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Suppose that $x$ is a number such that $f(x^−) < f(x^+)$ where $f(x^−) := \lim_{t\to x^−} f(t)$ and $f(x^+) := \lim_{t\to x^+} f(t)$.

a) What numbers cannot possibly be in the range of $f$? Let’s denote these numbers as $R_x$.

b) If $x$ satisfies $f(x^−) < f(x^+)$, why must there be a rational number in $R_x$?

My solution:

a) $f(x)$

b) Since $f(x^-) < f(x^+)$ and with any two real numbers such that $a < b$, there always exists a rational number between them (I think). Is there a theorem for this?

Looking to see if I have done this correctly and if there is a better solution for b).

kennytm
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Remy
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  • (b) ⇒ Rational number is dense. You need to show $R_x$ from (a) is an interval first though. – kennytm Oct 31 '16 at 18:13
  • Ok, I will attempt this – Remy Oct 31 '16 at 18:14
  • Well if f(x-) < f(x+) wouldn't the interval just be f(x) = (f(x-), f(x+))? What is there to left to show? – Remy Oct 31 '16 at 18:16
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    Do you have any more information of the function, like monotoneity? Because as it stands now it seems to me (a) is empty and (b) is thus ill-formed. – The Vee Oct 31 '16 at 18:17
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    I apologize. f is monotonically increasing – Remy Oct 31 '16 at 18:18
  • Then you have answered it fully with your comment and @kennytm's hint. It's not a very tricky question. – The Vee Oct 31 '16 at 18:19
  • Yeah, I just wanted to be sure – Remy Oct 31 '16 at 18:20
  • Since both limits exist, there will be a jump discontinuity, so the first answer will be an interval, and every interval will have both rational and irrational number – jnyan Oct 31 '16 at 18:28
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    In your comment, the interval doesn't have to be necessarily open – jnyan Oct 31 '16 at 18:30
  • I would like to add a further question. If we have $x_1$ < $x_2$ and f($x_1$-)<f($x_1$+) and f($x_2$-) < f($x_2$+) what is the relation between $R_{x_1}$ and $R_{x_2}$? I was thinking the min and max of $R_{x_2}$ > min and max of $R_{x_1}$,respectively. – Remy Oct 31 '16 at 18:33
  • Also, yeah that is true. – Remy Oct 31 '16 at 18:34
  • If monotonic function, then yes – jnyan Oct 31 '16 at 18:44
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    There's one more little twist. See my answer in a few minutes. – The Vee Oct 31 '16 at 18:50

1 Answers1

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The question is easy but since it is fundamental algebra, everything still needs to be proven.

It follows from the definition of a limit that for any sequence $x_k$ monotonically increasing to $x$, the sequence $f(x_k)$ approaches $f(x^-)$. Also, if the function is monotonically increasing, we know that $f(x_{k+1}) > f(x_k)$ for all $k$.

Let $u ≥ f(x^-)$. We prove that $u$ can not be an element of the range of $f$ restricted to $(-\infty, x)$: Let $k \in \mathbb{N}$, then $f(x_k) < f(x^-) ≤ u$. That means $\forall y < x_k, f(y) < u$. Since this holds for any $k \in \mathbb{N}$, we can take the union of all $\{y | y < x_k\}$ which is $\{y | y < x\}$:

$$\forall y < x, f(y) < u.$$

In complete analogy we find that for any $v ≤ f(x^+)$,

$$\forall y > x, f(y) > v.$$

Let now $w$ satisfies both the conditions, that is, $w \in [f(x^-), f(x^+)]$. We know that any $y < x$ and any $y > x$ have functional values which are strictly greater or strictly smaller than $w$.

