Two model theory meetings

October 19, 2014

José Iovino has asked me to help advertise the following:

At the 2015 Joint mathematics meetings (JMM), January 10-13, in San Antonio, TX, there will be two special sessions on model theory.

The first is Beyond First Order Model Theory, a special session of the ASL and the AMS. See here for the schedule, list of speakers, and abstracts.

The second is Model Theory and Applications. See here for schedule, speakers, and abstracts.

The first session will be preceded by a conference (with the same title) at the University of Texas, San Antonio. Additional details can be found here, the speakers are Will Boney (University of Illinois at Chicago), H. Jerome Keisler (University of Wisconsin – Madison), Michael Makkai (McGill),  Maryanthe Malliaris (University of Chicago), Paul Larson (Miami University), Chris Laskowski (University of Maryland), and Sebastien Vasey (Carnegie Mellon).


414/514 References on continuous nowhere differentiable functions

October 19, 2014

Just as the last two times I have taught 414/514, I am assigning a final project on the topic of continuous nowhere differentiable functions (see here and here for the previous times).

The project requires that you choose an example of a continuous nowhere differentiable function, and to write a report describing the function, indicating who first introduced it, and presenting complete proofs of its continuity and nowhere differentiability. Additional information relevant for context is highly encouraged.

I am including links to two encyclopedic references on the subject. Feel free to follow the arguments there closely if needed, or to consult other sources, but make sure that what you turn in is your own version of the details of the argument, and that full details (rather than a sketch) are provided.

  1. Johan Thim’s Master thesis (Continuous nowhere differentiable functions), written under the supervision of Lech Maligranda.
  2. A.N. Singh’s short book on The theory and construction of non-differentiable functions. (See here for a short review.)

As I mentioned before,

Please take this project very seriously (in particular, do not copy details from books or papers, I want to see your own version of the details as you work through the arguments). Feel free to ask for feedback as you work on it; in fact, asking for feedback is a good idea. Do not wait until the last minute.

The project should be typeset and is due Wednesday, December 17 (though I strongly encourage you to turn it in earlier).

Please contact me by email as soon as you have chosen the topic you are going to cover, and I’ll list it here, to avoid repetitions.


414/514 Homework 2 – Monotone and Baire one functions

October 10, 2014

This set is due in two weeks, on Monday, October 27, at the beginning of lecture.

1. Let f:[a,b]\to\mathbb R be increasing. We know that f(x-) and f(x+) exist for all x\in[a,b], and that f has at most countably many points of discontinuity, say t_1,t_2,\dots For each i let I_i,J_i be the intervals (f(t_i-),f(t_i)) and (f(t_i),f(t_i+)). Some of these intervals may be empty, but for each i at least one of them is not. (Here we follow the convention that f(a-)=f(a) and f(b+)=f(b).) Let \mathrm{lh}(I) denote the length of the interval I, and say that an interval (\alpha,\beta) precedes a point t iff \beta\le t.

Verify that \sum_i(\mathrm{lh}(I_i)+\mathrm{lh}(J_i))<+\infty and, more generally, for any x,

s(x):=\sum\{\mathrm{lh}(I_i)\mid I_i precedes f(x)\} +\sum\{\mathrm{lh}(J_i)\mid J_i precedes f(x)\}<+\infty.

Define a function f_0:[a,b]\to\mathbb R by setting f_0(x)=f(x)-s(x). Show that f_0 is increasing and continuous.

Now, for each n>0, define f_n:[a,b]\to\mathbb R so that f_n\upharpoonright[a,t_n)=f_{n-1}\upharpoonright[a,t_n), f_n(t_n)=f_{n-1}(t_n)+\mathrm{lh}(I_n), and f_n(x)=f_{n-1}(x)+\mathrm{lh}(I_n)+\mathrm{lh}(J_n) for all x\in(t_n,b]. Show that each f_n is increasing, and its only discontinuity points are t_1,\dots,t_n.

Prove that f_n\to f uniformly.

Use this to provide a (new) proof that increasing functions are in Baire class one.

2. Solve exercise 3.Q in the van Rooij-Schikhof book: If f:[a,b]\to\mathbb R is such that for all x, we have that f(x-) and f(x+) exist, then f is the uniform limit of a sequence of step functions. The approach suggested in the book is the following:

Show that it suffices to argue that for every \epsilon>0 there is a step function s such that |f(x)-s(x)|<\epsilon for all x.

To do this, consider the set A=\{x\in[a,b]\mid there is a step function s on [a,x] such that |f(t)-s(t)|<\epsilon for all t\in[a,x]\}.

Show that A is non-empty. Show that if a\le y\le x and x\in A, then also y\in A. This shows that A is an interval {}[a,\alpha) or {}[a,\alpha], with \alpha\le b. Show that in fact the second possibility occurs, that is, \alpha\in A. For this, the assumption that f(\alpha-) exists is useful. Finally, show that \alpha=b. For this, use now the assumption that f(\alpha+) exists.

