# College Math Teaching

## January 30, 2015

### Nilpotent ring elements

Filed under: advanced mathematics, algebra, matrix algebra, ring theory — Tags: , — collegemathteaching @ 3:12 am

I’ve been trying to brush up on ring theory; it has been a long time since I studied rings in any depth and I need some ring theory to do some work in topology. In a previous post, I talked about ideal topologies and I might discuss divisor toplogies (starting with the ring of integers).

So, I grabbed an old text, skimmed the first part and came across an exercise:

an element $x \in R$ is nilpotent if there is some positive integer $n$ such that $x^n = 0$. So, given $x, y$ nilpotent in a commutative ring $R$ one has to show that $x+y$ is also nilpotent and that this result might not hold if $R$ is not a commutative ring.

Examples: in the ring $Z_9, 3^2 =0$ so $3$ is nilpotent. In the matrix ring of 2 by 2 matrices,

$\left( \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right)$ and $\left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right)$ are both nilpotent elements, though their sum:

$\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$ is not; the square of this matrix is the identity matrix.

Immediately I thought to let $m, n$ be the smallest integers for $x^m =y^n = 0$ and thought to apply the binomial theorem to $(x+y)^{mn}$ (of course that is overkill; it is simpler to use $(x+y)^{m+n}$. Lets use $(x+y)^{m+n}$. I could easily see why $x^{m+n} = y^{m+n} =0$ but why were the middle terms ${m+n \choose k} x^{(m+n)-k}y^k$ also zero?

Then it dawned on me: $x^n=0 \rightarrow x^{n+k}=0$ for all $k \geq 0$. Duh. Now it made sense. ðŸ™‚

## January 28, 2015

### Prime Ideal Topology

Filed under: Uncategorized — collegemathteaching @ 11:38 am

Note: this is just a watered down version of the Zariski topology on the spectrum of a commutative ring. I got the idea from Steen and Seecbach’s book Counterexamples in Topology.

I am presenting the idea while attempting to use as little ring theory as possible, as some of my students have not had abstract algebra as yet.

Consider the integers $Z = \{....-3, -2, -1, 0, 1, 2, 3,... \}$. An ideal $I$ is a subset of $Z$ that consists of 0 and all multiples of a given integer. The smallest positive integer $k$ in an ideal is the generator of that ideal; we denote that ideal by $(k)$.

Examples: $(1) =Z$ since every integer is a multiple of 1, $(2) = \{ ...-6, -4, -2, 0, 2, 4, 6, ... \}$ and $(6) = \{ ... -12, -6, 0, 6, 12, 18, ... \}$.

An ideal is PRIME if it is generated by a prime number. Now if $p$ is a prime and $x \notin (p), y \notin (p)$ $xy \notin (p)$ because neither has $p$ as a prime factor. So the prime ideals are those ideals whose compliments are multiplicatively closed.

Consider $X$, the set of prime ideals of $Z$. That is, $X = \{ (0), (2), (3), (5), (7), (11), ... \}$ The elements (points) of $X$ are prime ideals. Yes, $(0)$ is a prime ideal because if $ab =0$ then $a = 0$ or $b = 0$. $(1)$ is not a prime ideal because $(1) = Z$ and a prime ideal cannot be all of $Z$.

Let’s create a basis for a topology on $X$: let $V_x = \{ I \in X, x \notin I \}$. We are indexing subsets of prime ideals by positive integers and zero. Now $V_0 = \emptyset$ as every ideal contains zero. $V_1 = X$ as no prime ideal contains $1$. Now if $x \in Z$, $x$ has a prime factorization which contains prime factors $p_1, p_2, ...p_k$, so $x \in \cap^k_{i=1} (p_i)$ so $V_x = X -\cup_{i=1}^k (p_i)$. That is, the open basis elements are those collections of prime ideals that have a finite complement (those generated by the prime factorization of $x$.

