point set topology | College Math Teaching

June 7, 2016

Pop-math: getting it wrong but being close enough to give the public a feel for it

Filed under: advanced mathematics, analysis, calculus, point set topology, topology — Tags: pop math, popular mathematics, space filling curves — collegemathteaching @ 9:36 pm

Space filling curves: for now, we’ll just work on continuous functions $f: [0,1] \rightarrow [0,1] \times [0,1] \subset R^2$ .

A curve is typically defined as a continuous function $f: [0,1] \rightarrow M$ where $M$ is, say, a manifold (a 2’nd countable metric space which has neighborhoods either locally homeomorphic to $R^k$ or $R^{k-1})$ . Note: though we often think of smooth or piecewise linear curves, we don’t have to do so. Also, we can allow for self-intersections.

However, if we don’t put restrictions such as these, weird things can happen. It can be shown (and the video suggests a construction, which is correct) that there exists a continuous, ONTO function $f: [0,1] \rightarrow [0,1] \times [0,1]$ ; such a gadget is called a space filling curve.

It follows from elementary topology that such an $f$ cannot be one to one, because if it were, because the domain is compact, $f$ would have to be a homeomorphism. But the respective spaces are not homeomorphic. For example: the closed interval is disconnected by the removal of any non-end point, whereas the closed square has no such separating point.

Therefore, if $f$ is a space filling curve, the inverse image of a points is actually an infinite number of points; the inverse (as a function) cannot be defined.

And THAT is where this article and video goes off of the rails, though, practically speaking, one can approximate the space filling curve as close as one pleases by an embedded curve (one that IS one to one) and therefore snake the curve through any desired number of points (pixels?).

So, enjoy the video which I got from here (and yes, the text of this post has the aforementioned error)

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April 10, 2015

Cantor sets and countable products of discrete spaces (0, 1)^Z

Filed under: advanced mathematics, analysis, point set topology, sequences — Tags: cantor sets, cantor spaces, compact sets, product topology — collegemathteaching @ 11:09 am

This might seem like a strange topic but right now our topology class is studying compact metric spaces. One of the more famous of these is the “Cantor set” or “Cantor space”. I discussed the basics of these here.

Now if you know the relationship between a countable product of two point discrete spaces (in the product topology) and Cantor spaces/Cantor Sets, this post is probably too elementary for you.

Construction: start with a two point set $D = \{0, 2 \}$ and give it the discrete topology. The reason for choosing 0 and 2 to represent the elements will become clear later. Of course, $D_2$ is a compact metric space (with the discrete metric: $d(x,y) = 1$ if $x \neq y$ .

Now consider the infinite product of such spaces with the product topology: $C = \Pi^{\infty}_{i=1} D_i$ where each $D_i$ is homeomorphic to $D$ . It follows from the Tychonoff Theorem that $C$ is compact, though we can prove this directly: Let $\cup_{\alpha \in I} U_{\alpha}$ be any open cover for $C$ . Then choose an arbitrary $U$ from this open cover; because we are using the product topology $U = O_1 \times O_2 \times ....O_k \times (\Pi_{i=k+1}^{\infty} D_i )$ where each $O_i$ is a one or two point set. This means that the cardinality of $C - U$ is at most $2^k -1$ which requires at most $2^k -1$ elements of the open cover to cover.

Now let’s examine some properties.

Clearly the space is Hausdorff ( $T_2$ ) and uncountable.

1. Every point of $C$ is a limit point of $C$ . To see this: denote $x \in C$ by the sequence $\{x_i \}$ where $x_i \in \{0, 2 \}$ . Then any open set containing $\{ x_i \}$ is $O_1 \times O_2 \times...O_k \times \Pi^{\infty}_{i=k+1} D_i$ and contains ALL points $y_i$ where $y_i = x_i$ for $i = \{1, 2, ...k \}$ . So all points of $C$ are accumulation points of $C$ ; in fact they are condensation points (or perfect limit points ).

(refresher: accumulation points are those for which every open neighborhood contains an infinite number of points of the set in question; condensation points contain an uncountable number of points, and perfect limit points are those for which every open neighborhood contains as many points as the set in question has (same cardinality).

