The 19-th annual meeting of BEST will be hosted at Boise State University during the weekend of March 27 (Saturday) – March 29 (Monday), 2010.

It is organized by Liljana Babinkostova, Andrés E. Caicedo, Masaru Kada, and Marion Scheepers (scientific committee), and Billy Hudson (social committee).

Contributed and invited talks will be held on Saturday, Sunday and Monday at the Department of Mathematics, Boise State University. The invited speakers curently include:

• Justin Moore (Cornell University)
• Frank Tall (University of Toronto)
• Toshimichi Usuba (University of Bonn)

The conference webpage is available here. Anyone interested in participating should contact the organizers as soon as possible by sending an email to best@math.boisestate.edu

There are three important deadlines regarding the conference:

• Lodging: The Hampton Inn & Suites is providing rooms at a reduced rate for BEST participants. To take advantage of the reduced rate, reservations must be made online by MARCH 12. After March 12 rooms will be available at prevailing rates.
• Financial support: Limited financial support is available to partially offset lodging expenses of up to eight participants, and to partially offset lodging plus airfare expenses of up to two participants. Please see the conference website for details on applying for support. The deadline for applying for financial support is MARCH 3.
• Abstracts: Atlas Conferences, Inc. is providing abstract services for BEST 19. The deadline for submitting an abstract for invited or contributed talk is MARCH 25. Abstracts can be submitted here, and viewed here.

1. Non-measurable sets

In these notes I want to present a proof of the Banach-Tarski paradox, a consequence of the axiom of choice that shows us that a naive understanding of the concept of volume can lead to contradictions. A good reference for this topic is the very nice book The Banach-Tarski paradox by Stan Wagon.

The result is one of a family of theorems indicating limitations of any reasonable notion of measure on the real numbers or in Euclidean space, and in this section I include a few examples. We will work with Lebesgue measure. The version of this measure for we denote Actually, we do not need to know much about Lebesgue measure. It suffices that has the following properties (and, really, we will only look at or ):

• is a measure, so is a function with domain a -algebra of subsets of and range and is -additive, i.e., if is a sequence of pairwise disjoint elements of the domain of (i.e., measurable sets), then

  

• If is the unit -cube, i.e.,

  

  then (Other than we do not really need this fact.)

• is non-atomic, in the sense that each singleton is measurable, and for all
• is invariant under the group of isometries of For example, if for measurable and let

  

  Then is measurable, and We say that is translation invariant. Also, if

  

  then is measurable, and In is not only translation invariant but also invariant under rotations and reflections. In fact, if

  

  for and measurable, then is measurable, and but we won’t be needing this fact.

is our formalization of the notion of length, similarly, and formalize the notions of area and volume.

Theorem 1 There is a subset of that is not measurable.

Proof: We present an argument due to Vitali. On say that iff This is an equivalence relation. Let be a choice set, so is a singleton for each -equivalence class Then

This is a countable union of disjoint measurable sets. By translation invariance (and monotonicity, a consequence of -additivity), if then each of the sets in the union has measure zero as well, which leads to the contradiction

Similarly,

and therefore, if then again a contradiction.

It follows that is not Lebesgue measurable.

Remark 1 Translation invariance is essential here. It is consistent with that Lebesgue measure can be extended to a measure defined on all subsets of

The example above used the axiom of choice explicitly, guaranteeing the existence of the set Another typical use of choice is the existence of a well-ordering of A set of measure zero is called null. Otherwise, it is non-null. Note that a non-null set needs not be measurable. The example below uses a bit more of the properties of and namely, Fubini’s theorem.

Theorem 2 No well-ordering of a non-null sets of reals is measurable (as a subset of ).

Proof: We claim that whenever is non-null and is a well-ordering of then is non-measurable. We proceed by contradiction. Say that is the least ordinal for which there is an enumeration of a non-null set such that is measurable.

For let and For let denote the unique such that

Note that is measurable: By Fubini’s theorem, for almost all and almost all both and are measurable. Since for any then must be measurable as well.

It suffices to show that for some is non-null and measurable. If so, is a measurable well-ordering of is order-type contradicting the minimality of

Now, for almost all is measurable. Thus if no is as claimed, then for almost all we have that is null. Hence,

But for almost all we have has positive measure. Contradiction.

