Question 1.Let be differentiable, with derivative . Suppose that . Show there exists such that .

Compare Question 1 with Rolle’s Theorem.

Question 2 (Darboux’s Theorem).Let be differentiable, with derivative . Show that has the intermediate value property.

Compare Question 2 with the Mean Value Theorem.

Question 3.Let be differentiable, with derivative . Show that there exists such that is continuous at .

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 satisfies the intermediate value property. Need 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.

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

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

Definition.A function has theIVT Propertyif, for all real numbers with and there exists such that .

So, using this terminology, the Intermediate Value Theorem is the statement that a continuous function has the IVT Property.

Definition.Let and let . We say that is aweak limitof at if

for all there exists such that . Naturally, we say thathas a weak limit somewhereif there exists such that is a weak-limit of at .We say that is an

accumulation pointof a real sequence if for all there is an infinite set such that for all we have .

**The Questions**

Question 1.

Show the existence or non-existence of a function which has the IVT Property but which is nowhere continuous.

Question 2.

Prove or disprove the claim that every function has a weak limit somewhere.

Question 3.

Find a real sequence such that every real number is an accumulation point of .

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 has the property that, for all with , the restriction of to the interval is surjective onto . Then 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 attains *all* real values, then the property is evidently satisfied.

Okay…. So what?

Lemma 2.Suppose that a function has the property that, for all with , the restriction of to the interval is surjective onto . Then is continuous nowhere.

Proof.Let be given, and consider, for example, . For any , by assumption restricted to the open interval is surjective onto . In particular, the image under of is not a subset of .

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 is , and the cardinality of is also . We may choose a bijection.

Also, let be the natural map sending each real number to its coset modulo , namely . Note that is surjective. Thus, the composition is also a surjection. Indeed, by the density of the rationals in the reals, each proper open interval contains a representative of each coset of in . That is, for each , there exists such that . A consequence of this is that the restriction to of is still surjective onto . Finally, the restriction of to is still surjective onto , which shows that is locally surjective, as required.

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.

]]>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….

]]>**(h)** is *henselian*, i.e. admits a nontrivial henselian valuation,

**(eh)** is *elementarily henselian*, i.e. every is henselian,

**(def) ** admits a definable nontrivial henselian valuation, and

**(-def) ** admits a -definable nontrivial henselian valuation.

By `definable’ we mean `definable in the language of rings’. Our paper is the latest work on this topic, which began with the work of Prestel and Ziegler who show, in [PZ78], that there exist t-henselian, nonhenselian fields. Subsequently, several more recent papers explore this topic and related issues of definable valuations; examples include [JK15] and [FJ15]. Our work builds on these three papers, and others. A more complete survey of the literature is available in our paper.

There are some trivial implications between the above properties:

and

Franziska and I investigate whether any other implications hold between these properties across the class of all fields of characteristic zero, and across the smaller classes

and,

for a prime;

where denotes the canonical henselian valuation on . Our main theorem gives the `complete picture’ for each of these classes

Main Theorem (Theorem 1.1, [AJ15])

- In the class the complete picture is

- For each prime , in the class the complete picture is

- Consequently, in the class the complete picture is simply given by (1) and (2), above.

Given the `trivial’ implications, (a) and (b) above, the proof of Theorem 1.1 boils down to showing the following:

- In the class the implication holds.
- For each prime , in the class the implications and hold.
- No implications hold other than the ones listed above and the `trivial’ ones.

The proof of (4) has two stages: first we show that using Theorem B from [JK15]; then we employ the Omitting Types Theorem to remove the parameter.

The proof of (5) is more tricky. The implication follows from basic properties of the canonical henselian valuation combined with some model-theoretic trickery. Franziska and I call these arguments `Sirince tricks’ because we wrote them down when we were on a research visit to the wonderful Nesin Mathematics Village, near Sirince, Turkey! The proof of the implication depends on some delicate arguments involving uniform definitions of canonical -henselian valuations.

Finally, the proof of (6) is a set of four counterexamples. The first (given by Prestel-Ziegler, [PZ78]) simply shows the existence (in equicharacteristic zero) of t-henselian nonhenselian fields. The second and third examples were given in [JK15] and are based on the construction in [PZ78]. Finally, the fourth example is based on a more sophisticated construction give in [FJ15] of t-henselian nonhenselian fields with various properties – we extend it to defectless fields of positive characteristic.