Examples: $V_2 = X - (2), V_8 = X - (2), V_{60} = X - \{(2) \cup (3) \cup (5) \}$.

What is a closed set in this topology? A set $C \subset X$ is closed if there exists some ideal $I$ (not necessarily prime) such that $C = \{ P \in X | I \subset P \}$. For example, $C = \{ (2), (3), (5) \}$ is a closed set because $X -C$ is open. And note $(30) \subset (2), (30) \subset (3), (30) \subset (5)$. Note: traditionally, the Zariski topology is defined in terms of closed sets.

Clearly a finite union of closed sets is closed and an arbitrary intersection of closed sets is closed.

Now this topology is irreducible, which means that every non-empty pair of open sets intersect as all of them contain $(0)$. Remember: $V_0 = \emptyset$ as every ideal contains 0. Hence, this topology is not Hausdorff nor is it $T_1$ ( a topology is $T_1$ if every two points lie in different open sets, though these sets may intersect each other). This does have the property of being $T_0$ in that, given two points, there is at least one open set that does not contain both of the points as $(p) \notin V_p$ for $p$ prime.

## January 25, 2015

### An interesting topological space

Filed under: advanced mathematics, topology — Tags: , — collegemathteaching @ 2:46 pm

I am teaching an undergraduate topology course this semester. While we are still going through basic set theory, I’ve been racing ahead and looking at examples. Yes, Counterexamples in Topology (Steen and Seebach) is an excellent reference. In fact, I have two copies! ðŸ™‚

One space that caught my eye is Alexandroff Square. Take the usual closed $[0,1] \times [0,1]$ square in the plane. Now if $(x,y)$ is a non-diagonal point, let a local basis be open segments of the form $\{x \} \times (y - \epsilon, y + \epsilon )$, that is, small open vertical line segments that miss the diagonal. Open sets that include diagonal points $(x,x)$ are open horizontal strips $[0,1] \times (x + \epsilon, x - \epsilon)$ MINUS a finite number of vertical line segments $\{x_i \} \times (x + \epsilon, x - \epsilon)$.

This topological space is compact, Hausdorff, regular (that is, $T_3$) and normal (that is, $T_4$)

(quick reminder: Hausdorff (or $T_2$) means that any two points lie in disjoint open sets, Regular means that a point and a disjoint closed set can be separated by mutually disjoint open sets, and normal means that two disjoint closed sets can be separated by mutually disjoint open sets.)

One unusual aspect (to me) about this topology is how different the open sets are; there should be a way of characterizing this property. In a sense, the collection of open sets isn’t homogeneous.

I’ve decided to play with a simpler example that is based on the Alexandroff square:

consider the closed interval $[-1,1]$ For all $x \neq 0$ use the discrete topology. For $0$ use the entire interval minus any finite set as a local basis. Then one obtains many of the same features of the Alexandroff square, for similar reasons.

So, what do I mean by “different types” of open sets for different points?
This might work: let $x, y \in X$. Now I’d say that the open neighborhoods for $x, y$ are similar if, for every open neighborhood $U_x$ containing $x$ there is a continuous bijection $f|U_x \rightarrow X$ where $f(x) = y$ and $f(U_x)$ is an open neighborhood of $y$. That is, $f|U_x$ is a homeomorphism.

Let’s turn to the Alexandroff square for a minute. If $p$ is not a diagonal point, choose $U_p$ to be vertical open line segment. Now let $q$ be a diagonal element; every open $U_q$ has an open subset which is homeomorphic to the standard 2-disk in the plane (usual topology). Hence no local homemorphism from $U_p$ to $U_q$ can exist.

## January 23, 2015

### Making a math professor happy…

Filed under: calculus, class room experiment, elementary mathematics — Tags: , — collegemathteaching @ 10:28 pm

Calculus III: we are talking about polar curves. I give the usual lesson about how to graph $r = sin(2 \theta)$ and $r = sin(3 \theta)$ and give the usual “if $n$ is even, the graph of $r = sin (n \theta)$ has $2n$ petals and if $n$ is odd, it has $n$ petals.

