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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be strictly monotonically increasing.

(i) Is $f$ not continuous at $p \in \mathbb{R}$, there exists a non-empty, open interval $(a_p, b_p) \subset \mathbb{R}$ such that $f(x)\leq a_p$ for all $x < p$ and $f(x) \geq b_p$ for all $x > p$.

(ii) The set of discontinuity points $$ \{ p \in \mathbb{R} | f \; \mbox{is not continuous at} \; p \}$$ is countable.

For (i) I could be way off, but I am picturing a graph with $p$ on the $x$-axis for which the value $f(p)$ on the $y$-axis is undefined. Am I correct to interpret the open interval $(a_p, b_p)$ as an interval on the $y$-axis which should be contained within the distance between two $f(x)$'s (one for $x<p$ and the other for $x>p$)? If that is so far correct, there could be 2 types of discontinuous points $p$, a jump or removable type. For the jump it would be easier to show that somehow the interval $(a_p, b_p)$ is smaller than the vertical jump... For a removable discontinuity $c$, I would think that the interval could contain just the $y$-axis value $\displaystyle \lim_{x\to c}f(x)$ but I don't really know how to express that the upper and lower bounds would be just above and below that...

With (ii) I am currently trying to understand a proof, what exactly does the notation $f(p-)$ or $f(p+)$ mean in this context? Is it simply the value when approached from the left or the right(respectively)?

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For (ii): your interpretation is correct. –  Shai Covo Dec 15 '10 at 23:33
    
For (i) you are close, but you cannot have a removable discontinuity as the domain is then not all of $\mathbb{R}$. For a jump you must still have a value of $f$ at $p$. It might be $f(p-)$ or $f(p+)$ or anywhere in between. Your last sentence in (i) is correct-there is no interval that includes only one point. –  Ross Millikan Dec 15 '10 at 23:39
    
@Ross, Ok so would it be a good step to define $\epsilon = \mbox{min}(|\frac{f(p)-f(p-)}{2}|,|\frac{f(p)-f(p+)}{2}|)$? Then I could say $a_p = f(p)- \epsilon$ and $b_p = f(p) + \epsilon$ (since there are an infinite amount of real numbers between real numbers this would not be just one point). I guess I would have to make exception cases since $f(p)$ might be equal to either $f(p-)$ or $f(p+)$, but otherwise is it more or less the right idea? –  ghshtalt Dec 16 '10 at 0:38
    
@user3711: This will work in the case $f(p)$ is strictly between $f(p-)$ and $f(p+)$. Patching it up for the other case is not hard. Do you know that $f(p-)$ and $f(p+)$ are well defined? Wikipedia says so, but no justification is given. –  Ross Millikan Dec 16 '10 at 0:58
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@user3711: When you are done, might I suggest posting your solution as an answer? You can request comments on your write-up that way, and eventually you can accept. –  Arturo Magidin Dec 16 '10 at 2:38

1 Answer 1

up vote 1 down vote accepted

Revised answer:

(i)$x < p \Rightarrow f(x) < f(p)$ (because $f$ is strictly monotonically increasing).

Because this set of values is bounded from above there exists a supremum, namely $f(p-)$.

Likewise, $x > p \Rightarrow f(x) > f(p)$ (because $f$ is strictly monotonically increasing).

Because this set of values is bounded from below there exists an infinimum, namely $f(p+)$.

Let $a_p = f(p-)$ and $b_p = f(p+)$.

$\Rightarrow f(x) < a_p \forall x < p$ and $f(x) > b_p \forall x >p$.

Furthermore, $p$ discontinuous $\Rightarrow f(p-) \neq f(p+)$, $f$ strictly monotonically increasing $\Rightarrow f(p+) - f(p-) > 0 \Rightarrow (a_p, b_p) \neq \emptyset$.

(ii)For every point of discontinuity $p_i$, there exists (after (i)) an interval $(a_{p_i}, b_{p_i})$. These intervals are by construction disjoint. Because $\mathbb{R}$ is dense, in every interval there exists an element from $\mathbb{Q} \Rightarrow$ the set of discontinuity points has cardinality at most equal to that of $\mathbb{Q} \Rightarrow$ the set of discontinuity points is countable.

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My comment about well-defined meant that you use f(p-) but are you sure there the limit exists? I am pretty sure, but haven't proved it. I was haunted by the vision of a function that has a jump discontinuity at every rational, so that there are points arbitrarily close to p that are across another jump. But the jumps have to get small quickly so the total converges to something finite. I think then you can show that f(p-) exists. –  Ross Millikan Dec 16 '10 at 23:35
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@Ross: $f(p-)$ is the supremum of the values $f(x)$ for $x < p$. This set of values is bounded above by $f(p)$, so the supremum exists. (Similarly for $f(p+)$ with infima.) –  Pete L. Clark Dec 16 '10 at 23:57
    
OK, I was thinking of a limit, but not just the sup. But you are right, the supremum is enough for this purpose. –  Ross Millikan Dec 17 '10 at 0:12
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Why do you not simply set $a_p=f(p-)$ and $b_p=f(p+)$ in all cases? No need to complicate matters if $f(p)=f(p-)$ or $f(p)=f(p+)$. By the way: Compactness is something else; being able to find a real between any two reals is denseness ("${\mathbb R}$ is dense in itself", sometimes one says); note that ${\mathbb Q}$ also has this property, and in fact there is a rational between any two reals, which may help with part (ii). –  Andres Caicedo Dec 17 '10 at 6:26
    
Oh, I guess you need to justify that $f(p-)$ and $f(p+)$ exist. This is just the supremum axiom. –  Andres Caicedo Dec 17 '10 at 6:27

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