Remark 2 The argument shows that, even if there is an extension of Lebesgue measure to all subsets of the completion of is not defined on all of

2. Tarski’s circle-squaring problem

In order to understand the statement of the Banach-Tarski result, it is convenient to first explain why it is stated for rather than the plane.

Theorem 3 (Banach, von Neumann) For or there is a finitely additive “measure” extending Lebesgue measure, defined on all of and invariant under isometries.

I won’t prove this result here. It is a consequence of a more general extension theorem, related to the fact that the group of isometries of the plane is amenable. Its proof requires choice. An obvious consequence of this result is that if a (measurable) figure in the plane has some area and it is cut out into finitely many pieces that are then rearranged into another measurable figure then

In contrast, the Banach-Tarski paradox allows us to divide a sphere the size of a pea into pieces that, when rearranged, make up a sphere the size of the sun. Clearly, this precludes the existence of an extension of Lebesgue measure in as in the Banach-von Neumann result. On the other hand, a version of the paradox can be obtained for . In order to discuss this, I need a few notions.

Definition 4 Two sets in are equidecomposable if one can be partitioned into finitely many disjoint sets (called pieces) that can be rearranged (via isometries) to form a partition of the other.

There is a well-known particular instance of equidecomposability that has been studied in some detail in the context of classical Euclidean geometry.

Definition 5 Two polygons in are equidissectable if, allowing overlap on boundaries, one can be split into finitely many triangles that can be rearranged (via isometries) into the other.

Sometimes one says that the polygons are congruent by dissection.

Theorem 6 (Bolyai-Gerwien) Two polygons are equidissectable iff they have the same area.

Proof: I only present a sketch. One direction is obvious. For the other, first one argues that a polygon is dissectable into triangles. Then, that any triangle is equidissectable with a square. This is done in two stages. First, one easily shows that a triangle is equidissectable with a rectangle. Some care is then needed to see that a rectangle is equidissectable with a square; but this is classical, it only goes a bit beyond constructing the geometric mean of two given numbers. Finally, one shows that two squares are equidissectable with a single square, and then induction completes the proof. The argument for two squares amounts to one of the well-known proofs of the Pythagorean theorem.

The version of the Banach-Tarski paradox in is a generalization of the following strengthening of the Bolyai-Gerwien result:

Theorem 7 (Banach-Tarski) Two polygons in are equidecomposable iff they have the same area.

The proof (that I omit) follows the same outline as the paradox in .

Tarski then asked for the possibility of extending Theorem 7 to other regions. In particular, he asked whether the circle and a square of the same area are equidecomposable. This question remained open until 1990, when M. Laczkovich solved it affirmatively, in his paper Equidecomposability and discrepancy; a solution of Tarski’s circle-squaring problem, Journal für die reine und angewandte Mathematik, 404 (1990), 77-117. Among his results, I highlight:

Theorem 8 (Laczkovich)

1. Any two polygons of the same area are equidecomposable using only translations.
2. Let be a piecewise smooth Jordan curve for which there are two positive constants such that the curvature at each point of is between and Then the domain enclosed by is equidecomposable to a square via translations alone.

The number of pieces required in the decomposition is rather large. Laczkovich computes that about pieces are required in the equidecomposition of an arbitrary isosceles right triangle and a square. In contrast, 5 pieces are required for the Banach-Tarski paradox in .

The pieces of the decomposition are also not explicitly definable; in particular, Dubins,Hirsch, and Karush showed that if only Jordan domains are used, then the circle and the square are not equidecomposable.

3. Paradoxical group actions

The full statement of the Banach-Tarski paradox in is as follows:

Theorem 9 Any two bounded subsets of with nonempty interior are equidecomposable.

We will prove a weaker result, namely, that a sphere is equidecomposable with two copies of itself. Recall that denotes the unit sphere in The argument takes three steps. First, we talk about paradoxical group actions and show that the free group on two generators, acts paradoxically on itself. This is then used to show Hausdorff’s paradox, that there is a countable set such that acts paradoxically on We conclude by showing the Banach-Tarski paradox itself, that acts paradoxically on

Recall that a group action is a map from a group into the group of bijections of a set into itself, such that:

• where is the identity of
• For all and

To ease readability, we will follow the usual convention of writing rather than and or even rather than Also, given and write or for

Definition 10 Let the group act on the set We say that acts paradoxically on iff there are pairwise disjoint subsets of

and corresponding elements of

such that

It is easy to check that if acts paradoxically on we can assume moreover that the pieces satisfy that see Fact 1. This explains the name: Note that then we have that is “equidecomposable” with two copies of itself, since and are both equidecomposable with where we are abusing notation, using subsets of rather than of and elements of rather than isometries.