2. $C$ is totally disconnected (the components are one point sets). Here is how we will show this: given $x, y \in C, x \neq y,$ there exists disjoint open sets $U_x, U_y, x \in U_x, y \in U_y, U_x \cup U_y = C$ . Proof of claim: if $x \neq y$ there exists a first coordinate $k$ for which $x_k \neq y_k$ (that is, a first $k$ for which the canonical projection maps disagree ( $\pi_k(x) \neq pi_k(y)$ ). Then
$U_x = D_1 \times D_2 \times ....\times D_{k-1} \times x_k \times \Pi^{\infty}_{i=k+1} D_i$ ,

$U_y = D_1 \times D_2 \times.....\times D_{k-1} \times y_k \times \Pi^{\infty}_{i = k+1} D_i$

are the required disjoint open sets whose union is all of $C$ .

3. $C$ is countable, as basis elements for open sets consist of finite sequences of 0’s and 2’s followed by an infinite product of $D_i$ .

4. $C$ is metrizable as well; $d(x,y) = \sum^{\infty}_{i=1} \frac{|x_i - y_i|}{3^i}$ . Note that is metric is well defined. Suppose $x \neq y$ . Then there is a first $k, x_k \neq y_k$ . Then note

$d(x,y) = \frac{|x_k - y_k|}{3^k} + \sum^{\infty}_{i = k+1} \frac{|x_i - y_i|}{3^i} \rightarrow |x_k -y_k| =2 = \sum^{\infty}_{i=1} \frac{|x_{i+k} -y_{i+k}|}{3^i} \leq \frac{1}{3} \frac{1}{1 -\frac{2}{3}} =1$

which is impossible.

5. By construction $C$ is uncountable, though this follows from the fact that $C$ is compact, Haudorff and dense in itself.

6. $C \times C$ is homeomorphic to $C$ . The homeomorphism is given by $f( \{x_i \}, \{y_i \}) = \{ x_1, y_1, x_2, y_2,... \} \in C$ . It follows that $C$ is homeomorphic to a finite product with itself (product topology). Here we use the fact that if $f: X \rightarrow Y$ is a continuous bijection with $X$ compact and $Y$ Hausdorff then $f$ is a homeomorphism.

Now we can say a bit more: if $C_i$ is a copy of $C$ then $\Pi^{\infty}_{i =1 } C_i$ is homeomorphic to $C$ . This will follow from subsequent work, but we can prove this right now, provided we review some basic facts about countable products and counting.

First lets show that there is a bijection between $Z \times Z$ and $Z$ . A bijection is suggested by this diagram:

which has the following formula (coming up with it is fun; it uses the fact that $\sum^k _{n=1} n = \frac{k(k+1)}{2}$ :

$\phi(k,1) = \frac{(k)(k+1)}{2}$ for $k$ even
$\phi(k,1) = \frac{(k-1)(k)}{2} + 1$ for $k$ odd
$\phi(k-j, j+1) =\phi(k,1) + j$ for $k$ odd, $j \in \{1, 2, ...k-1 \}$
$\phi(k-j, j+1) = \phi(k,1) - j$ for $k$ even, $j \in \{1, 2, ...k-1 \}$

Here is a different bijection; it is a fun exercise to come up with the relevant formulas:

Now lets give the map between $\Pi^{\infty}_{i=1} C_i$ and $C$ . Let $\{ y_i \} \in C$ and denote the elements of $\Pi^{\infty}_{i=1} C_i$ by $\{ x^i_j \}$ where $\{ x_1^1, x_2^1, x_3^ 1....\} \in C_1, \{x_1^2, x_2 ^2, x_3^3, ....\} \in C_2, ....\{x_1^k, x_2^k, .....\} \in C_k ...$ .

We now describe a map $f: C \rightarrow \Pi^{\infty}_{i=1} C_i$ by

$f(\{y_i \}) = \{ x^i_j \} = \{y_{\phi(i,j)} \}$

Example: $x^1_1 = y_1, x^1_2 = y_2, x^2_1 = y_3, x^3_1 =y_4, x^2_2 = y_5, x^1_3 =y_6,...$

That this is a bijection between compact Hausdorff spaces is immediate. If we show that $f^{-1}$ is continuous, we will have shown that $f$ is a homeomorphism.

