# Understanding an argument in Intermediate Value Theorem

• I
• Hall
Yes, it does imply that at every distance from ##f(c)## there is a ##f(x)## for an ##x## in the interval.

#### Hall

We have to prove:
If ##f: [a,b] \to \mathcal{R}## is continuous, and there is a ##L## such that ##f(a) \lt L \lt f(b)## (or the other way round), then there exists some ##c \in [a,b]## such that ##f(c) = L##.

Proof: Let ##S = \{ x: f(x) \lt L\}##. As ##S## is a set of real numbers and non-empty, therefore we can assume ##\sup S = c##.

CASE 1: Here is the standard argument "if ##f(c) \gt L##, then by continuity ##f(c-h) \gt L## for some small ##h##". How does continuity imply that? Is it like this:
if ##f(c) \gt L##, and for some small ##h## if ##f(c-h) \lt L##, then we have
\begin{align*} f(c) - f(c-h) \gt 0 \\ \lim_{h \to 0} (f(c) - f(c-h)) \gt 0 && \textrm{and by continuity, we have} \\ f(c) - f(c) \gt 0 \\ 0 \gt 0 && \textrm{which is absurd} \\ \end{align*}
Therefore, for small ##h## ##f(c-h) \gt L##.

This is just a simple application of the definition of continuity of $f$ at $c$.

By definition, $f$ is continuous at $c$ iff for every $\epsilon > 0$ there exists $\delta > 0$ such that if $|x - c| < \delta$ then $|f(x) - f(c)| < \epsilon$.

If $f(c) > L$ we can therefore make the particular choice of $\epsilon = f(c) - L > 0$ and conclude that if $0 < h < \delta$ then $$f(c - h) > f(c) - (f(c) - L) = L.$$

• WWGD
pasmith said:
This is just a simple application of the definition of continuity of $f$ at $c$.

By definition, $f$ is continuous at $c$ iff for every $\epsilon > 0$ there exists $\delta > 0$ such that if $|x - c| < \delta$ then $|f(x) - f(c)| < \epsilon$.

If $f(c) > L$ we can therefore make the particular choice of $\epsilon = f(c) - L > 0$ and conclude that if $0 < h < \delta$ then $$f(c - h) > f(c) - (f(c) - L) = L.$$
Why ##|f(c-h) -f(c)| = f(c) - f(c-h)##?

Hall said:
Why ##|f(c-h) -f(c)| = f(c) - f(c-h)##?

We have $$|f(c) - f(c-h)| < f(c) - L \Leftrightarrow f(c) - (f(c) - L) < f(c - h) < f(c) + (f(c) - L)$$ We don't care about the upper bound here; all we need to know is that $f(c - h)$ is strictly greater than $L$.

• Hall
@pasmith I took some time to carefully understand your completely formal validation of that argument.

Sir, I would like to request your wise opinion on the following explanation, though a bit informal, of that argument:
If ##f(c) \gt L##, we can safely write ##f(c) - L = \varepsilon##. Since, the definition of continuity says
"A function is continuous at point ##x## if the difference between ##f(x)## and ##f(x + \delta)## (where ##\delta## is an infinitesimal increment) is ##f(x) - f(x+\delta)##"
(Cauchy's Course de Analyse, section 2.2)

So, decrease ##c## by an amount ##delta## such that ##f(c) - f(c-\delta) = \varepsilon/2##. And on coupling it with the previous relation on L, we get
##f(c-\delta) - L = \varepsilon/2##
##f(c-\delta) \gt L##.

But it seems to me that when I assumed there exists a delta, which upon subtracted from c, which would produce a distance of ##epsilon/2##, is in itself an application of IVT.

This definition of continuity
\begin{align*} \textrm{ for every epsilon > 0 there exits a delta > 0 such that}\\ |x-c| \lt \delta \implies |f(x) - f(c) | \lt \epsilon \end{align*}
Doesn't it imply that at every distance from ##f(c)## there is a ##f(x)## for an ##x## in the interval?

You can also see this from the perspective of connectedness. Continuous maps preserve connectedness. If there was no such c with f(c)=L, the image set would have a disconnection at L. And your set S will need to be bounded above in order to be guaranteed having an upper bound; the set ##(3, \infty)## has no ( Real) upper bound.

• Hall