Re: constant and locally constant

In article
<40ee2c83-6b57-41c5-82a8-57d618743c90@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
water <waterloo2005@xxxxxxxxx> wrote:

Exercise 1. Show that a function that is locally constant at each
point of an interval [a,b] is constant on [a,b].

A calculus student might "solve" this by spotting that such a function
would have to
have a zero derivative everywhere. Thus, by a familiar calculus
principle, the function
is constant. But the proof of that principle requires a compactness
argument anyway,
and the exercise should be attempted from first principles.

Above is copied from a paper.

I think "a familiar calculus principle" is that f is constant when f
has zero derivative everywhere.

Why does the principle need compactness?

It doesn't "need" compactness, as the other replies show.
However, the simplest known proof does use compactness...
(1) A continuous function on a closed interval achieves a max and min
there (using compactness)
(2) At a local max or min, a differentiable function has derivative 0
(from the definition of derivative)
(3) Rolle's Theorem, easy from (1) and (2)
(4) Mean Value Theorem, using (3).
(5) If a differentiable function is not constant, then by (4) it would
have a point with nonzero derivative.

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
.

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