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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?

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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. –  user1296727 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
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3 Answers

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 $y \geq b$.

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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....)

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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 $\sup S_b$.

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Was there something of this which was unclear from my post? –  user1296727 Jul 16 '12 at 0:17
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@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. –  user1296727 Jul 16 '12 at 0:26
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