# Show that a function that is locally increasing is increasing?

A function $f : \mathbb{R} \to \mathbb{R}$ is locally increasing at a point $x$ if there is a $\delta > 0$ such that $f(s) < f(x) < f(t)$ whenever $x-\delta < s < x < t < x+\delta$.

Show that a function that is locally increasing at every point in $\mathbb{R}$ must be increasing, i.e., $f(x) < f(y)$ for all $x < y$.

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This seems like an obvious application of the definition of "compact" to the interval $[x,y]$. – coffeemath Oct 22 '12 at 10:11
Adding to coffeemath's point: remember the property (maybe definition, depends how you go about it) of compact sets in $\mathbb{R}$ that "every open covering has a finite subcovering". That should allow you to "globalize" the local statement. – Guilherme Freitas Oct 23 '12 at 8:43

HINT: Suppose that $f$ is locally increasing at every point of $\Bbb R$, but $f$ is not an increasing function; then there are $a,b\in\Bbb R$ such that $f(a)\ge f(b)$. Let $B=\{x\in[a,b]:f(a)\ge f(x)\}$; $B\setminus\{a\}$ is non-empty, since $b\in B$, and is bounded below by $a$, so $b$ has a greatest lower bound $x_0\in[a,b]$.
Show that $x_0\in B$, and consider two cases:
1. $x_0=a$. Use the hypothesis that $f$ is locally increasing at $a$ to get a contradiction.
2. $x_0>a$. Then $f(x)>f(x_0)$ for each $x\in(a,x_0)$. Use the fact that $f$ is locally increasing at (what point?) to get a contradiction.
Alternatively, if you’re familiar with the open cover definition of compactness, you can let $[a,b]$ be any closed interval and use the hypothesis that $f$ is locally increasing at every point to get a cover $\mathscr{U}$ of $[a,b]$ by open intervals on each which $f$ is increasing, then use compactness of $[a,b]$ to get a finite subcover and show directly from the existence of this finite subcover that $f$ must be increasing on $[a,b]$. Since $a\le b$ are arbitrary, this shows that $f$ is increasing.
Clarification: if $x_0 \in B$, a subset of the interval not containing a, how can case 1 hold? – Zhanfeng Lim Dec 14 '15 at 4:25