Given a group acting on a set write to denote that that can be partitioned into finitely many pieces,

and that there are (not necessarily distinct) elements of such that, letting for then the are pairwise disjoint and partition

It is easy to verify that is an equivalence relation.

Fact 1 Given a group acting on a set we have that acts paradoxically on iff there are disjoint sets such that and

Proof: Verify that the argument for the Schröder-Bernstein theorem gives that if then

A trivial example of a paradoxical action is given by the group acting on via the action for an infinite set. This action is paradoxical, since any infinite is in bijection with two copies of itself. A more interesting example, essential to the argument, is as follows:

Theorem 11 acts paradoxically on itself by left multiplication.

Proof: Let be generators of and consider the following 4 subsets:

• 
• 
• and
• 

Note that are disjoint and partition Moreover,

and

This gives the result, as

The key tool we use to show that acts paradoxically on the sets in that interest us is the following result. Say that an action of a group on a set is without nontrivial fixed points iff the only element of that fixes a point of is the identity

Theorem 12 If acts on without nontrivial fixed points, then acts paradoxically on

Proof: Fix an action of on without nontrivial fixed points. Using the axiom of choice, let be a choice set picking an element of each orbit Note that for any there is a (unique) such that and therefore there is a such that In fact, there is a unique such Otherwise, there are and such that

and therefore Since is a choice set, then But then since the action of is without nontrivial fixed points.

Let be a paradoxical partition of so Let and By our observation on the paragraph above, and Since we immediately get

4. The Hausdorff paradox

In this section, I prove the following result:

Theorem 13 There is a countable set such that the natural action of the group of isometries of on is paradoxical.

In light of the results from last section, it suffices to find two isometries such that the group they generate is free, and acts on without nontrivial fixed points, for some appropriate countable set

There are many ways of doing this. Following Wagon’s suggestion, let be the counterclockwise rotation around the -axis by an angle of The matrix associated to is easily found to be:

Note that is the transpose of

Now let be the counterclockwise rotation around the -axis by an angle of As with it is easy to see that the matrix associated to is:

As with

Lemma 14 The group generated by is free.

Proof: I sketch the argument. Let be a reduced word in the alphabet We need to show that if is nontrivial (as a word), then it is not the identity (as an isometry). Since iff we may assume that ends in either or We say that is valid.

I claim that

for some integers with and In particular, this shows that does not map into itself, so

The claim is proved by induction on the length of The result is clear if since and is just the first column of that has the required form with and

Suppose now and we know the result for all shorter lengths. We have that or for some valid word We have that for some appropriate integers

Then

where and in the first case, or and in the second case.

In both cases, we have that are integers, and that All that remains is to argue that is not divisible by 3. For this, we consider the word such that or or or Note that is valid in the first and fourth cases. In the second and third cases, is valid unless it is the empty word. In any case, note that for some integers with we have

and unless is the empty word, in which case

From the formulas above, we see respectively that with or with or or In all cases, that now follows from the induction hypothesis, and we are done.

The Hausdorff paradox follows immediately: It suffices to show that there is a countable set such that the natural action of the group on is without nontrivial fixed points. But for any matrix other than the identity, either fixes no vector in or else it fixes exactly two, of the form and This is because either is not an eigenvalue of or else it is an eigenvalue of multiplicity 1: Note that if has multiplicity 3 as an eigenvalue, and it cannot have multiplicity 2, because is the product of the eigenvalues of

Now take as the set of all vectors in that are fixed by some Since is countable, and each contributes at most two vectors to is countable.

5. The Banach-Tarski paradox

We now conclude the proof of the following result:

Theorem 15 (Banach-Tarski) The group of isometries of acts paradoxically on

As mentioned earlier, there is a stronger form of the paradox where any two bounded sets with nonempty interior and equidecomposable. Several additional strengthenings are possible. For example, only 5 pieces are needed to witness the equidecomposition (and 4 do not suffice). Let be the group of isometries of Very recently, Trevor Wilson, then an undergraduate at Caltech, showed that the equidecomposition can be carried out continuously, i.e., say that and are bounded and with nonempty interior. Then there is a partition

and continuous functions

for such that

whenever and and

For the strong version of the Banach-Tarski paradox and the fact that precisely 5 pieces are needed, see Wagon’s book. For the continuous version, see Trevor Wilson, A continuous movement version of the Banach-Tarski paradox: A solution to de Groot’s problem, The Journal of Symbolic Logic, 70 (3), 2005, 946-952.

