal to 0.7, but the limit from above seems to be equal to 0.8.

 

[zenterix]

What does it say about the convention for choosing $0.7\bar 0$ or $0.6\bar 9$ as the decimal expansion for $0.7$?

Missed this message earlier :)

Here is the exact wording:

 

[PeroK]

I mean actually formally prove using an epsilon-delta proof.

From simply looking at the graph and the function definition, I would say the limit exists for every $x$ except at places such as $0.8$. The reason is that the limit from below exists and is equal to 0.7, but the limit from above seems to be equal to 0.8.

By the convention you quoted above, $0.8$ is not a valid decimal expansion. It should be $0.7\bar9$.

 

[zenterix]

By the convention you quoted above, 0.8 is not a valid decimal expansion. It should be 0.79¯.

Indeed, I've spent a few hours on this problem and have not yet understood exactly what is meant by the passage I quoted. I understand that $0.7\bar{9}$ and $0.8\bar{0}$ are equal. It's more murky what it means that "we will always use the one ending in 9's". Does this mean that the notation "0.8" is eliminated from use? What we would normally refer to as 0.8 is in this problem always referred to as $0.7\bar{9}$?

It's weird, because the graph from the solution manual shows numbers such as .7, .6, .8, etc.

It also goes back to my question on why the intervals in that graph are open at such points as $0.67$, $0.7$, $0.87$.

So yes, you are probably right, I shouldn't use $0.8$; when I wrote the previous message, as I wrote it I had the dilemma of whether to write $0.8$ or $0.7\bar{9}$.

Let me try again:

The limit exists for every $a \in \mathbb{R}$ except at points where x has a decimal expansion that ends in $7\bar{9}$, e.g. $0.7\bar{9}$. At such points, the limit from below exists and equals f(a). The limit from above, however, seems to equal a. For example, for $x=0.7\bar{9}$, the limit from below is $0.7$ and the limit from above is $0.7\bar{9}$

 

[anuttarasammyak]

I would say the limit exists for every $x$ except at places such as $0.8$.

So we may say the limit f(x) exist except for x=0.abc...def8 where digits a,b,c,..,d,e,f are not seven. There are countable infinite points.

 

[zenterix]

So we may say the limit f(x) exist except for x=0.abc...def8 where digits a,b,c,..,d,e,f are not seven.

Not quite. See my updated version above. The limit exists except where $x$ is of the form $0.abc...hij7\bar{9}$, ie it ends in $7\bar{9}$

 

[anuttarasammyak]

According to the rule may we say
$f(0.abc...hij7\bar{9})=0.abc...hij7$ because all the digits after the first 7 should be replaced to zero or deleted and $f(0.abc...hij8)=0.abc...hij8$ ?

The rule and general relation $0.8=0.7\bar{9}$ seem not compatible.

 

[PeroK]

Indeed, I've spent a few hours on this problem and have not yet understood exactly what is meant by the passage I quoted. I understand that $0.7\bar{9}$ and $0.8\bar{0}$ are equal. It's more murky what it means that "we will always use the one ending in 9's". Does this mean that the notation "0.8" is eliminated from use? What we would normally refer to as 0.8 is in this problem always referred to as $0.7\bar{9}$?

Yes. I would have expected the opposite. The one without the 9's. I think he's trying to be too clever.

It's weird, because the graph from the solution manual shows numbers such as .7, .6, .8, etc.

Exactly, even Spivak caught himself out with his convention. Note that $0$ is still okay, as there's no alternative!

The limit exists for every $a \in \mathbb{R}$ except at points where x has a decimal expansion that ends in $7\bar{9}$, e.g. $0.7\bar{9}$. At such points, the limit from below exists and equals f(a). The limit from above, however, seems to equal a. For example, for $x=0.7\bar{9}$, the limit from below is $0.7$ and the limit from above is $0.7\bar{9}$

Yes, I think that's right. The left-hand limit of $f$ at $0.7\bar 9$ equals $0.7\bar0$. Although, Spivak has tripped himself up now, because that's not a valid decimal. It should be $f(0.7\bar 9) = 0.6\bar 9$.

Leaving that to one side, the right-hand limit is $0.8$, which must be written $\equiv 0.7\bar 9$.

The limit, therefore, does not exist at $x = 0.7\bar 9$.

 

[zenterix]

According to the rule may we say
$f(0.abc...hij7\bar{9})=0.abc...hij7$ because all the digits after the first 7 should be replaced to zero or deleted and $f(0.abc...hij8)=0.abc...hij8$ ?

