"Extreme Value Theorem"
by math_explorer, Sep 17, 2013, 1:57 PM
Calculus is for calculating: you learn how to differentiate stuff, while you can prove theorems by claiming they're "outside the scope of this book." So what happens when you learn analysis and try to patch the holes and make everything rigorous?
---
After seeing this, I found my copy of "baby Rudin" (thesadistic affectionate and apparently well-known nickname for Principles of Mathematical Analysis) and found the ridiculous counterexample ![\[ f(x) = \begin{cases} x^2 \sin(1/x) &\mbox{if } x \neq 0 \\ 0 &\mbox{if } x = 0 \end{cases} \]](//latex.artofproblemsolving.com/6/5/2/6527bc53afa3c7f787e7f66e9cb6379a4d8abf09.png)
in about a minute. I'm finally beginning to appreciate this book.
A couple conversations about what motivated him to ask this, we decided that everything he wanted was a few deductions away from the "Extreme Value Theorem", which is what my old calculus textbook (Larson, Hostetler, Edwards, 8th ed.) states as "A continuous function defined on a closed interval attains a minimum and a maximum on the interval." The calculus textbook continues,
Well, it's obviously in the scope of Rudin, so that's not an excuse I can use now. How do I prove this?
How exactly does this get proved in Rudin?
After thinking about it I feel like there should be a simpler way, one that's less circuitous and more intuitive, even if it doesn't generalize to that infinite-dimensional space of superpseudorandommorphisms from a strange Greek letter to a stranger Cyrillic letter with a star superscript that really matters 2.5 years into grad school.
Seriously, what is "a set for which every open cover has a finite subcovering" supposed to be? I had to invoke the not-at-all-common-or-obvious terms "cover" and "subcovering" to make the definition look short. I have no idea what it means, except that it ends up being equal to closed and bounded in all the metric spaces that I've considered seriously so far. (I could probably end the post right here and wait for pythag011 to comment with a witty rejoinder, but that wouldn't be any fun.)
Okay, let's try this.
---
Real numbers. How are they defined? Wikipedia lists a few alternatives: they can be axiomatically defined as "the unique Archimedean complete totally ordered field up to isomorphism"; they can also be constructed as Dedekind cuts, infinite decimal expansions, or equivalence classes of Cauchy sequences in the rationals. This is not really important.
You're probably familiar with real numbers enough. You can add, subtract, multiply, divide, and compare them, i.e. they're an ordered field. The only other thing we need is the least-upper-bound property: if we have a set of real numbers that are bounded above, there's a least upper bound, a bound that "perfectly fits" the set. Precisely, there is a "least-upper-bound" so that the entire set is less than or equal to it, and no smaller bound has the same property.
(The rationals don't have this property, since
has no rational that bounds it perfectly. For any bound, you can find a smaller bound between it and that beastly irrational 
.)
Boundedness. How is it defined? This is easy; a set is bounded if there's a fixed number that's greater than the distance between any two points. There are many alternative but trivially equivalent definitions.
Closedness. How is it defined? A set is closed if it "contains its boundary" --- there are no fuzzy edges where you can get really close but can't actually reach it. Rigorously: a closed set
contains all its limit points, that is, any point such that for any distance 
, it's closer than distance to a point in .
I'm not even going to define "open". I don't need it. It's confusing enough that one set can be closed, open, both, or neither. Poor English.
Compactness. How is it defined? I don't care, I'm going to work with what we have, even if it's terribly incorrect in general. For the rest of this post I hereby decree that a set is compact if it is closed and bounded.
---
We can try proving the Extreme Value Theorem now, actually; it will show a rather intuitive path. So: we have a continuous function![$f : [a, b] \to \mathbb{R}$](//latex.artofproblemsolving.com/4/b/a/4baf2d51eb4d8376d7a95126b727c893b588c70d.png)
. Since we want to use continuity, our imagined proof will end with taking a limit of 
. In both cases it's pretty clear what we want the limit to be equal to, so here's what we end up with:
First, we prove is bounded on ![$[a, b]$](//latex.artofproblemsolving.com/d/a/2/da2e551d2ca2155b8d8f4935d2e9757722c9bab6.png)
(otherwise where do we get a maximum?) If it isn't, then we can pick a sequence of points 
in so that 
as 
. Then we ???!!!???!!!, so we have 
so that 
, which is impossible as this limit should be defined and equal to 
(and thus finite).
Next, we prove's range is closed on . Suppose we know that for some value 
, we can pick 
so that 
is arbitrarily close to (i.e. is "on the boundary"). More precisely, we can pick a sequence of points in so that 
as . Then we ???!!!???!!!, so we have so that 
, so by continuity 
. Thus the boundary points of 's range are in 's range.