Proof: In view of the Hausdorff paradox, it suffices to show that, with the group of isometries of we have that whenever is countable. To see this, it suffices to check that there is a rotation such that are pairwise disjoint. because if that’s the case, letting

we have

To find note that there is a line that goes through the origin and misses Let be the set of angles such that for some there is a point such that where is the rotation around of radians.

Clearly, is countable, so there is an angle not in and we can take In effect, we have that for all But then for all as well, and we are done.

Typeset using LaTeX2WP. Here is a printable version of this post.

Here is the final exam, and here are the solutions.

This is the last homework assignment of the term: Assume Evaluate the cardinal number the size of the set of all  functions

In this set of notes I want to sketch Gödel’s proof that is consistent with the other axioms of set theory. Gödel’s argument goes well beyond this result; his identification of the class of constructible sets eventually led to the development of inner model theory, one of the main areas of active research within set theory nowadays.

A good additional reference for the material in these notes is Constructibility by Keith Devlin.

1. Definability

The idea behind the constructible universe is to only allow those sets that one must necessarily include. In effect, we are trying to find the smallest possible transitive class model of set theory.

is defined as

where for limit, and where

The first question that comes to mind is whether this definition even makes sense. In order to formalize this, we need to begin by coding a bit of logic inside set theory. The recursive constructions that we did at the beginning of the term now prove useful.

I won’t provide full details, but I expect that what follows allows you to fill in the gaps without difficulties.We begin by coding formulas inside set theory. For this, we begin by coding variables. What I mean is, recall that our official list of variables is even though we usually write We need to assign a set to each variable in a way that whenever we want to talk about the variable, we can instead talk about the set. Then we will do the same for all the logical symbols, and use this to code formulas.

One simple way of coding variables is letting the ordered pair code the variable Clearly, there is a formula

that holds of iff codes a variable, namely, states that is an ordered pair, that its first coordinate is 0, and its second coordinate is a natural number.

(Later, we will code some symbols with natural numbers. It is in order to distinguish between these symbols and the variables, that we use rather than simply to code )

It will prove convenient to use constants, to represent sets. We have a constant for each set . We want to code each constant by a set. One way could be to simply let the set act as the constant Unfortunately, this would be ambiguous, as when we have coded a formula by a set , we wouldn’t know whether is coding or the constant , for example. There are other issues as well. For this reason, we choose to code the constant by the ordered pair As before, there is a formula

which holds of iff codes a constant for some

Now that we have coded variables and constants, we can code formulas. We will represent the logical symbols as follows:

and other symbols, are treated as abbreviations.

We begin by coding atomic formulas. Remember that these are formulas of the form or where are either constants or variables. We simply code them by a finite sequence and, in general, we will use finite sequences to represent formulas. The sequence coding where is either or would be where codes , codes , codes , etc. Formally, we see that there is a formula

That holds iff codes an atomic formula: says that is a function with domain that and, for

To code arbitrary formulas, remember that to each we can associate a parsing sequence consisting of the subformulas of listed in such a way that whenever a formula is listed, all its subformulas have been listed before, and that if appears in a parsing sequence, then it is indeed a formula. We can then define

by saying that there is a function with domain some natural number and such that and, for all we have that

or there is a such that

where is the formula coded by (which we can take simply to mean that is a sequence of length that and ), or there are such that

where are the formulas coded by (meaning, is a sequence of length 5, and ), or there is , and a variable such that

where is the formula coded by (meaning, is a sequence of length 3, and ). Any function witnessing we can call a parsing witness for We can then define a class function

that, whenever holds, assigns to the set of free variables of the formula coded by Remember that this means that is a formula of two free variables, that for every there is a unique such that holds, and that if , then is as stated and, as usual, we write rather than The definition of is by recursion, using a parsing witness for . More carefully, we define by recursion, so that if fails, then Otherwise,

• If codes then is if exactly one of is a variable, and that variable is or if and or otherwise.
• If codes and codes then I.e., if and is a formula, then
• If codes and code then
• If codes and codes then

In order to do this, we use a parsing witness for One checks that, no matter what parsing witness is used, is always computed the same way.