What about $y = x$? It turns out that the above lines tell nothing of this case. The value of $f(x)$ may coincide with either of the bounds or can lie inbetween (imagine the signum function at zero). Luckily for us, this is just a single value out of a whole interval. So we exclude it and we're done:

$$R_x = [f(x^-), f(x^+)] \setminus \{f(x)\}.$$

Depending on whether $f(x)$ lies strictly between the two bounds, $R_x$ is either a half-open interval or a union of two such intervals. In either case, we use the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$ to prove that a rational number must lie within the set $R_x$.

The Vee
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  • That's interesting! I did not think about that. I have one final question if you wouldn't mind – Remy Oct 31 '16 at 18:57
  • @JohnH Sure, what is it? – The Vee Oct 31 '16 at 18:58
  • Let E = {x$\in\mathbb{R}$ : f(x−) < f(x+)}. Explain why E is at most a countable set. – Remy Oct 31 '16 at 18:58
  • So I think this means the set of all x's where the limits of f(x) are not equal from the left and right – Remy Oct 31 '16 at 18:59
  • NB I updated my answer, it's actually the closed interval minus one point. – The Vee Oct 31 '16 at 19:03
  • closed because the limiting values from the left and right are open, thus $R_x$ contains those two points, correct? – Remy Oct 31 '16 at 19:03
  • Nope, if $f(x)$ lies strictly between then the endpoints are there. If the function is continuous from the left or right, it becomes open at one point. The formula covers both cases. – The Vee Oct 31 '16 at 19:05
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    @JohnH The second question can be figured out from the answer here: http://math.stackexchange.com/questions/978823. In short, if $E$ was uncountable then you would have such union. – The Vee Oct 31 '16 at 19:07
  • I am still a little confused with the limiting points from the left and right. So we have no idea if they are both open, one open one closed, or both closed? – Remy Oct 31 '16 at 19:09
  • Both can't be open at the same time. The value of $f(x)$ is unique, it can't be below $f(x^-)$ and can't be above $f(x^+)$, but it can be equal to one of them. If $f(x) = f(x^-)$, then $R_x = (f(x^-), f(x^+)]$. If it's equal to $f(x^+)$, it's similarly closed from below and open from above. If it's equal to neither, then $R_x = [f(x^-), f(x)) \cup (f(x), f(x^+)]$. – The Vee Oct 31 '16 at 19:11
  • So we know for sure $R_x$ cannot be both f(x-) and f(x+)? Shouldn't the interval be open then? Since we don't know which one is open or closed? – Remy Oct 31 '16 at 19:14
  • It has both in the last case. And there's no "we don't know": we do know, it only depends on the value of $f(x)$ which is specific to every instance of the problem. When you have a particular function $f$ and a particular point $x$ where it has a jump, then you can evaluate $f(x)$ and know exactly which of the cases will take place. – The Vee Oct 31 '16 at 19:16
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    Alright, that makes sense. – Remy Oct 31 '16 at 19:17
  • If you're still here, I am having trouble applying the thread you gave to my problem since they're working with intervals [a,b] and I just have single points x, which for all we know, could be infinitely many. – Remy Oct 31 '16 at 19:30
  • @JohnH That answer only gives an idea. You could do the same as they do there but with the intervals $R_x, x \in E$ to obtain a rational point per each point of $E$. Now because of the monotoneity of $f$ no two of the intervals have an intersection, so the rational numbers are guaranteed to be different. But that means you have an injection $E \to \mathbb{Q}$, which is not possible if $E$ is uncountable. – The Vee Oct 31 '16 at 19:35
  • I thought the set of x values were the values along the x-axis in which the limiting values from the left and right of f(x) were not equal, not points IN $R_x$? – Remy Oct 31 '16 at 19:37
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    That's right. But for each such $x$ (that is, $x \in E$), you have its corresponding $R_x$ from which you can pick a rational point. So taking $x$ from $E$, you know there's $R_x$ disjoint with the other $R_x$'s, you take a $q \in R_x \cap \mathbb{Q}$, you make a mapping $E \to \mathbb{Q} : x \mapsto q$. – The Vee Oct 31 '16 at 19:40