The first part yes, and this is why the intervals are closed on the right.

About the second part ($f(0.abc...hij8)=0.abc...hij8$) I am not sure. This number ends in a string of zeros, so it is equal to another number that ends in $7\bar{9}$. On the line $y=x$, at $x=0.7\bar{9}$, there must be an open ball. What happens just to the right of $x=0.y\bar{9}$ is probably an interesting thing.

I would imagine that the first numbers would be numbers without 7's in their decimal expansion. Then again, not matter what number you choose as the first one to the right, you can always find a number with a 7 in its decimal expansion (actually you can always find infinite intervals of such numbers). However, I think it's possible to show that the limit from above at $a=0.7\bar{9}$ is $a$ itself.

I'm interested in seeing what a formal proof of this looks like.

 

[anuttarasammyak]

You may polish up the idea
$$f(0.8-\epsilon)=0.7$$
$$f(0.8+\epsilon)>0.8$$

 

[PeroK]

I'm interested in seeing what a formal proof of this looks like.

What I suggest is using the usual convention of $0.8$ instead of $0.7\bar 9$. The rule of "replacing all subsequent digits by $0$" cuts across that convention.

Now, you want a formal proof that:

a) $\lim_{x \rightarrow 0.8} f(x)$ does not exist. In particular that $\lim_{x \rightarrow 0.8_-} f(x) = 0.7$, and $\lim_{x \rightarrow 0.8_+} f(x) = 0.8$.

b) $\lim_{x \rightarrow 0.5} f(x) = 0.5$. And, since $f(0.5) = 0.5)$, $f$ is continuous at $x = 0.5$.

You might also want to check $x = 0.7$ and $x = 0.81$

Finally, you need a technique to extend these results to any $x$.

 

[PeroK]

Hint: try to sketch out a proof first. Don't dive into $\epsilon-\delta$ until you know what you are trying to do.

 

[zenterix]

Hint: try to sketch out a proof first. Don't dive into $\epsilon-\delta$ until you know what you are trying to do.

Not sure if you saw this, but this was my initial sketch and attempt above. It is convoluted. I'm actually going to come back to this problem at a later time, I've spent too long on it today!

Proposition: $\lim\limits_{x\to a}f(x)$ exists for any number $a$ that is not on the right (closed) end of one of the aforementioned intervals. These are numbers with a decimal expansion that ends in $7\bar{9}$

Proof:

My approach is to consider three cases:

1) $a$'s decimal expansion ends in $6\bar{9}$ (alternatively, ends in a digit 7 with only zeros after), ie $a$ is a point on the $y=x$ line but is also part of an interval containing numbers that all have the same value $f(x)=f(a)$
2) $a$'s decimal expansion contains at least one digit 7, followed by at least one non-zero digit
3) $a$'s decimal expansion contains no digit 7 and does not end in $6\bar{9}$ , ie $a$ is a single point on the $y=x$ line

Let me try to prove case 1).

$\forall \epsilon>0$, consider the interval $A=(a,a+\delta_1)$, $\delta_1>0$.
Let $p_1$ be the position in which the first digit 7 is located in the decimal expansion of $a$ (tenths is 1, hundredths is 2, etc). Let $p_2=p_1+1$.

If we choose $\delta_1=\frac{1}{10^{p_2}}$ then $|x-a|<min(\delta_1, \epsilon) \implies f(x)=a \implies |f(x)-a|=0<\epsilon$

Now consider the interval $B=(a-\delta_1, a)$.

By assumption, $a$ has one digit 7 followed by zeros.

There does not exist a number $n<a$, such that there is no number between $n$ and $a$ with no digit 7 in its decimal expansion. In other words, for any number $n<a$, there is always a number $n<n_1<a$ such that $n_1$ has no digit 7 in its decimal expansion. (I guess I would need to prove this).

Call $a-n_1=\delta_2<\delta_1$

What can we now say about $|f(x)-a|$ when $x \in (a-\delta_2,a)$?

