## Extra Maths: Analysis, Darboux’s Theorem

As before, these questions are aimed at first-year or second-year undergraduates studying a course in Real Analysis.

Question 1. Let $f:[a,b]\longrightarrow\mathbb{R}$ be differentiable, with derivative $f'$. Suppose that $f'(a)>0>f'(b)$. Show there exists $c\in(a,b)$ such that $f'(c)=0$.

Compare Question 1 with Rolle’s Theorem.

Question 2 (Darboux’s Theorem). Let $f:\mathbb{R}\longrightarrow\mathbb{R}$ be differentiable, with derivative $f'$. Show that $f'$ has the intermediate value property.

Compare Question 2 with the Mean Value Theorem.

Question 3. Let $f:[a,b]\longrightarrow\mathbb{R}$ be differentiable, with derivative $f'$. Show that there exists $c\in[a,b]$ such that $f'$ is continuous at $c$.

This is very difficult! See this article. Nevertheless, I would like to expand on this argument in a subsequent post. Watch this space.

Question 4. Suppose that $g:\mathbb{R}\longrightarrow\mathbb{R}$ satisfies the intermediate value property. Need $g$ admit an anti-derivative?

Actually, this one is straightforward, so I’ll give the argument right away.

Proof. Certainly not. In fact we have just seen that every derivative (so, a function that admits an anti-derivative) is continuous somewhere. However, there are plenty of functions that satisfy the intermediate value property which are continuous nowhere. For example, there are functions whose graphs are dense in the plane. $\square$

Remark. For an in-depth discussions of the continuity of derivatives, see this article.

## Extra Maths: Analysis, Accumulation

Here are three fun Analysis questions, really aimed at my students who are studying their first/second course on Real Analysis, but hopefully it’s of wider interest as well.

We begin at the beginning, with a couple of definitions.

Definition. A function $f:\mathbb{R}\longrightarrow\mathbb{R}$ has the IVT Property if, for all real numbers $a,b,x$ with $a and $\min\{f(a),f(b)\} there exists $\xi\in[a,b]$ such that $f(\xi)=x$.

So, using this terminology, the Intermediate Value Theorem is the statement that a continuous function $\mathbb{R}\longrightarrow\mathbb{R}$ has the IVT Property.

Definition. Let $f:\mathbb{R}\longrightarrow\mathbb{R}$ and let $a,l\in\mathbb{R}$. We say that $l$ is a weak limit of $f$ at $i$ if
for all $\varepsilon>0$ there exists $b\in(a-\varepsilon,a+\varepsilon)\setminus\{a\}$ such that $f(b)\in(l-\varepsilon,l+\varepsilon)$. Naturally, we say that $f$ has a weak limit somewhere if there exists $a,l\in\mathbb{R}$ such that $l$ is a weak-limit of $f$ at $a$.

We say that $l\in\mathbb{R}$ is an accumulation point of a real sequence $(a_{n})$ if for all $\varepsilon>0$ there is an infinite set $N\subseteq\mathbb{N}$ such that for all $n\geq N$ we have $|a_{n}-l|<\varepsilon$.

The Questions

Question 1.
Show the existence or non-existence of a function $f:\mathbb{R}\longrightarrow\mathbb{R}$ which has the IVT Property but which is nowhere continuous.

Question 2.
Prove or disprove the claim that every function $f:\mathbb{R}\longrightarrow\mathbb{R}$ has a weak limit somewhere.

Question 3.
Find a real sequence $(a_{n})$ such that every real number $l\in\mathbb{R}$ is an accumulation point of $(a_{n})$.

Where should we start? Perhaps my favourite of the three is the first one: to show the existence or non-existence of a function with the IVT property but which is nowhere continuous. There are several approaches we might take, beginning with the following lemma.

Lemma 1. Suppose that a function $f:\mathbb{R}\longrightarrow\mathbb{R}$ has the property that, for all $a,b$ with $a, the restriction of $f$ to the interval $(a,b)$ is surjective onto $\mathbb{R}$. Then $f$ has the IVT property.

Hopefully the proof of this lemma is quite clear: the IVT property asks, of such a function restricted to such an interval, that it attain certain values. So if $f$ attains all real values, then the property is evidently satisfied.

Okay…. So what?

Lemma 2. Suppose that a function $f:\mathbb{R}\longrightarrow\mathbb{R}$ has the property that, for all $a,b$ with $a, the restriction of $f$ to the interval $(a,b)$ is surjective onto $\mathbb{R}$. Then $f$ is continuous nowhere.