We can then define a formula describing satisfiability. This is a formula

which holds iff codes a formula in the language of set theory (i.e., is the only non-logical symbol in , and the free variables of are exactly as shown), is a tuple of elements of , and, letting be the formula resulting from substituting in each free occurrence of by we have

(We may also write or even the admittedly ambguous for )

This definition is by recursion on the complexity of of course. More precisely, the definition refers to a parsing witness for and implements the inductive clauses of Tarski’s definition of truth in models. As before, there ends up being no difference, no matter what parsing witness is used. I omit additional details, but it should be routine at this point to write down in detail both the formula and the formula

The point of all this is that, now that we have a single formula which captures the notion of satisfiability, then we can define by

Note that now that we have a formula describing it follows that “” is also expressible by a formula, i.e., is indeed a class. Clearly, then, if is a set, then so is as

and this is a set, by comprehension.

Remark 1 We have taken advantage in this section of the fact that in set theory, we can directly use functions. Gödel’s incompleteness theorem for Peano Arithmetic begins in very much the same form, by coding logic inside number theory. But Gödel has the additional problem that, in number theory, we do not have functions but only numbers, and so he first needs to device a way of coding finite sequences by numbers.

Once this is done, there is still the problem that there does not seem to be a reasonable way of coding arbitrary models by numbers, so instead Gödel works not with satisfiability but with provability. The possibility of using models directly is another advantage of working within set theory.

Remark 2 On the other hand, although we have a formula that we can use to code all statements of the form for a set and a formula, we do not have a formula that holds of iff codes a formula that is true in This is a somewhat subtle point related to both a theorem of Tarski usually called “undefinability of truth” and to Gödel’s incompleteness theorem.

Although I won’t prove this fact, informally, the issue is that expanding the definition of for a formula with alternations of quantifiers requires itself alternations, so there is no formula that (provably in ) works for all And this is, ultimately, because is a proper class rather than a set.

2. Basic properties of

In this section we present some basic properties of and of the constructible hierarchy that we will require in the proof that The basic facts listed in the theorem below, and the fact that is a model of .

Theorem 1

 

 

 

1. For all we have
2. For all is transitive. Therefore, is transitive.
3. For all If then On the other hand, if is infinite,
4. Whenever then
5. For all In particular, contains all the ordinals.

 

Proof: We begin by arguing simultaneously by induction that, for all is transitive and for all

This is clear from the definition if is and from the definition and induction, if is a limit ordinal. Suppose now the result for and argue for For this, note that (by induction) it suffices to show that and that is transitive.

Let Then since is transitive. But then

This proves that

Now, if then so and, therefore, This shows that is transitive.

Now we show that for all by induction. Again, this is immediate from the definitions if is 0 or a limit ordinal. At successor ordinals, we have

We argue by induction that if This immediately gives as well that Suppose, then, that (this clearly holds for ). Let Then say

a finite set, since itself is finite. But then

This shows that and we already have the other containment.

On the other hand, if for infinite, then

where the first equality follows from noting that each element of is determined by a formula in the language of set theory (and there are only countably many of them) and a finite tuple of elements of

At limit stages we have that

where the sum indicates the size of a (possibly infinite) disjoint union.

We still need to check that for infinite. This is a consequence of the next item, that if This is because, in particular, it follows that and then of course

To prove the result, we only need to argue that The rest follows by induction. But

By induction, we have and a further inductive argument (looking for the least counterexample) gives that, in fact, But then

That follows immediately, and this completes the proof.

Before we prove that we need a particular case of the reflection theorem:

Lemma 2 Suppose that is a formula. Then there are arbitrarily large ordinals such that

for all tuples

In the situation of the lemma, we say that is reflected down from to .

Proof: This can be established by an argument very similar to the one we used in the notes on the Löwenheim-Sk\o lem theorem. Namely, for and each of its subformulas consider Sk\o lem functions from Let be the set of these Sk\o lem functions. Recall that if is closed under then for any we have that iff

Now, given any we describe an iterative procedure: Let be the closure of under Let be least such that In general, given consider its closure under and let be least such that Now note that if then

is closed under and therefore has the required property. Since was arbitrary, we are done.