Question: “does that mean it is impossible to have a graph with 6 petals then”? ðŸ™‚

Yes, one can have intersecting petals and one try: $r = |sin(3 \theta) |$. But you aren’t going to get it without a trick of some sort.

## January 17, 2015

### Convergence of functions and nets (from advanced calculus)

Sequences are a very useful tool but they are inadequate in some astonishingly elementary applications.

Take the case of pointwise convergence of functions (studied earlier in this blog here and here. )

Let’s look at the very simple example: $f(x) = 0$ for all $x \in R$. Yes, this is just the constant function.

Now consider a set $A$ which consists of all functions $g(x) = 0$ if $x \in F$ where $F$ is some finite set of points, and $g(x) =1$ otherwise. Remember $|F| = k 0$ there exists $m$ such that for all $k \geq m$ we have $|h_k(x) -h(x) | < \epsilon$. Note: the $m$ can vary with $x$.

Now recall that $A \subset R^R$. Now what topology are we using in $R^R$, since this is a product space? For pointwise convergence, we use the product topology in which the open sets are the usual open sets for a finite number of values in $R$ and the entire real line for the remaining values.

If this seems strange, consider the easier case $R^N$ where $N$ represents the natural numbers. Then we can view elements of $R^N$ as sequences, each of which takes a real value. And the open sets here will be the “sequences” $(U_1, U_2, U_3, ....U_k, ...)$ where the $U_i$ are copies of the real line, except for a finite number of indices in which case the $U_i$ can be some arbitrary open set in the real line.

For $R^R$, we index by the real numbers instead of by $N$.

So, in terms of pointwise convergence, this means for every $x \in R$ the $\epsilon_x$ is associated with the open set in the product topology where the real line factor associated with that value of $x$ has the usual real line open sets and the remaining factors of $R$ just have the whole real line; in short, they don’t really matter when figuring out of this function converges AT THIS VALUE OF $x$.

So with the product topology in place, look at our $f(x) = 0$ for all $x$. Given ANY open set $U \subset R^R, f \in U$, we see that $U \cap A \ne \emptyset$. For example, if $h(x) = 1$ for $x \ne 0$, $h(0) = 0$ and $U$ is the open set which corresponds to $(-\delta, \delta)$ in the $0$ factor and the real line elsewhere, then $h \in U \cap A$. So, we conclude that $f$ is in the topological closure of $A$.

But THERE IS NO SEQUENCE IN $A$ which converges to $f$. In fact, it is relatively easy to see that if $g_i$ is any sequence in $A$ and if $g_i \rightarrow g$, then $g$ is zero in at most a countable number of points. That is, if we use the Lebesgue integral, $\int^b_a g(x) dx = b-a.$ (of course, $g$ might not be Riemann integrable).

So sequences cannot reach a point in the closure. For the experts: this shows that $R^R$ in the product topology is not a first countable topological space; that is, its topology has no countable neighborhood basis. This also implies that it is not a metric space (or, more precisely, can’t be made into a metric space).

But I am digressing. The point is that, in situations like this, we want another tool to take the place of a sequence. That will be called a net.

Nets and Directed Sets
Roughly, a net is a “sequence like” thing that can be indexed by an uncountable set. But this indexing set needs to have a “direction like” quality to it. So, what works are “directed sets”.

A directed set is a collection of objects $a_{I}$ with a relation $\preccurlyeq$ that satisfy the following properties:

1) $a_I \preccurlyeq a_I$ (reflexive property)

2) If $a_I \preccurlyeq a_J$ and $a_J \preccurlyeq a_K$ then $a_I \preccurlyeq a_K$ (transitive property)

3) Given $a_I$ and $a_J$ there exists $a_K$ where $a_J \preccurlyeq a_K$ and $a_I \preccurlyeq a_K$ (direction)

Note: though a directed set could be an ordered set (say, $R$ with the usual order relation) or a partially ordered set (say, subsets ordered by inclusion), they don’t have to be. For example: one can form a directed set out of the complex numbers by declaring $w \preccurlyeq z$ if $|w| \leq |z|$. Then note $i \preccurlyeq 1$ and $1 \preccurlyeq i$ but $1 \ne i$.

