Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In exercise 49 of Spivak's Calculus, a function $h$ is termed to be increasing at any point $a$ if there exists a $\delta > 0$ such that

$$ a - \delta < x < a \implies h(x) < h(a) $$ $$ a < x < a - \delta \implies h(a) < h(x) $$

and the reader is asked to prove that a function which is increasing at all points in some interval is increasing on that interval.

I was able to show this using Heine-Borel...

(Proof: Let $x, y \in I$, $x<y$, where all members of $I$ are increasing w.r.t. $h$. The collection of open intervals $(t - \delta_t, t + \delta_t), t \in [x,y]$, covers $[x, y]$, and therefore has a finite open subcover $\mathcal{C}$; for any two successive $c_i, c_{i+1}$ (the center-points of open intervals in $\mathcal{C}$) we have some $\gamma$ in the overlap of their covers, which says that $h(c_i) < h(c_{i+1})$. Chaining them together, we have

$$ h(x) \leq h(c_1) <h(c_2) < \cdots < h(c_n) \leq h(y) $$

so that $h(x) < h(y)$.)

...but Spivak never introduced Heine-Borel. He suggests

Prove [the result] by considering for each $b$ in $[0, 1]$ the set $S_b$ = $\{x: h(y) \geq h(b)$ $\forall y \in [b, x]\}$ (Hint: Prove that $S_b = \{x: b \leq x \leq 1\}$ by considering $\sup(S_b)$).

I admit that I don't see what he's getting at. Does somebody else know what he means?

share|cite|improve this question
I don't understand the problem, what do you mean by "prove that a function which is increasing at all points in some interval is increasing on that interval."? – user38268 Jul 15 '12 at 23:47
$S_b$ is bounded above and nonempty, therefore has a sup. Get a contradiction by assuming that sup is strictly less than 1. – GEdgar Jul 15 '12 at 23:48
Could you tell me the page? – Pedro Tamaroff Jul 15 '12 at 23:52
@PeterTamaroff This is the first. ed. I'm working out of; it's p. 189 there. – Chris Jul 15 '12 at 23:54
In my edition page 189 is Uniform Continuity. Could you be more specific? I cannot find it. – Pedro Tamaroff Jul 15 '12 at 23:57

For the sake of the OP, I'll translate what my version says:

A function $f$ is increasing on $a$ if there is a $\delta >0$ such that

$$f(x)>f(a)\text{ if } a<x<a+\delta$$

$$f(x)<f(a)\text{ if } a-\delta<x<a$$ Observe this doesn't mean $f$ is increasing on $(a-\delta,a+\delta)$.

$(a)$ Assume that $f$ is continuous on $[0,1]$ and that $f$ is increasing on $a$ for all $a$ in $[0,1]$. Prove $f$ is increasing in $[0,1]$. (Convince yourself there is something to be proven). Hint: For $0<b<1$ show the minimum of $f$ on $[b,1]$ must lie on $b$.

$(b)$ Prove part $(a)$ without the supposition that $f$ is continuous, considering for each $b$ in $[0,1]$ the set $$S_b=\{ x:f(y)\geq f(b) \text{ for all } y \in[b,x]\}$$ Hint: Prove that $S_b=\{x:b\leq x \leq 1\}$ considering $\sup S_b$.

share|cite|improve this answer
Was there something of this which was unclear from my post? – Chris Jul 16 '12 at 0:17
@user1296727 yes. – user38268 Jul 16 '12 at 0:24
@BenjaLim Reading your above comment, I am sorry that I did not clarify. By increasing on an interval, I meant that for arbitrary $x, y \in I$, $x < y \implies h(x) < h(y)$. The other term I defined at the top; I assumed the distinction was clear. – Chris Jul 16 '12 at 0:26
up vote 0 down vote accepted

GEdgar suggested that this be proven by contradiction; we first will show that

Lemma: $S_b := \{x: \forall y \in [b,x] (h(b) \leq h(y))\} = \{x: b \leq x \leq 1\}$.

Proof: Suppose that there exists a $b$ for which this is not true; we then have $\sup(S_b) = \lambda < 1$. Then we must have $h(\lambda) > h(b)$, which says - since $\forall y < \lambda$ we have $h(y) \geq h(b)$ - that $\lambda$ is in $S_b$. (If we didn't have $h(\lambda) > h(b)$, we would have either $h(\lambda) \leq h(b)$, so that there would exist some stretch below $\lambda$ where the function would be less than $h(\lambda)$, and thus less than $h(b)$, by hypothesis.) But $\lambda \in S_b$ says that there exists some stretch above $\lambda$ where $h$ is always larger than $h(\lambda)$, and so $\lambda$ cannot be the supremum of $S_b$, which is a contradiction.

Given the lemma, we have shortly that $h$ is increasing, since $x < y$ says $y \in S_x$, etc.

(And crap, but do I feel stupid but for asking this question....)

share|cite|improve this answer

Based on Peter's post, I would assume you are trying to prove Part b.

b) Let $\alpha =\sup S_{b}$. If $b \leq y < \alpha$, then there is some $x$ in $S_{b}$ with $y < x$. Therefore $f(y) \geq f(b)$. Moreover, since $f$ is increasing at $\alpha$, we have $f(\alpha) > f(x)$ for $x < \alpha$ sufficiently close to $\alpha$, so $f(\alpha) > f(b)$. This shows that $\alpha$ is actually in $\sup S_{b}$. Now if $\alpha < 1$, there would be $\delta > 0$ such that $f(x) > f(\alpha)$ for $\alpha < x < \alpha +\delta$. This shows that all such $x$ are in $S_{b}$, contradicting the fact that $\alpha = \sup S_{b}$. So $\alpha=\sup S_{b} =1$. So $f(y) \geq f(b)$ for all $y \geq b$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.