If $x$ has no digit 7 in its decimal expansion, then $f(x)=x \implies |f(x)-a|=|x-a|<\delta_2$

If $x$ has one or more digits 7 in its decimal expansion, then

$$a-\delta_2<f(x)\leq x<a \implies -\delta_2<f(x)-a<0<\delta_2\tag{1}$$

$$\implies |f(x)-a|<\delta_2<\delta_1<\epsilon$$

Therefore,

$$0<a-x<\delta_2 \implies |f(x)-a|<\epsilon$$

Also

$$0<x-a<\delta_2 \implies |f(x)-a| <\epsilon$$

$$\implies |x-a|<\delta_2 \implies |f(x)-a| < \epsilon$$

Ie, for $a$ values that satisfy case 1), we have that $\lim\limits{x \to a} f(x) = a$

Note: in $(1)$ I claimed that $a-\delta_2<f(x)\leq x$. This is only valid in the very particular case being considered, namely that $a-\delta_2$ has no digit 7's in its decimal expansion, and therefore $f(a-\delta_2)=a-\delta_2$. For any $x \in (a-\delta_2,a)$, x either has no digit 7's in its decimal expansion, such that $f(x)=x>a-\delta_2$ or it does have one or more digit 7's in which case $f(x)$ is the value x with zeros after the first digit 7 in the decimal expansion. This must be larger than $a-\delta_2$, which I think I can prove by contradiction, but I won't do so because this post is too long as it is, maybe separately I can show this.

 

[PeroK]

I didn't see it. Looks like you dived right in! Let me look at this problem myself first.

 

[PeroK]

@zenterix I'm not sure you're proof quite works. Take, for example, $a = 0.800005$ If we subtract $0.00001$ from this we get $x = 0.799995$. And $f(x)$ is not 'close' to $f(a)$.

So, the problem is that we have numbers that are close together than have function values further apart.

If we take any decimal expansion, then either:

1) It terminates with infinitely many zeroes.

2) It terminates with infinitely many nines.

3) Neither of these.

We can sort out 1) and 2).

For 3) the problem is that we could get arbitrarily long (although terminating) strings of zeroes or nines. This stops us controlling the relationship between $|x - a|$ and $|f(x) - f(a)|$.

An idea:

If neither 1) nor 2) apply we can always find a pair of consecutive digits that lie between $01$ and $98$. Assume these are digits $k +1$ and $k +2$. Now, we can add or subtract anything less than $\frac 1{10^{k+2}}$ without changing the first $k$ digits.

 

[zenterix]

Take, for example, a=0.800005 If we subtract 0.00001 from this we get x=0.799995. And f(x) is not 'close' to f(a).

The proof I presented was for the specific case of a number $a$ ending in $6\bar{9}$ (originally I specified this as ending in $7\bar{0}$). These are points at the left end of one of the infinite intervals composing the graph of $f$.

$a=0.800005$ isn't part of the available values in this partial proof.

 

[PeroK]

The proof I presented was for the specific case of a number $a$ ending in $6\bar{9}$ (originally I specified this as ending in $7\bar{0}$). These are points at the left end of one of the infinite intervals composing the graph of $f$.

$a=0.800005$ isn't part of the available values in this partial proof.

Okay, I see that now. The argument for a number like $a = 0.6$ is:

1) If $x > a$, then $x \ge f(x) > a = f(a)$, so the right-hand limit exists and is equal to $f(a)$

2) If $x < a$, then for $a - \frac 1 {10^n} < x < a$, $x = 0.5999 \dots 9 d_{n+1} d_{n+2} \dots$ and we see that $a - \frac 1 {10^n} < f(x) < a$. So, the left-hand limit exists and is equal to $f(a)$.

Similarly, the right and left hand limits exist for $a = 0.8$, but are different.

If $a$ has a $7$ (which is not followed by all zeros or all nines), then we find the next non-zero digit and that gives us an interval less than $a$ on which $f(x) = f(a)$. Again, $f$ is continuous at $a$.

The tricky case is where $a$ has no digit seven. That's where you need the trick of finding two digits that I described above.

I'm not sure it's worth spending too much time on this problem.

 

[zenterix]

Let me try and fill in the parts that you left implicit for the case of an $a$ with a decimal expansion with no digit $7$ in it, ending in zeros, e.g. $a=0.6$.

If $x>a$ then $x\geq f(x)>a=f(a)$.

This means that $f(x)$ is always between $x$ and $a$, so $f(x)-a$ is between 0 and $x-a$, ie $x-a \geq f(x)-a>0$.

Therefore,

if $a$ has a decimal expansion that contains no digit 7 and ends in all zeros, then $\forall \epsilon>0$, if $x>a$ and $0<x-a<\epsilon$ then $|f(x)-a|<\epsilon$

$\implies$ if $a$ has a decimal expansion that contains no digit 7 and ends in all zeros, then $\lim\limits_{x \to a_+} f(x)=a$

If $x<a$

Let $\delta=\frac{1}{10^n}$, where $n$ is the first position where the trailing zeros start in the decimal expansion

Then $0<a-x<\delta \implies 0<a-\delta<x<a$

Note that $f(a-\delta)=a-\delta$, since $a-\delta$ has no digit 7 in its decimal expansion. Therefore

$$0<f(a-\delta)=a-\delta<f(x) \leq x < a$$
$$\implies 0<a-f(x)<\delta$$

Here we'd actually need to prove $a-\delta<f(x)$, but let's assume this is true.