Now a net is a map from a directed set into a space. It is often denoted by $x_I$ ($I$ is an element in the index set, which is a directed set). So, a real valued net indexed by the reals is, well, an object in $R^R$.

Now given a set $U$ in a topological space, we say that a net $x_I$ is eventually in $U$ if there is an index $J$ such that, for all $K \succcurlyeq J$, $x_K \in U$ and we say that $x_I$ is eventually in $U$. We say that a net $x_J \rightarrow x$ if for all open sets $U, x \in U$ we have $x_J$ is eventually in $U$.

Now getting back to our function example: we CAN come up with a net in $A$ that converges to our function $f$; we merely have to be clever at how we choose our index set though. One way: make a directed set $g_I \in A$ by declaring $g_I \preccurlyeq g_J$ if $g^{-1}_I (0) \subset g^{-1}_J(0)$. Now if we take any neighborhood of $f$ in the product topology, (remember that this consists of the product of a finite number of the usual open set in the real line with an infinite number of copies of the real line), we have elements of this net eventually in this open set, namely the functions which are zero for the values of $R$ that correspond to those open sets. (see here for a couple of ways of doing this)

This demonstrates the usefulness of nets. Note that trying to use a “sequence idea” by just starting with a function that is zero at exactly one point and then going to a function that is zero at two points, three points,…can only get you to a function that is zero at a countable number of points, which is NOT in $A$. That is, one “leaves $A$ prior to getting to where one wants to go, which is a function that is zero at all points of the real line.

On the other hand, a directed set can “start” at an uncountable number of elements of $A$ to begin with and get to being eventually in any basic open set containing $f$ in a finite number of steps. Of course, one must allow for an uncountable number of sequence like paths to get into any of the uncountable number of basic open sets, but each path consists of only a finite number of steps.

### Math professors: should you leave academia for the private sector?

Filed under: editorial — Tags: , — collegemathteaching @ 3:55 am

Maybe, maybe not. Here is a post from someone who did.

## January 16, 2015

### Power sets, Function spaces and puzzling notation

I’ll probably be posting point-set topology stuff due to my being excited about teaching the course…finally.

Power sets and exponent notation
If $A$ is a set, then the power set of $A$, often denoted by $2^A$, is a set that consists of all subsets of $A$.

For example, if $A = \{1, 2, 3 \}$, then $2^A = \{ \emptyset , \{1 \}, \{ 2 \}, \{3 \}, \{1, 2 \}, \{1,3 \}, \{2, 3 \}, \{1, 2, 3 \} \}$. Now is is no surprise that if the set $A$ is finite and has $n$ elements, then $2^A$ has $2^n$ elements.

However, there is another helpful way of listing $2^A$. A subset of $A$ can be defined by which elements of $A$ that it has. So, if we order the elements of $A$ as $1, 2, 3$ then the power set of $A$ can be identified as follows: $\emptyset = (0, 0, 0), \{1 \} = (1, 0, 0), \{ 2 \} = (0,1,0), \{ 3 \} = (0, 0, 1), \{1,2 \} = (1, 1, 0), \{1,3 \} = (1, 0, 1), \{2,3 \} = (0, 1, 1), \{1, 2, 3 \} = (1, 1, 1)$

So there is a natural correspondence between the elements of a power set and a sequence of binary digits. Of course, this makes the counting much easier.

The binary notation might seem like an unnecessary complication at first, but now consider the power set of the natural numbers: $2^N$. Of course, listing the power sets would be, at least, cumbersome if not impossible! But there the binary notation really shows its value. Remember that the binary notation is a sequence of 0’s and 1’s where a 0 in the i’th slot means that element isn’t an element in a subset and a 1 means that it is.

