Understanding an argument in Intermediate Value Theorem

[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 $h$ $f(c-h) \gt L$.

 

[pasmith]

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 $$f(c - h) > f(c) - (f(c) - L) = L.$$

 

[Hall]

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 $|f(c-h) -f(c)| = f(c) - f(c-h)$?

 

[pasmith]

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?

 

[WWGD]

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.