We've shown that

if

1) $a$ has a decimal expansion that contains no digit 7 and ends in all zeros
2) $x<a$
2) $\forall \epsilon>0$ we choose a $\delta=\frac{1}{10^n}$ such that $\delta<\epsilon$ and where $n$ is larger than the first position where the trailing zeros start in the decimal expansion

then

$0<a-x<\delta \implies |f(x)-a|<\delta<\epsilon$

$\implies \lim\limits_{x \to a_-} f(x) = a$

 

[PeroK]

The difference between us is that you have generated $\delta$ based on $\epsilon$ that you can't justify. And, I've justified the solution without formalising it with $\delta$ based on $\epsilon$.

Here we'd actually need to prove $a-\delta<f(x)$, but let's assume this is true.

This is effectively what you need to prove. Otherwise your $\epsilon-\delta$ is just window-dressing!

 

[PeroK]

PS I'm not sure what solution Apostol would expect, but producing an explicit $\epsilon-\delta$ proof in this case would be seriously complex. It's better to look for the ideas that control the difference in function values. That's what I would expect as a valid solution.

 

[zenterix]

I'm not sure what solution Apostol would expect

It's Spivak actually, and he's not expecting an epsilon-delta proof in this case; he's simply asking where the limit exists, not a proof. This epsilon-delta madness is the product of my curiosity.

 

[zenterix]

The difference between us is that you have generated $\delta$ based on $\epsilon$ that you can't justify. And, I've justified the solution without formalising it with $\delta$ based on $\epsilon$.This is effectively what you need to prove. Otherwise your $\epsilon-\delta$ is just window-dressing!

Proving that $a-\delta<f(x)$ seems like a side-proof in the main proof, not the main thing to be proved.

Here's an attempt at a proof:

Proposition: $b<x \implies f(b) \leq f(x)$

Proof: Assume $0<b<x<1$.

If $x$ contains no digit 7 in its decimal expansion then $f(x)=x>b \geq f(b) \implies f(x)>f(b)$

If $x$ contains a digit 7 in its decimal expansion then it occurs first in the decimal expansion at position $p_1$

When we compare the decimal expansions of $x$ and $b$, there is a leftmost (ie most significant) position at which the expansions differ, with the digit in $b$ being smaller than the digit in $x$ at this position. Call this position $p_2$.

If $p_1 \geq p_2$ then $f(b)\leq b<f(x)\leq x$, because $f$ contains the same digit in $p_2$ as x.

If $p_1<p_2$ then $b$ contains a digit 7 at position $p_1$. If it's the first digit 7 in $b$ then $f(b)=f(x)$.

If its not the first digit 7 then there is a position $p_3<p_1$ with a 7, and since $p_3<p_2$ it must be the case that $x$ also has a digit 7 at position $p_3$. But this contradicts our assumption that the first $7$ in $x$ occurs at $p_1>p_3$.

Therefore, in any case, $f(b) \leq f(x)$.

 

[PeroK]

Proving that $a-\delta<f(x)$ seems like a side-proof in the main proof, not the main thing to be proved.

Here's an attempt at a proof:

Proposition: $b<x \implies f(b) \leq f(x)$

The main proof as you call it is potentially important, as it formalises the argument. But, without the side proof it means nothing.

That proposition must be false. The problem is that a small reduction in $x$ after a string of zeroes may cause a long string of nines. Using the double digit idea solves this problem.

 

[zenterix]

The main proof as you call it is potentially important, as it formalises the argument. But, without the side proof it means nothing.

That proposition must be false. The problem is that a small reduction in $x$ after a string of zeroes may cause a long string of nines. Using the double digit idea solves this problem.

The proposition should actually be with a $\leq$, ie $a-\delta \leq f(x)$. Is this what you mean?

 

[PeroK]

The proposition should actually be with a $\leq$, ie $a-\delta \leq f(x)$. Is this what you mean?

You're right. But, I don't see how that proposition helps.

 

[zenterix]

This is effectively what you need to prove. Otherwise your ϵ−δ is just window-dressing!

It's the proof of this, which was a step in the main proof. It's required for the latter.