We are now ready for the main result of this section:

Theorem 3 (Gödel)

Proof: Extensionality: This follows from transitivity of and extensionality in

Pairing: If then they belong to some Then

Union: If let be such that Then is definable in

Infinity:

Foundation: This follows from transitivity of and foundation in Given there is in a with Since is transitive,

Power set: Given we can let

Note that is a set, as it is the range of a function Let and

Then and i.e.,

Note there is no reason to suspect that is actually the true

Comprehension: Let let be a formula, and let be a tuple of elements of Write for the tuple of constants for in Let

We need to show that It is here that Lemma 2 is useful: Pick large enough so and for any

iff

Then

Replacement: Let be a formula and let be a tuple of sets in such that

Let We need to find a such that

since replacement for then follows by comprehension. Using replacement in , let the unique such that first appears in Then

and we are done.

Choice: Note that since is a model of the formalization of logic inside set theory can be carried out in It then makes sense to talk, for any of i.e., the set that in satisfies the definition of One can easily verify by induction the following two claims:

1. For all
2. For all

To prove that choice holds in begin by fixing some sufficiently definable enumeration of all (codes for) formulas without constant symbols (so the enumeration is in ). Given let:

• be the least such that
• be the least natural such that is definable over using and some parameters.

We now define with the properties that is a well-ordering of for all and the orders end-extend each other, i.e, and if and then The construction is carried out inside

Once this is done, we can then define a global well-ordering of all of by setting, for

iff either or else say, and

The definition of the orderings is carried out by recursion in and if is limit, Finally, given we define as follows: Given set

iff

• Either or else
• but or
• and but the tuple of parameters in used to define by means of the formula precedes the corresponding tuple for in the lexicographic ordering of tuples induced by

It is straightforward to check that the orderings so defined are indeed well-orderings, and we are done.

Remark 3 One can organize the proof of Comprehension in somewhat more carefully, to avoid the appeal to the axiom of choice implicit in the use of Sk\o lem functions in the proof of Lemma 2. Note that the argument above can then be formalized in since the proof that choice itself holds in does not use the axiom of choice in This establishes Gödel’s result that, if is consistent, then so is

3. The condensation lemma

In this section we sketch the fundamental condensation lemma, the last preliminary result needed in the proof of in The condensation lemma is in turn a consequence of the following basic result:

Lemma 4 (Mostowski collapsing lemma) If is a set and satisfies the axiom of extensionality, then there is a unique transitive set and a unique map that is an isomorphism between and Moreover, is the identity on any transitive subset of

Proof: Suppose that is as wanted. Then, for any with we have Also, if then there must be some such that and we have This means that

By induction on the rank of it follows that there is a unique map defined by recursion by means of the equation above. Letting be the range of it also follows that is unique.

By extensionality, it is routine to check that so defined is injective, and therefore a bijection between and In effect, suppose is of least rank such that there is some distinct from and such that

If there is some then, by definition of we have so (by definition of again) we must have with Since we have But then contradicts the minimality of the rank of A similar contradiction is obtained if

By definition of as the range of we have that is indeed transitive. It remains to verify that is an -isomorphism. By definition, for implies Suppose now that and Since is injective, it follows that and we are done.

That is the identity for any that is already transitive is now immediate.

is usually called the collapsing map, and is the collapse of

The condensation lemma extends the above when is an elementary substructure of an initial segment of by concluding that the transitive set it collapses to, is also an initial segment of

Theorem 5 (Condensation lemma) Suppose is a limit ordinal, and i.e., is an elementary substructure of Then there is a unique and a unique map such that is an isomorphism.

Proof: Begin by checking that To do this, argue by induction on that for all since every element of is definable. In particular, if there is nothing to prove. Now suppose

Note that satisfies the axiom of extensionality, and therefore so does The existence and uniqueness of now follows from Mostowski’s collapsing lemma. Let be the collapsing map.

We want to show that for some But this follows from checking that the construction of can be carried out inside with being the output. Hence, i.e.,

Then and one checks that for all It follows that Since each ordinal of is for some ordinal it is clear that

4. holds in

We conclude the course by showing:

Theorem 6 (Gödel)

Proof: We argue inside (or, if you wish, within the theory ). Let be a cardinal, and let be a subset of There is some such that and it suffices to show that the least such is strictly smaller than Because then and this set has size by Theorem 1.

To see this, let be large enough and a limit ordinal, so that Let be an elementary substructure of of size and such that Let be the transitive collapse of and let be the collapsing map. Note that as and is the identity on since is transitive and Since has size so does so has size Since we are done.