Thus,'s range is closed and bounded.
Q?!?E?!?D
Of course, this is not at all a proof, since there are two glaring red gaps here. What could we prove that fills them?
In both cases, we have a bunch of points, and we want some point so that a subsequence of 
converges to . And that's all we need to prove now! (Essentially the same thing is proved as 2.37 in Rudin, but it doesn't seem to get used much.)
Lemma. Let be an infinite sequence of points in a closed interval . Then there exists a subsequence 
that converges to a point in .
Proving this is also rather intuitive. We just want to find a place where there are a lot of points nearby. So we cut in half, into ![$[a, (a+b)/2]$](//latex.artofproblemsolving.com/1/f/c/1fc7d34b8b1d8a3c4150d0fb46e0c7dcd577155d.png)
and ![$[(a+b)/2, b]$](//latex.artofproblemsolving.com/1/a/1/1a1a7ec5f2bd8bc310e3811a4a12adf12528575f.png)
. At least one of these intervals contains infinitely many points of the sequence .
And we repeat! Keep cutting the interval with infinitely many points in half, and keep picking the half that still has infinitely many points. We get an infinite sequence of intervals, each of which is one of the halves of the previous one, and each of which contains infinitely many terms of the sequence.
How do we pick? Well, we really want it to be in all of those intervals, and the least-upper-bound property comes in useful again: if the intervals are ![$[a_i, b_i]$](//latex.artofproblemsolving.com/b/c/0/bc04bd6abf483862782136e613e9a11f028e0529.png)
, we pick it to be the least-upper-bound of all the 
's. So would be at least as large as every , and since each 
is an upper bound of all of the 's, this value of is at most as large as every . So is in all the intervals.
Then to get a sequence that converges to, we just build a sequence of 
s by starting with a random , then picking the next one from the next interval that comes after all the previous 's; since there are always infinitely many choices, such a choice always exists. And we're done!
---
chaotic_iak wrote:
The derivative of a differentiable function on an interval of the real line is continuous.
Is the statement true?
Is the statement true?
After seeing this, I found my copy of "baby Rudin" (the
![\[ f(x) = \begin{cases} x^2 \sin(1/x) &\mbox{if } x \neq 0 \\ 0 &\mbox{if } x = 0 \end{cases} \]](http://latex.artofproblemsolving.com/6/5/2/6527bc53afa3c7f787e7f66e9cb6379a4d8abf09.png)
in about a minute. I'm finally beginning to appreciate this book.A couple conversations about what motivated him to ask this, we decided that everything he wanted was a few deductions away from the "Extreme Value Theorem", which is what my old calculus textbook (Larson, Hostetler, Edwards, 8th ed.) states as "A continuous function defined on a closed interval attains a minimum and a maximum on the interval." The calculus textbook continues,
Quote:
Although [it] is intuitively plausible, a proof of this theorem is not within the scope of this text.
Well, it's obviously in the scope of Rudin, so that's not an excuse I can use now. How do I prove this?
How exactly does this get proved in Rudin?
- (2.18) We define "closed" and "open" topologically in terms of neighborhoods and limit points in a massive definition cluster
- (2.32) We define compact sets as sets for which every open cover has a finite subcovering
- (2.35) We prove that closed subsets of compact sets are compact, along with a bunch of other random facts nearby;
- (2.40) we prove that
-cells (closed -dimensional rectangles) are compact... - (2.41) ...and show that all this implies the Heine-Borel theorem: "compact" is equivalent to "closed and bounded" in
. - (4.8) Two chapters later, we prove that continuous functions invert open sets to open sets
- (4.14) and that this implies they map compact sets to compact sets
- (4.16) so finally continuous real functions on compact metric spaces (closed intervals, in particular) attain the supremum (and infimum) of the values they attain, i.e. have a maximum.
-
.After thinking about it I feel like there should be a simpler way, one that's less circuitous and more intuitive, even if it doesn't generalize to that infinite-dimensional space of superpseudorandommorphisms from a strange Greek letter to a stranger Cyrillic letter with a star superscript that really matters 2.5 years into grad school.
Seriously, what is "a set for which every open cover has a finite subcovering" supposed to be? I had to invoke the not-at-all-common-or-obvious terms "cover" and "subcovering" to make the definition look short. I have no idea what it means, except that it ends up being equal to closed and bounded in all the metric spaces that I've considered seriously so far. (I could probably end the post right here and wait for pythag011 to comment with a witty rejoinder, but that wouldn't be any fun.)
Okay, let's try this.