Since a subset of the natural numbers is defined by its list of elements, every subset has an infinite binary sequence associated with it. We can order the sequence in the usual order 1, 2, 3, 4, ….
and the sequence 1, 0, 0, 0…… corresponds to the set with just 1 in it, the sequence 1, 0, 1, 0, 1, 0, 1, 0,… corresponds to the set consisting of all odd integers, etc.

Then, of course, one can use Cantor’s Diagonal Argument to show that $2^N$ is uncountable; in fact, if one uses the fact that every non-negative real number has a binary expansion (possibly infinite), one then shows that $2^N$ has the same cardinality as the real numbers.

Power notation
We can expand on this power notation. Remember that $2^A$ can be thought of setting up a “slot” or an “index” for each element of $A$ and then assigning a $1$ or $0$ for every element of $A$. One can then think of this in an alternate way: $2^A$ can be thought of as the set of ALL functions from the elements of $A$ to the set $\{ 0, 1 \}$. This coincides with the “power set” concept as set membership is determined by being either “in” or “not in”. So, the set in the exponent can be thought of either as the indexing set and the base as the value each indexed value can take on (sequences, in the case that the exponent set is either finite or countably finite), OR this can be thought of as the set of all functions where the exponent set is the domain and the base set is the range.

Remember, we are talking about ALL possible functions and not all “continuous” functions, or all “morphisms”, etc.

So, $N^N$ can be thought of as either set set of all possible sequences of positive integers, or, equivalently, the set of all functions of $N$ to $N$.

Then $R^N$ is the set of all real number sequences (i. e. the types of sequences we study in calculus), or, equivalently, the set of all real valued functions of the positive integers.

Now it is awkward to try to assign an ordering to the reals, so when we consider $R^R$ it is best to think of this as the set of all functions $f: R \rightarrow R$, or equivalently, the set of all strings which are indexed by the real numbers and have real values.

Note that sequences don’t really seem to capture $R^R$ in the way that they capture, say, $R^N$. But there is another concept that does, and that concept is the concept of the net, which I will talk about in a subsequent post.

## January 12, 2015

### Thoughts prior to next semester

Filed under: point set topology, topology — Tags: — collegemathteaching @ 3:20 am

Next semester: I have a usual load of two calculus courses: calculus III (polar coordinates up to the three main integral theorems…or merely Stokes theorem), “applied calculus II” (the follow on to the course sometimes called “business calculus” or “brief calculus”) and an undergraduate course in topology.

The latter will be fun, but a major time suck, and it will be challenging to teach.

Here is where the challenge comes: though I got my Ph. D. in 1991, I have NEVER taught our topology course. But I have published many papers in this area; I really know the stuff well. But I’ve never tried to explain it to non-research mathematicians.

So, how does one teach something that someone has “done” for 26 years but has never tried to explain to a beginner?

Then there is the challenge of the course itself.

On one hand, there is some basic stuff that one needs to know to advance in mathematics (e. g. the point set and the metric stuff).
On the other hand: the more interesting stuff (the tori, Klein bottles, Mobius bands) is really a collection of “parlor tricks” unless it is presented mathematically, and one needs the basic stuff to do a proper job.

Add to that the non-intuitive notation and jargon:

So: I’ll probably start with some set theory, some metric space theory (say, at the epsilon-delta calculus level at first), some topology of $R^n$ and then go to the more abstract stuff, with a lot of examples.

## January 9, 2015

### Bad notation drove me nuts….(and still does)

Filed under: advanced mathematics, calculus, topology — Tags: , — collegemathteaching @ 8:36 pm

I remember one of my first classes in algebraic topology. The professor was talking about how to prove that $\pi_1(S^1) = Z$. For those who might be rusty: I am talking about the fundamental group of the circle, which is a group structure put on the set of continuous maps of the circle into the circle, where the maps all contain a set “base point” and two maps are equivalent if there is a homotopy (continuous deformation) between the two.