We have thus established that, if is consistent, then so is In particular, this shows that one cannot refute in On the other hand, in 1963, Paul Cohen devised a very powerful method, known as forcing that, in particular, can be used to show that if is consistent, then so are and This provides a concrete example of a natural mathematical statement (namely, ) verifying the incompleteness theorem for set theory.

Typeset using LaTeX2WP. Here is a printable version of this post.

Here is quiz 8.

Problem 1 asks to find the power series for .

Method 1: Using basic algebra, expand the cube:

This gives as a sum of powers of . Since there is only one way of doing this, this is the series we were looking for (and we have ).

Method 2: Let . Then we have:

Using that

we have

Problem 2 asks to find the power series for .

Perhaps the easiest way to proceed is to note that

and therefore

Problem 3 asks to find the radius of convergence of the series

As usual, we look at and , which we find by replacing with in the previous formula, so we have

We then take the quotient and simplify:

and this last expression converges to as . To see this, note that

By the ratio test, the series converges if and diverges if . Note that is the same as or i.e., .

Typeset using LaTeX2WP. Here is a printable version of this post.

Laurie Cavey, Wed. December 9, 2:40-3:30 pm, MG 120.

Developing Students’ Understanding of Mathematical Definitions: Why Bother?

Definitions are a fundamental part of doing mathematics, yet studies indicate that many students struggle to learn and apply definitions. In fact, many instructors wonder (myself included) how students can misapply definitions that are so clearly stated. Part of the issue is that a student’s previous mathematical experiences influence how she thinks, even when encountering a new idea that is seemingly unrelated. Not knowing what these experiences might entail, it can be difficult to know how to help students develop a better understanding of a particular definition. So, why bother? I will provide a brief overview of the research in this area including an instructional strategy (student generated examples) that may influence the way we think about developing students’ understanding of definitions.

Marion Scheepers, Wed. December 2, 2:40-3:30 pm, MG 120.

Cryptography

Online shopping and banking, Wireless communication and remote control devices have become common place. Nontrivial computing power and scanning devices of high power have become readily available. This creates an environment in which information in transit can be easily accessed or changed by unknown parties.

Cryptography is the main tool used to keep information secure. In this talk we will give a brief, motivated, outline of some of the mathematical foundations of cryptography. We also give an example to illustrate that mere possession of a good crypto-system does not guarantee security – one must also use it right.

Happy Thanksgiving!

Here is quiz 7.

Problem 1 asks to write out the first few terms of the following series to show how the series starts, and then find the sum of the series:

Let denote the sum of the first terms of the series. Then, for example,

and

The series is geometric, i.e., it has the form In this case, and Since the series converges, and adds up to

Problem 2 asks to explain why the following series converges or diverges, and if it converges, find its sum:

The series diverges. Perhaps the easiest way of checking this is by using the -th term test: If a series converges, then So, if then diverges. In this case,

so the series diverges.

Problem 3 asks to express the following number as the ratio of two integers:

To do this, simply note that

and that (except for the first term) this is a geometric series,

with and Hence, the series adds up to

Typeset using LaTeX2WP. Here is a printable version of this post.

Here is quiz 6.

Problem 1 is Exercise 7.7.64 from the book. It asks to determine the convergence of

There are several ways of approaching this problem.

Method 1: Begin by noting that

and so

can be evaluated by using the substitution so which transforms the integral into

that can be solved with the trig. substitution so and

Now we return to the problem:

We have found the value of the integral, so in particular, it converges.

Method 2: First,

We use comparison for both and

If Since

we have by comparison that is finite.

If Since

we have by comparison that is also finite.

Hence, converges.

Method 3: As before,

The change of variables transforms into

and

which of course is the same as

So all we need to do is to show that converges, and for this we can use comparison, because

and

Problem 2 is an easier version of Exercise 7.7.63 from the book. It asks to determine the convergence of

The problem here is that we cannot find so we are forced to use comparison. A natural thing to try would be to compare with

Since we have

The problem is that has an asymptote at 0. However, we can split the integral as

The first integral is finite, simply because it is the integral of a continuous function on some finite interval, there is nothing improper here. The second converges by comparison with the -integral with

It follows that the whole integral converges.

A slightly different approach is to use limit comparison rather than direct comparison:

It follows that converges because does; and we have to treat

as above. We cannot just compare with (which happens to diverge), because the limit comparison test requires that the functions that we compare are continuous, and is not continuous at 0.

Typeset using LaTeX2WP. Here is a printable version of this post.