---
Real numbers. How are they defined? Wikipedia lists a few alternatives: they can be axiomatically defined as "the unique Archimedean complete totally ordered field up to isomorphism"; they can also be constructed as Dedekind cuts, infinite decimal expansions, or equivalence classes of Cauchy sequences in the rationals. This is not really important.
You're probably familiar with real numbers enough. You can add, subtract, multiply, divide, and compare them, i.e. they're an ordered field. The only other thing we need is the least-upper-bound property: if we have a set of real numbers that are bounded above, there's a least upper bound, a bound that "perfectly fits" the set. Precisely, there is a "least-upper-bound" so that the entire set is less than or equal to it, and no smaller bound has the same property.
(The rationals don't have this property, since
has no rational that bounds it perfectly. For any bound, you can find a smaller bound between it and that beastly irrational 
.)Boundedness. How is it defined? This is easy; a set is bounded if there's a fixed number that's greater than the distance between any two points. There are many alternative but trivially equivalent definitions.
Closedness. How is it defined? A set is closed if it "contains its boundary" --- there are no fuzzy edges where you can get really close but can't actually reach it. Rigorously: a closed set
contains all its limit points, that is, any point such that for any distance 
, it's closer than distance to a point in .I'm not even going to define "open". I don't need it. It's confusing enough that one set can be closed, open, both, or neither. Poor English.
Compactness. How is it defined? I don't care, I'm going to work with what we have, even if it's terribly incorrect in general. For the rest of this post I hereby decree that a set is compact if it is closed and bounded.
---
We can try proving the Extreme Value Theorem now, actually; it will show a rather intuitive path. So: we have a continuous function
![$f : [a, b] \to \mathbb{R}$](http://latex.artofproblemsolving.com/4/b/a/4baf2d51eb4d8376d7a95126b727c893b588c70d.png)
. Since we want to use continuity, our imagined proof will end with taking a limit of 
. In both cases it's pretty clear what we want the limit to be equal to, so here's what we end up with:First, we prove
is bounded on ![$[a, b]$](http://latex.artofproblemsolving.com/d/a/2/da2e551d2ca2155b8d8f4935d2e9757722c9bab6.png)
(otherwise where do we get a maximum?) If it isn't, then we can pick a sequence of points 
in so that 
as 
. Then we ???!!!???!!!, so we have 
so that 
, which is impossible as this limit should be defined and equal to 
(and thus finite).Next, we prove
's range is closed on . Suppose we know that for some value 
, we can pick 
so that 
is arbitrarily close to (i.e. is "on the boundary"). More precisely, we can pick a sequence of points in so that 
as . Then we ???!!!???!!!, so we have so that 
, so by continuity 
. Thus the boundary points of 's range are in 's range.Thus,
's range is closed and bounded.Q?!?E?!?D
Of course, this is not at all a proof, since there are two glaring red gaps here. What could we prove that fills them?
In both cases, we have a bunch of points
, and we want some point so that a subsequence of 
converges to . And that's all we need to prove now! (Essentially the same thing is proved as 2.37 in Rudin, but it doesn't seem to get used much.)Lemma. Let
be an infinite sequence of points in a closed interval . Then there exists a subsequence 
that converges to a point in .Proving this is also rather intuitive. We just want to find a place where there are a lot of points nearby. So we cut
in half, into ![$[a, (a+b)/2]$](http://latex.artofproblemsolving.com/1/f/c/1fc7d34b8b1d8a3c4150d0fb46e0c7dcd577155d.png)
and ![$[(a+b)/2, b]$](http://latex.artofproblemsolving.com/1/a/1/1a1a7ec5f2bd8bc310e3811a4a12adf12528575f.png)
. At least one of these intervals contains infinitely many points of the sequence .And we repeat! Keep cutting the interval with infinitely many points in half, and keep picking the half that still has infinitely many points. We get an infinite sequence of intervals, each of which is one of the halves of the previous one, and each of which contains infinitely many terms of the sequence.
How do we pick
? Well, we really want it to be in all of those intervals, and the least-upper-bound property comes in useful again: if the intervals are ![$[a_i, b_i]$](http://latex.artofproblemsolving.com/b/c/0/bc04bd6abf483862782136e613e9a11f028e0529.png)
, we pick it to be the least-upper-bound of all the 
's. So would be at least as large as every , and since each 
is an upper bound of all of the 's, this value of is at most as large as every . So is in all the intervals.Then to get a sequence that converges to
, we just build a sequence of 
s by starting with a random , then picking the next one from the next interval that comes after all the previous 's; since there are always infinitely many choices, such a choice always exists. And we're done!
May 2020
January 2020