Understanding an argument in Intermediate Value Theorem

• I
• Hall
In summary: 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

1. What is the Intermediate Value Theorem (IVT)?

The Intermediate Value Theorem is a mathematical theorem that states that if a continuous function has two values, a and b, at two different points in its domain, then it must also have every value between a and b at some point in its domain.

2. How is the IVT used to understand arguments?

The IVT is often used in mathematical proofs to show that a function must have a certain value at some point in its domain. It helps to support arguments and conclusions by providing a logical framework for understanding the behavior of continuous functions.

3. What are the key components of an argument using the IVT?

An argument using the IVT typically involves identifying a continuous function, stating the values a and b at two different points in its domain, and then using the IVT to show that there must be a point in the domain where the function takes on a value between a and b.

4. What are some common misconceptions about the IVT?

One common misconception is that the IVT can be used for all functions, when in fact it only applies to continuous functions. Another misconception is that the IVT guarantees the existence of a specific value, when in reality it only guarantees the existence of a value between two given points.

5. How can the IVT be applied in real-world situations?

The IVT can be applied in many real-world situations, such as in economics to model supply and demand, in physics to analyze motion, and in chemistry to understand reaction rates. It provides a powerful tool for understanding the behavior of continuous functions in various contexts.

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