 #1
zenterix
 66
 20
 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 xaxis 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 nonzero 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 ##xa<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 ##an_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=xa<\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,
$$0<ax<\delta_2 \implies f(x)a<\epsilon$$
Also
$$0<xa<\delta_2 \implies f(x)a <\epsilon$$
$$\implies xa<\delta_2 \implies f(x)a < \epsilon$$
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.