3. Show that if f:[a,b]\to\mathbb R is the pointwise limit of a sequence of functions f_1,f_2,\dots, then there is a dense subset X\subseteq [a,b] where the convergence is in fact uniform.

4. (This is example 1.1 in Andrew Bruckner’s Differentiation of real functions, CRM monograph series, AMS, 1994. MR1274044 (94m:26001).) We want to define a function f:[0,1]\to\mathbb R. Let C be the Cantor set in {}[0,1]. Whenever (a,b) is one of the components of the complement of C, we define f(x)=(2(x-a)/(b-a))-1 for x\in[a,b]. For x not covered by this case, we define f(x)=0. Verify that f is a Darboux continuous function, and that it is discontinuous at every point of C.

Verify that f is not of Baire class one, but that there is a Baire class one function that coincides with f except at (some of) the endpoints of intervals [a,b] as above.

Verify that f is in Baire class two.


414/514 Simple examples of Baire class one functions

October 6, 2014

Recall that a real-valued function f defined on an interval I is (in) Baire class one (\mathcal B_1) iff it is the pointwise limit of continuous functions.

Examples are continuous functions, of course, but functions in \mathcal B_1 do not need to be continuous. An easy example is the function f:[0,1]\to\mathbb R given by f(x)=0 if x\ne 1 and f(x)=1 if x=1. This is the pointwise limit of the functions f_n(x)=x^n. By the way, an easy modification of this example shows that any function that is zero except at finitely many points is in \mathcal B_1.

Step functions are another source of examples. Suppose that a=x_0<x_1<\dots<x_{n-1}<x_n=b and that s:[a,b]\to\mathbb R is constant on each (x_i,x_{i+1}). Then s is the pointwise limit of the functions s_k, defined as follows: Fix a decreasing sequence \epsilon_k converging to 0, with \epsilon_k\le 1/k and 2\epsilon_k<x_{i+1}-x_i for all i. Now define \hat s_k as the restriction of f to

\displaystyle \{x_0,x_1,\dots,x_n\}\cup\bigcup_{i=0}^{n-1}[x_i+\epsilon_k,x_{i+1}-\epsilon_k],

and let s_k:[a,b]\to\mathbb R extend \hat s_k by joining consecutive endpoints of the components of its domain with straight segments.

An important source of additional examples is the class of derivatives. Suppose f:\mathbb R\to\mathbb R and f(x)=g'(x) for all x. This is the pointwise limit of the functions f_n(x) given by

f_n(x)=\displaystyle\frac{g\left(x+\frac1n\right)-g(x)}{\frac1n}.

This simple construction does not quite work if f is defined on a bounded interval (as x+1/n may fall outside the interval for some values of x). We can modify this easily by using straight segments as in the case of step functions: Say f:[a,b]\to\mathbb R. For n large enough so 1/n<b-a, define f_n(x) as above for x\in[a,b-1/n], and now set f_n(b)=f(b) and extend f_n linearly in the interval {}[b-1/n,b].

Additional examples can be obtained by observing, first, that \mathcal B_1 is a real vector space, and second, that it is closed under uniform limits (the latter is not quite obvious). This gives us, for instance, that all monotone functions are in \mathcal B_1, since monotone functions are the uniform limit of step functions on bounded intervals: Given an increasing f:[a,b]\to\mathbb R, let f_n(x)=\lfloor nf(x)\rfloor/n. It follows that all functions of bounded variation are in \mathcal B_1, since any such function is the difference of two increasing functions.

Another interesting source of examples is characteristic functions. Given X\subseteq\mathbb R, the function \chi_X is in \mathcal B_1 iff X is both an \mathbf F_\sigma and a \mathbf G_\delta set.

On the other hand, \chi_{\mathbb Q} is not in \mathcal B_1, since it is discontinuous everywhere while Baire class one functions are continuous on a comeager set.


Comparability of cardinals from Zorn’s lemma

September 26, 2014

One of the basic consequences of the axiom of choice is that any two sets are comparable, that is, there is an injection from one into the other. The standard argument for this uses that choice is equivalent to the well-ordering theorem: One can prove (without choice) that any two well-ordered sets are comparable, and the well-ordering theorem states that any set is well-orderable.

If for some (foolhardy) reason (say, one is teaching an analysis or algebra course) one is interested in the result, but wants to avoid discussing the theory or well-orders, it seems desirable to have a proof based directly on Zorn’s lemma.

A few weeks ago, Sam Coskey and I found ourselves discussing such a proof. It turns out the argument, though not as well-known as may be expected, dates back at least to Chaim Samuel Hönig, and his short note Proof of the well-ordering of cardinal numbers, Proc. Amer. Math. Soc., 5, (1954), 312. MR0060558 (15,690a). It goes as follows: Let the sets A and B be given. Consider the collection of all partial injections f from A into B. That f is partial means that there is a C\subseteq A such that f:C\to B. Order this collection by extension: f\le f' iff f\subseteq f'. This poset satisfies the conditions of Zorn’s lemma, so it admits maximal elements. One easily verifies that, if f is maximal, then A is the domain of f, or B is its range. In either case, this gives us an injection from one of the sets into the other.