He remarked that he hoped “it was clear” that the circle was NOT simply connected.

That confused the heck out of me, because I had fallen into the trap of confusing the circle with a disk bounded by the circle!

Remember, for years, I had heard things like “the area of a circle is”…when in fact, the circle has area zero. The disk in the plane bounded by the circle has an area though.

So, when I teach, I try to point out bad or inconsistent notation. Example: $sin^2(x)$ means $(sin(x))^2$ rather than $sin(sin(x))$ as the notation $f^2$ might suggest. But $sin^{-1}(x)$ means $arcsin(x)$ and NOT $csc(x) = \frac{1}{sin(x)}$. But…$\frac{d^2 y}{dx^2}$ means $\frac{d}{dx}(\frac{dy}{dx})$.

And please, don’t even get me started about $dx$ that appears in integrals. I remember a student asking me about that when we did “integration by substitution”: “we never used the $dx$ for anything up until now!” he said…correctly.

This blog describes many of the things that I am thinking about at the moment. Currently, I am thinking about “wild knots”, which are embeddings of the circle into 3 space which cannot be deformed (by a deformation of space) into a smooth embedding of the circle.

Here are two examples of knots that can’t be deformed into a smooth knot:

Now the term “knot” implies that an embedding is present; the space that is being embedded is a circle. Of course, one might confuse a particular embedding with the equivalence class of equivalent embeddings; some old time authors distinguished the two concepts. Most modern ones (myself included) don’t.

Now I am interested in knots that are formed by the embedding of two “arcs”, each of which is non-wildly embedded (not wild is called “tame”).

In the case of arcs, authors sometimes mean “the arc itself” and in other cases mean “the embedding of an arc” (e. g. “arcs in 3 space”). Yes, there are some arcs that are so pathologically embedded that there is no deformation of space that takes the arc to a smooth arc. Unfortunately, the term “arc” can mean “the underlying space” or “the embedding”.

This will be one focus of my research in 2015: I hope to show that a knot that has one wild point (roughly speaking: one point that can never be assigned a tangent vector) that is the union of two tamely embedded arcs is never determined by its compliment. That might sound like gibberish, but in 2014 I proved that a knot that is an infinite product of knots (which are converging to a single wild point) has a complement which is homeomorphic to the complement of a knot that is wild at ALL of its points.

Of course THINKING that I can prove something and proving it are two different things. I remember spending two years trying to prove something that was false (I published the counter example) and, for part of my Ph.D. thesis, I attempted to prove something that turned out to be false; of course the counterexample came over 20 years after my attempt.

### Poincare Conjecture and Ricci Flow

My area of research, if you can say that I still have an area of research, is geometric topology. Yes, despite everything, I’ve managed to stay moderately active.

One big development in the past decade and a half is the solution to the Poincare Conjecture and the use of Ricci Flow to solve it (Perelman did the proof).

As far as what the Poincare Conjecture is about:

(If you’ve had some algebraic topology: the Poincare Conjecture says that an object that has the same algebraic information as the 3 dimensional sphere IS the three dimensional sphere, topologically speaking).

Now the proof uses Ricci Flow. Yes, to understand what Ricci flow is about, one has to understand differential geometry. BUT it you’ve had some brush with vector calculus (say, the amount that one teaches in a typical “Calculus III” course), one can get some intuition for this concept here.

Watch the video: it is fun. ðŸ™‚

Now when you get to the end, here is what is going on: instead of viewing a space (such as, say, the 2-d sphere) as being embedded in a larger space, one can talk about the space as being intrinsic; that is, not “sitting in” some ambient space. Then every point can be assigned some intrinsic curvature, and Ricci flow works in that setting.

Of course, one CAN always find a space to isometrically embed your space in (Nash embedding theorem) and still pretend that the space is embedded somewhere else; some “first course in differential topology” texts do exactly that.

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