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The extreme value and intermediate value theorems are two of the most important theorems in calculus. They are generally regarded as separate theorems.

However, there is a very natural way to combine them:

Let $f:[a,b] \to \mathbb{R}$ be a continuous mapping. Then $f([a,b]) = [c, d]$ where $c \leq d$.

Two questions:

$(1)$ Why isn't this combination more common?

$(2)$ Can we prove this in one go (ie., without first proving the IVT and EVT and deducing this as a corollary)?

From the perspective of topology, this theorem follows from the fact that the continuous image of a connected and compact space is connected and compact. I'm looking for a "calculus" style proof (ie., using an $\epsilon-\delta$ argument, least upper bound property and/or sequences).

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  • $\begingroup$ @bof Perhaps, but I'm asking if there's a clever way to kill two birds with one stone. In other words, instead of proving this by having a proof of the IVT and then having a proof of the EVT beneath it, can we "combine" both proofs in some manner? $\endgroup$ – MathematicsStudent1122 Dec 24 '16 at 9:08
  • $\begingroup$ I am supposing that you are looking for a way to teach this at an introductory, elementary level, not a way to prove it for yourself. $\endgroup$ – DanielWainfleet Dec 24 '16 at 10:23
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    $\begingroup$ One small point: in an arbitrary ordered field the mean value theorem and the intermediate value theorem are equivalent and in turn equivalent to most of the interesting results in 1-variable calculus. Your version of the theorem would also be equivalent, but I don't know how one could prove it directly. $\endgroup$ – Glitch Dec 24 '16 at 10:45
  • $\begingroup$ @user254665 I am not a mathematics instructor, but part of this question is definitely pedagogical. These theorems are, after all, taught universally in intro calculus courses; depending on the rigour of the course, they may also be also proven. This way of presenting the theorems is perhaps more "compact" and not too difficult for students to understand. $\endgroup$ – MathematicsStudent1122 Dec 24 '16 at 10:59
  • $\begingroup$ What do you gain by combining them? Suppose you can prove to your calculus students, in one fell swoop, that the continuous image of a compact connected set is compact and connected. They still won't kinow that the continuous image of a compact set is compact, nor that the continuous image of a connected set is connected. $\endgroup$ – bof Dec 26 '16 at 8:00
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Definition: A subset $I \subset \mathbb{R}$ is said to be connected if for every $a,b \in I$, $[a, b] \subset I$.

Theorem 1: A set of real numbers is connected if and only if it is an interval.


Theorem 2: The image of a closed interval $[a, b]$ under a continuous function is connected. Moreover, this interval is closed.

Discussion: The first part of Theorem 2 is the IVT, except that you don't restrict your focus to just the endpoints; the 'moreover' statement is the EVT.

You want to use one basic technique to derive Theorem 2. The EVT part can be regarded as just "mopping up" loose ends. The basic technique is that both the IVT and EVT are 'joined at the hip' with the construction of the reals.

Proof

Let $f$ denote the continuous function and $s_0$ and $t_0$ be any points in the interval. Let $y_0$ be a number between $f(s_0)$ and $f(t_0)$; we must find a number $k_0$ satisfying $f(k_0) = y_0$.

Divide the closed interval formed by $s_0$ and $t_0$ in half using the midpoint $x_0 = .5(s_0+t_0)$. If $f(x_0) = y_0$, we are done. If not, there are three possible cases:

$\tag 1 f(x_0) \text{ is less than both } f(s_0) \text { and } f(t_0)$

$\tag 2 f(x_0) \text{ is between both } f(s_0) \text { and } f(t_0)$

$\tag 3 f(x_0) \text{ is greater than both } f(s_0) \text { and } f(t_0)$

Regardless of the case, we are forced to choose one of the two intervals, and after renaming its points, that $f$ is defined on this interval $[s_1, t_1]$ and that our number $y_0$ is between $f(s_1)$ and $f(t_1)$.

So we can continue this process, and if necessary we can intersect all the nested closed intervals $[s_k,t_k]$ obtaining a singleton set containing a number $k_0$. Using the continuity of $f$ and reductio ad absurdum, you must conclude that $f(k_0) = y_0$; a "$\varepsilon \text{/} \delta$" argument is not necessary here.

For the second part of the theorem, let us first show that the image of $f$ is bounded from above. If it isn't, we can bisect the interval $[a,b]$ and a point in the left or right side can be found so that its image under $f$ is greater than $1$. We can continue this, and on the $n$ time we are looking for
$f$ to be greater than $n$. Again, intersecting these nested intervals results in a singleton point, call it $x_H$. It is easy to see that $f(x_H) \gt n$ for all $n \ge 1$, and that is impossible.

The argument to show that $f$ is bounded from below is identical.

Finally, we have to show that both $inf(f([a,b])$ and $sup(f([a,b])$ are the images of points in the domain of $f$. Since the proof follows the above 'bisecting down to a singleton that works' method, we leave the details to the reader.


Bolzano–Weierstrass Theorem: Let $(x_n)$ be a bounded sequence contained in an interval $[a, b]$. A subsequence of $(x_n)$ can be found that converges to some point in $[a,b]$.

Proof

It is trivial exercise to find a subsequence if the sequence only takes on values from a finite set. To start you can select any point in $x_k$ in the interval $[a, b]$. We bisect the interval and select either the left or right side so long as it has an infinite number of points in common with $(x_n)$. We can also select an $x_l$ in this interval with $l \gt k$,

We continue bisecting to infinity, and intersecting the nested closed interval we reach a singleton real number $x_0$ while constructing a subsequence. It is easy to see that this new sequence converges to $x_0$.

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