Proof. Let $a\in\mathbb{R}$ be given, and consider, for example, $\varepsilon=1$. For any $\delta>0$, by assumption $f$ restricted to the open interval $(a-\delta,a+\delta)$ is surjective onto $\mathbb{R}$. In particular, the image under $f$ of $(a-\delta,a+\delta)$ is not a subset of $(f(a)-1,f(a)+1)$. $\square$

Ah-ha! Now it’s clear why we’re interested in these sorts of function. Such a locally surjective function‘ would be a nowhere continuous function which nevertheless has the IVT property. The big question now is: does any locally surjective function exist?

Well, YES. Such functions do exist, and there are several ways to find them. First we give a slightly abstract construction, which might seem a little unsatisfactory.

Method 1. Recall that the cardinality of $\mathbb{R}$ is $2^{\aleph_{0}}$, and the cardinality of $\mathbb{R}/\mathbb{Q}$ is also $2^{\aleph_{0}}$. We may choose a bijection

$F:\mathbb{R}/\mathbb{Q}\longrightarrow\mathbb{R}$.

Also, let $\phi:\mathbb{R}\longrightarrow\mathbb{R}/\mathbb{Q}$ be the natural map sending each real number $r$ to its coset modulo $\mathbb{Q}$, namely $r+\mathbb{Q}$. Note that $\phi$ is surjective. Thus, the composition $F\circ\phi:\mathbb{R}\longrightarrow\mathbb{R}$ is also a surjection. Indeed, by the density of the rationals in the reals, each proper open interval $(a,b)$ contains a representative of each coset of $\mathbb{Q}$ in $\mathbb{R}$. That is, for each $r\in\mathbb{R}$, there exists $q\in\mathbb{Q}$ such that $r+q\in(a,b)$. A consequence of this is that the restriction to $(a,b)$ of $\phi$ is still surjective onto $\mathbb{R}/\mathbb{Q}$. Finally, the restriction of $F\circ\phi$ to $(a,b)$ is still surjective onto $\mathbb{R}$, which shows that $F\circ\phi$ is locally surjective, as required.$\square$

Wonderful, that was easy. But it wasn’t particularly constructive. Perhaps it felt a little unsatisfactory to rely on the cardinalities of these sets. Is there a better way?

That’s where I’ll leave it for the moment. To be continued!

In the meantime, try to construct a locally surjective function by using the decimal expansion of each real number.

## Today was the first edition of Geometry Cafe’…

Today was the first edition of Geometry Cafe’ – a little lunchtime group to discuss various geometrical bits and bobs. I spoke, mostly about areas and volumes of various standard solids and plane figures.

For example, I went through a cute proof of the formula for the volume of a sphere, more or less rigorous, and without any calculus. The main tool is Cavalieri’s Principle.

Although the group wasn’t too well attended (more lecturers than students in the audience), I think I will continue it. There were several suggestions of topics, including Platonic Solids, some important plane curves, conic sections, knots and braids, non-Euclidean Geometry….

## Henselianity in the language of rings, with Franziska Jahnke

I would like to write about a new paper ([AJ15]) which Franziska Jahnke and I have written and put on the arXiv. It’s called Henselianity in the language of rings. In it we investigate the relationship between the following four properties of a field $K$:

(h) $K$ is henselian, i.e. $K$ admits a nontrivial henselian valuation,

(eh) $K$ is elementarily henselian, i.e. every $L\equiv K$ is henselian,

(def) $K$ admits a definable nontrivial henselian valuation, and

($\emptyset$-def) $K$ admits a $\emptyset$-definable nontrivial henselian valuation.

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## Puzzle corner: Solution to Sylvy’s puzzle #4

Cross-reference to puzzle page on my website

In this post I want to discuss the solution to Puzzle #4. Let’s recall the puzzle: it was to find an uncountable chain in $(\mathcal{P}(\mathbb{N}),\subseteq)$.

## Puzzle corner: Sylvy’s puzzle #5

Cross-reference to puzzle page on my website

Here is my next puzzle! Sorry that it’s been greatly delayed.

I mentioned previously the problem of constructing the midpoint of a given circle, using only ruler and compass. One student solved it nearly immediately (congratulations!), so here I offer a more general problem.

#### Sylvy’s puzzle #5

Given a parabola in the plane, construct its vertex using straightedge-and-compass construction (sometimes called ruler-and-compass’ construction).

