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  • sylvyanscombe 2:23 pm on February 3, 2019 Permalink | Reply
    Tags: analysis   

    Three Analysis Questions: 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<b and \min\{f(a),f(b)\}<x<\max\{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<b, 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<b, 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.

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  • sylvyanscombe 8:24 pm on October 3, 2018 Permalink | Reply
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    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….

     
  • sylvyanscombe 5:52 pm on December 20, 2015 Permalink | Reply
    Tags: , valued fields   

    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.

    (More …)

     
  • sylvyanscombe 11:48 am on December 18, 2015 Permalink | Reply
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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).

    Solution

    (More …)

     
  • sylvyanscombe 6:31 pm on November 10, 2015 Permalink | Reply
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    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).

    (More …)

     
  • sylvyanscombe 10:00 am on November 10, 2015 Permalink | Reply
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    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

    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.

     
  • sylvyanscombe 9:16 pm on November 4, 2015 Permalink | Reply
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    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!!

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

    Have fun!

     
  • sylvyanscombe 10:40 pm on October 28, 2015 Permalink | Reply
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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!

     
  • sylvyanscombe 9:56 pm on October 28, 2015 Permalink | Reply
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    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.

     

     
  • sylvyanscombe 9:38 pm on October 28, 2015 Permalink | Reply
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    Sylvy’s Weekly Puzzle #3, Solution (part I) 

    Cross-reference to puzzle page on my website

    I’m really pleased at the interest the third puzzle has generated! Or maybe I just haven’t been able to stop talking about it… 🙂 Anyway, here is the first part of the solution:

    The question asked us to investigate the world about which Sonic runs around in a certain special level. Let’s call that world \mathcal{S}. I want to describe \mathcal{S} as a topological suface. If you’ve not come across this notion before, you might like to think of a few key examples:

    • the (real) plane \mathbb{R}^{2},
    • the open unit disk D^{2}=\{(x,y)\in\mathbb{R}^{2}\;|\;x^{2}+y^{2}<1\}, and
    • the sphere S^{2}=\{(x,y,z)\in\mathbb{R}^{3}\;|\;x^{2}+y^{2}+z^{2}=1\}.

    Slightly more exotic examples:

    • the Möbius band,
    • the Torus,
    • (variations on the theme…) multi-handled tori, and
    • the Klein bottle.

    Perhaps now isn’t the time for a complete exposition of topological surfaces (okay, I may have got part-way through drafting one…!), that may come later. For the present I just want to say a few words about the Euler characteristic (which is denoted \chi) of a surface, and the formula that goes with it:

    V-E+F=\chi\qquad(\star_{1}).

    The first time I ever heard about this it was in the context of the Platonic solids, as follows.

    • Let’s begin by thinking of a cube. A cube has 8 corners (we’re going to call these vertices), 12 edges, and 6 faces. If V denotes the number of vertices, E the number of edges, and F the number of faces; then we have V-E+F=8-12+6=2.
    • Next, let’s think of a tetrahedron. In this case there are 4 vertices, 6 edges, and 4 faces; thus we have V-E+F=4-6+4=2.

    (Can you see where this is going?)

    • Now let’s think of an octahedron. Here there are 6 vertices, 12 edges, and 8 faces; we have V-E+F=6-12+8=2.

    You can check that the same formula

    V-E+F=2\qquad(\star_{2})

    holds for the other Platonic solids – but it doesn’t stop there. What about other solids, for example the Archimedean solid the rhombicuboctahedron?

    (I picked this one just because it has such a beautiful name.) The formula holds here too: we have V=24, E=48, F=26. Therefore (\star_{2}) holds for this solid:

    V-E+F=24-48+26=2.

    In fact:

    The formula (\star_{2}) holds for all polyhedra that are topologically equivalent to the sphere \mathcal{S}^{2}.

    Although I won’t properly define `topological equivalence’ here, let me just say that – roughly speaking – two shapes A and B are topologically equivalent if one can be transformed into the other by continuous shrinking, stretching, bending, twisting, etc. (discontinuous transformations such as tearing or joining are not allowed). Each of the polyhedra discussed above is topologically equivalent to the sphere.

    Let’s now think about the surface \mathcal{S} that sonic explores. It is given to us equipped with a vertices/edges/faces structure so we can calculate V-E+F. On \mathcal{S}, each face has four edges and each vertex is the endpoint of four edges. We don’t seem to know how many faces there are, so let n be the number of faces. Then there are 2n edges and n vertices. [Why?] Thus $V-E+F=0$. This shows that \mathcal{S} is not topologically equivalent to a sphere. This is a negative answer to the first part of the puzzle.

    On the other hand, \mathcal{S} could be a torus, as shown by the following map for one of these levels:

    [Map of a Special Stage in Sonic 3, copyright Sega (fair use of image for educational purposes)]

    How does this show that \mathcal{S} could be a Torus? You can see quite easily that the programmers have joined the left edge to the right edge and joined the top edge to the bottom edge. Thus, if Sonic walks off the left edge of the map, he will just pass round to the right edge (without noticing!).  This shape is quite easily seen to be a Torus. This is a positive answer to the second part of the puzzle.

    In the third part, I asked about whether \mathcal{S} could be a Klein bottle, and I’ll write a second post with the answer.

    [Edit: perhaps it’s unfair to use the map of the level that I found online to argue that \mathcal{S} might be a Torus. In fact the map (plus the description of how the edges are `obviously’ joined together) shows that \mathcal{S} definitely is a Torus. If you didn’t have the map of the level then I think it’s reasonable to simply argue that a Torus can admit a vertex/edges/faces structure as seen in the level (i.e. with `square’ faces joined in fours at vertices). We’ll address the question of whether \mathcal{S} must be a Torus in the second part of this answer, when I’ll also discuss the Klein bottle.]

     
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