A natural extension of the idea allows us to recover that the class of cardinals is not just linearly ordered, but in fact well-ordered, but an additional use of the axiom of choice is needed now (namely, in the form: The product of non-empty sets is non-empty). Suppose first that \mathcal C is a set. We argue that one of the members of \mathcal C injects into all others. The proof is essentially the same as before: Let A\in \mathcal C, and consider the family of all sequences (f_B\mid B\in\mathcal C) such that for some A'\subset A and all B, we have that f_B:A'\to B is injective. This is a partial order under coordinatewise inclusion. Again, Zorn’s lemma applies, so there is a maximal element \vec f=(f_B\mid B\in \mathcal C); call A' the common domain of all the f_B. If A'=A, we are done. Else (and this is where the additional use of choice comes in), for some B, the range of f_B is B: Otherwise, we can pick a'\in A\setminus A' and a sequence (b_B\mid B\in\mathcal C) such that, for all B, b_B\in B\setminus f_B[A']. But then, setting f'_B=f_B\cup\{(a',b_B)\}, we see that (f'_B\mid B\in\mathcal C) contradicts the maximality of \vec f. The result follows: Letting B be such that f_B is onto, we see that B injects into A', and A' injects into all sets in \mathcal C.

Finally, consider a (proper) class \mathcal C. Again, fix A\in\mathcal C. Let \mathcal C' be the collection of subsets of A equipotent to sets in \mathcal C. Since \mathcal C' is a set, the previous analysis applies, and we can find a C\in \mathcal C that injects into all members of \mathcal C that inject into A. It follows that C actually injects into all members of \mathcal C. Otherwise, there is a B\in\mathcal C that C does not inject into. But then B itself injects into C, and therefore into A. But this means that C injects into B after all.


Coming attractions

September 26, 2014

cs


414/514 Homework 1 – The reals

September 11, 2014

This set is due in two weeks, on Friday September 26, at the beginning of lecture.

1. Recall that in the construction of the reals via Dedekind cuts, a real is simply the left set of a pair (A\mid B) in a cut of \mathbb Q, that is, a “real” is a set A\subset \mathbb Q that is non-empty, bounded above, closed to the left (meaning, if x\in A, y\in\mathbb Q, and y<x, then y\in A), and such that A has no maximum. We also have a copy \mathbb Q^* of \mathbb Q inside \mathbb R, given by the identification q\mapsto q^*:=\{t\in\mathbb Q\mid t<q\}. We left a few details to be verified when we presented this construction.

Let r be a real (in the sense just described). Define carefully the real -r (meaning, describe -r as a specific set of rationals, and verify that it is a real in the sense under discussion), and verify that -r+r=0, and that -r is the only real with this property.

Define carefully the product rs of reals r and s, and verify that the distributive property holds.

Check that \mathbb R is Dedekind-complete, that is, any cut of \mathbb R is realized. (S0, ignoring the formal difference between \mathbb Q and \mathbb Q^*, this version of \mathbb R is the Dedekind-completion of \mathbb Q, and this gives us that it is also the Dedekind completion of itself. )

2. More generally, define the Dedekind completion of a dense order, and verify its existence and uniqueness (up to isomorphism). In particular, the field \mathbb R(x) of  rational functions admits a completion, call it \hat{\mathbb R}(x). Can we extend the addition operation on \mathbb R(x) so it is defined in all of \hat{\mathbb R}(x) and makes it into an abelian group? Can we extend the order so \hat{\mathbb R}(x) is in fact an ordered group? What, if any, is the problem trying to extend multiplication?

3. Recall that in the construction of the reals via Cauchy sequences, a real is an equivalence class of Cauchy sequences of rationals, under the equivalence relation that states that two Cauchy sequences q_0,q_1,\dots and r_0,r_1,\dots are equivalent iff q_0,r_0,q_1,r_1,\dots is a Cauchy sequence.

Verify that this is indeed an equivalence relation, and that, given equivalent sequences \vec q and \vec r, the resulting interleaving sequence is equivalent to both. Verify that the (pointwise) definitions of addition and multiplication make sense, and that the resulting set equipped with these operations is indeed a field. Define carefully the ordering relation, and prove that it gives us a field ordering. Finally, verify that the resulting ordered field is indeed Dedekind complete.

4. Recall the construction of the reals described in Street’s paper An efficient construction of real numbers. His short note makes many claims that are not proved there. Provide as many of the missing details as possible.

5. Given a linear order (X,<), in the order topology the open sets are (by definition) those subsets of X that are union of (bounded or unbounded) open intervals in X. Show that a linear order (X,<) is order isomorphic to \mathbb R iff the following three properties are verified:

  • X has no first or last elements.
  • X is connected, that is, we cannot write X=A\cup B where A and B are open, nonempty, and disjoint.
  • X is separable, that is, there is a countable subset of X that is dense in X.

 

 


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