## Sylvy’s Weekly Puzzle #3, Solution (part II)

Cross-reference to puzzle page on my website

In this post I want to carry on discussing the solution to puzzle #3. Recall that $\mathcal{S}$ denotes the world that Sonic explores in the Special Stage’. We are thinking of $\mathcal{S}$ as a topological space, upto topological equivalence.

We have already taken a look at the cases of the sphere and the torus, and we had shown that $\mathcal{S}$  cannot be a sphere and that $\mathcal{S}$ could be a torus. In fact we saw the map’ of the level: so it really IS a torus! But the question is: what can we say about $\mathcal{S}$ using only the topological information that sonic can gather?’

The final part of puzzle #3 asked us to see if $\mathcal{S}$ could be a Klein Bottle. A Klein bottle is the topological space obtained by taking a square and gluing together the opposite edges, as in the following diagram on the left. Note that the blue pair are glued together with a twist, while the red pair are glued straight’.

Gluing the edges of a square to form a Klein Bottle

Well, the answer is yes’. Let $m,n\in\mathbb{N}$ be such that $m$ is even and $n$ is odd. Let $C$ be a  rectangle which is divided into an $m\times n$ chessboard, coloured as usual so that white squares and black squares alternate. I think of the bottom’ or horizontal’ edges as being $m$-squares long, and the side’ or vertical’ edge as being $n$-squares long. Below on the right is a picture for the case $m=5,n=6$.

even-by-odd chessboard

Gluing the edges of this chessboard together in the way described above results in a Klein bottle which is tiled in the usual alternating black and white’ pattern that Sonic sees all around him. This shows that $\mathcal{S}$ could be a Klein bottle.

Note that we’ve only taken into account topological information. In later posts I want to tackle the geometric’ question.

## Sylvy’s weekly puzzle #4a

Cross-reference to puzzle page on my website

It seems that my first-year students (who form a large part of the target audience for my puzzles) are unfamiliar with the countable’ and uncountable’ terminology that I used in puzzle #4. …oops!

Instead of introducing it now, I’ll give a completely different puzzle which was originally shown to me by a friend (more on this later).

Sylvy’s weekly puzzle #4a

Show by means of a geometrical drawing’ that

$\arctan(1)+\arctan(2)+\arctan(3)=\pi$.

deadline: 11am Monday 9th of November

how to hand-in: send typed solution by email or put hand-written solution into the folder on my office door

prize: I’m very pleased to announce that the winner of this puzzle will get a book of mathematical puzzles!!

Have fun!

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## Sylvy’s weekly puzzle #4

Cross-reference to puzzle page on my website

This is an old one, but good fun.

Sylvy’s weekly puzzle #4

Let $\mathcal{P}(\mathbb{N})$ denote the powerset of the natural numbers. For clarity, we will consider 0 to be a natural number. Then the pair

$(\mathcal{P}(\mathbb{N}),\subseteq)$

is a partial order: the (weak) ordering is reflexive, antisymmetric, and transitive. But it is very definitely not a total order: given $X,Y\in\mathcal{P}(\mathbb{N})$ we may have neither $X\subseteq Y$ nor $Y\subseteq X$. A chain is a subset $\mathcal{C}$ of $\mathcal{P}(\mathbb{N})$ on which $\subseteq$ is a total order. For example:

$\mathcal{C}_{0}=\big\{\emptyset,\{0\},\{0,1,\},\{0,1,2\},...\big\}$

is a chain, but note that $\mathcal{C}_{0}$ is countable.

This week’s problem is to find an uncountable chain, that is a subset $\mathcal{C}$ of $\mathcal{P}(\mathbb{N})$ which is totally ordered by $\subseteq$ and uncountable.

Deadline: 10am Thursday 5th of November

(you can either send your solution by email or put a hand-written solution into the folder on my office door)

Prize: I’m very please to anounce that the winner will get a book of mathematical puzzles!!

Solution class: there will be a class at 10am on Thursday 5th of November in my office to go over the solution to this puzzle.

The webpage for these puzzles is http://anscombe.sdf.org/puzzle.html

Have fun!

## Two old-ish Terry Tao posts about model theory

I can’t get enough of Terry Tao’s blog What’s New, he writes so much and so brilliantly. Anyway, I found two really nice old-ish posts about model theory:

• this which talks about completeness, compactness, zeroth-order logic, and Skolemisation;
• and this which talks about nonstandard analysis, notions of elementary convergence’ and elementary completion’, countable saturation, compactness and saturation re-written from an analytical point-of-view, and the Szemerédi regularity lemma (something I always want to know more about).

I don’t have time to write anything more now, but later I will get back to it.

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