Let $D$ be an interval. Le $ f: D \rightarrow \mathbb{R}$ be a monotone increasing function.

Show that:

$f$ is continuous in a $\in$ $D$ if and only if

$$\sup f\{ x < a\mid x \in D\} = \inf f\{x > a \mid x \in D \}$$

Remark: We don't know what left-hand limit and right-hand limit is. I know the definition of sup and inf, but unfortunately I don't really have an idea how to solve this. I think that we can use the $\varepsilon$-$\delta$ criterion.

Edit : $ \text{“} \Longrightarrow \text{''}$ is clear. I only need a hint for $\text{“} \Longleftarrow \text{''}$.

  • $\begingroup$ Do you have $\sup f\{ x<a |a\in D\}$ or it must be $\sup f\{ x<a |x\in D\}$? $\endgroup$ – Shashi Dec 4 '17 at 19:36
  • $\begingroup$ I'm terribly sorry. Of course it has to be $ x \in D $ $\endgroup$ – RukiaKuchiki Dec 4 '17 at 19:44
  • 1
    $\begingroup$ @AngeloRendina sorry, I deleted the comment $\endgroup$ – ℋolo Dec 4 '17 at 20:01
  • $\begingroup$ Use intermediate value theorem. $\endgroup$ – fleablood Dec 4 '17 at 20:26
  • $\begingroup$ And note that sup f(<) = inf f(>) = f(a). $\endgroup$ – fleablood Dec 4 '17 at 20:26

You asked for a hint. Here is the hint:

$\sup f\{x<a\mid x \in D\} - \varepsilon$ is not an upper bound for $f\{x<a\mid x \in D\}$ and $\inf f\{x<a\mid x \in D\} + \varepsilon$ is not an lower bound for $f\{x>a\mid x \in D\}$

===== fulll proof below =====

(I'll use that notation $f(x>a) :=f\{x > a\mid x\in D\}$ and the same for $f(x< a)$. It's easier to type.)

Suppose $\sup f(x< a) = \inf f(x > a)$. Let $M = \sup f(x< a) = \inf f(x > a)$

It's easy to prove $M = f(a)$. If $ f(a) < M$ then there is a $c\in f(x < a)$ so that $f(a) < f(c) \le M$ contradicting $f$ is increasing. Likewise $f(a) > M$ is impossible for similar reasons.

Let $\varepsilon > 0$. $f(a) - \varepsilon$ is not an upper bound of $f(x< a)$ so there is $c< a$ so that $f(a)- \varepsilon < f(c) \le f(a)$. As there are $c': c < c' < a$ and $f(c) < f(c') \le f(a)$ because $f$ is strictly monotonically increasing, we know $f(c) < f(a)$.

Likewise $f(a) + \varepsilon$ is not a lower bound of $f(x>a)$ so there is $d > a$ so that $f(a) < f(d) < M + \varepsilon$.

Let $\delta = \min(a-c, d-a)$.

Then if $|x-a| < \delta$ then $c \le a-\delta < x < a+\delta \le d$. As $f$ is monotonically increasing $f(a)- \varepsilon < f(c) < f(x) < f(d)< f(a) + \varepsilon$ or in other words $|f(x) - f(a)| < \varepsilon$.

Thus $f$ is continuous.

  • $\begingroup$ Thank you to much. Thanks to your hint and some of your proof I solved this exercise. I understand your complete reasoning. :)+ $\endgroup$ – RukiaKuchiki Dec 5 '17 at 19:23

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