There are several key open questions. In my opinion, perhaps the most prominent is: can our results be extended to equicharacteristic ?

**[AJ15]** Sylvy Anscombe and Franziska Jahnke. *Henselianity in the language of rings*. Manuscript, 2015. arXiv

**[FJ15]** Arno Fehm and Franziska Jahnke. *On the quantifier complexity of definable canonical henselian valuations*. Mathematical Logic Quarterly, 61(4-5):347-361, 2015.

**[JK15]** Franziska Jahnke and Jochen Koenigsmann. *Defining coarsenings of valuations*. To appear in the Proceedings of the Edinburgh Mathematical Society, 2015. arXiv

**[PZ78]** Alexander Prestel and Martin Ziegler. *Model-theoretic methods in the theory of topological fields*. Journal für die reine und angewandte Mathematik, 299(300):318-341, 1978.

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 .

First we note that and have the same cardinality, i.e. . There are several well-known bijections that are often used to show this, but as I was writing this post I was reminded that none of the bijections that I’m familiar with are *explicit* in the sense of being given by some sort of `formula’. Here is a stackexchange discussion of this point (in particular, see the answer by Thomas Andrews).

Anyway, for the sake of this puzzle, we just need some bijection .

Now, for each , let be the preimage (under ) of the set of rational numbers in the interval .

**Proposition ** is an uncountable chain

This doesn’t need too much proof. Let be real numbers. If is such that then also . Thus .

Furthermore, there exists a rational number such that [why?]. Therefore , and so . This shows that , as required.

]]>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.

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

I want to give a rough description of what it means to perform a `straightedge-and-compass’ construction. We work in the Euclidean plane . You may be familiar with using a ruler and pair of compasses to construct geometrical drawings. For example, if you are given two points you can use the compasses to draw a circle with centre and radius .

Further, you can then construct a regular hexagon inside that circle such that the edge-length of the hexagon is equal to the radius of the circle. If you’ve not done it before, try it! It’s quite fun!

The issue arises: what shapes can one draw in this way? And which points in the plane can arise as intersections of circles/lines that have been constructed in this way?

We start with two points , as above. What can we do with ? We can draw a unique straight line which passes through , we can draw a circle with centre and radius , and we can draw a circle with centre and radius .

These three `circle-lines’ intersect in various points: draw them! The two circles intersect in two points; the line intersects in two points, one of which is ; and the line intersects in two points, one of which is .

Anyway, we get four new points. Using these new points, together with in various ways, we can construct other circle-lines. Then we can consider the intersections of *those*. And so on. All of these points that we obtain are said to be *constructible*. The key point is that there must be some *finite* sequence of points and circle-line constructions to get from to .

For simplicity (and so that we don’t have to worry about our choice of and ), we usually set to be the origin and set to be .

**Deadline: **11am Thursday 3rd of December

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

**Prize: **something chocolatey? I’m open to suggestions…

**Solution: **I’ll post the solution soon after the deadline.

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

Have fun!

]]>In this post I want to carry on discussing the solution to puzzle #3. Recall that denotes the world that Sonic explores in the `Special Stage’. We are thinking of 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 *cannot be* a sphere and that *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 using only the topological information that sonic can gather?’

The final part of puzzle #3 asked us to see if 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’.

Well, the answer is `yes’. Let be such that is even and is odd. Let be a rectangle which is divided into an chessboard, coloured as usual so that white squares and black squares alternate. I think of the `bottom’ or `horizontal’ edges as being -squares long, and the `side’ or `vertical’ edge as being -squares long. Below on the right is a picture for the case .

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 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.

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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

.

**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!!

**webpage:** http://anscombe.sdf.org/puzzle.html

Have fun!

]]>This is an old one, but good fun.

**Sylvy’s weekly puzzle #4**

Let denote the powerset of the natural numbers. For clarity, we *will* consider 0 to be a natural number. Then the pair

is a *partial order*: the (weak) ordering is reflexive, antisymmetric, and transitive. But it is very definitely **not** a *total order*: given we may have neither nor . A *chain* is a subset of on which **is** a total order. For example:

is a chain, but note that is countable.

This week’s problem is to find an *uncountable* chain, that is a subset of which is totally ordered by 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!

]]>- 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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