But that isn’t hard to do. Let $U \subset C$ be open; $U = U_1 \times U_2 \times U_3.... \times U_{m-1} \times \Pi^{\infty}_{k=m} C_k$ . Then there is some $k_m$ for which $\phi(k_m, 1) \geq M$ . Then if $f^i_j$ denotes the $i,j$ component of $f$ we wee that for all $i+j \geq k_m+1, f^i_j(U) = C$ (these are entries on or below the diagonal containing $(k,1)$ depending on whether $k_m$ is even or odd.

So $f(U)$ is of the form $V_1 \times V_2 \times ....V_{k_m} \times \Pi^{\infty}_{i = k_m +1} C_i$ where each $V_j$ is open in $C_j$ . This is an open set in the product topology of $\Pi^{\infty}_{i=1} C_i$ so this shows that $f^{-1}$ is continuous. Therefore $f^{-1}$ is a homeomorphism, therefore so is $f$ .

Ok, what does this have to do with Cantor Sets and Cantor Spaces?

If you know what the “middle thirds” Cantor Set is I urge you stop reading now and prove that that Cantor set is indeed homeomorphic to $C$ as we have described it. I’ll give this quote from Willard, page 121 (Hardback edition), section 17.9 in Chapter 6:

The proof is left as an exercise. You should do it if you think you can’t, since it will teach you a lot about product spaces.

What I will do I’ll give a geometric description of a Cantor set and show that this description, which easily includes the “deleted interval” Cantor sets that are used in analysis courses, is homeomorphic to $C$ .

Set up
I’ll call this set $F$ and describe it as follows:

$F \subset R^n$ (for those interested in the topology of manifolds this poses no restrictions since any manifold embeds in $R^n$ for sufficiently high $n$ ).

Reminder: the diameter of a set $F \subset R^n$ will be $lub \{ d(x,y) | x, y \in F \}$
Let $\epsilon_1, \epsilon_2, \epsilon_3 .....\epsilon_k ...$ be a strictly decreasing sequence of positive real numbers such that $\epsilon_k \rightarrow 0$ .

Let $F^0$ be some closed n-ball in $R^n$ (that is, $F^)$ is a subset homeomorphic to a closed n-ball; we will use that convention throughout)

Let $F^1_{(0) }, F^1_{(2)}$ be two disjoint closed n-balls in the interior of $F^0$ , each of which has diameter less than $\epsilon_1$ .

$F^1 = F^1_{(0) } \cup F^1_{(2)}$

Let $F^2_{(0, 0)}, F^2_{(0,2)}$ be disjoint closed n-balls in the interior $F^1_{(0) }$ and $F^2_{(2,0)}, F^2_{(2,2)}$ be disjoint closed n-balls in the interior of $F^1_{(2) }$ , each of which (all 4 balls) have diameter less that $\epsilon_2$ . Let $F^2 = F^2_{(0, 0)}\cup F^2_{(0,2)} \cup F^2_{(2, 0)} \cup F^2_{(2,2)}$

To describe the construction inductively we will use a bit of notation: $a_i \in \{0, 2 \}$ for all $i \in \{1, 2, ...\}$ and $\{a_i \}$ will represent an infinite sequence of such $a_i$ .
Now if $F^k$ has been defined, we let $F^{k+1}_{(a_1, a_2, ...a_{k}, 0)}$ and $F^{k+1}_{(a_1, a_2,....,a_{k}, 2)}$ be disjoint closed n-balls of diameter less than $\epsilon_{k+1}$ which lie in the interior of $F^k_{(a_1, a_2,....a_k) }$ . Note that $F^{k+1}$ consists of $2^{k+1}$ disjoint closed n-balls.

Now let $F = \cap^{\infty}_{i=1} F^i$ . Since these are compact sets with the finite intersection property ( $\cap^{m}_{i=1}F^i =F^i \neq \emptyset$ for all $m$ ), $F$ is non empty and compact. Now for any choice of sequence $\{a_i \}$ we have $F_{ \{a_i \} } =\cap^{\infty}_{i=1} F^i_{(a_1, ...a_i)}$ is nonempty by the finite intersection property. On the other hand, if $x, y \in F, x \neq y$ then $d(x,y) = \delta > 0$ so choose $\epsilon_m$ such that $\epsilon_m < \delta$ . Then $x, y$ lie in different components of $F^m$ since the diameters of these components are less than $\epsilon_m$ .

Then we can say that the $F_{ \{a_i} \}$ uniquely define the points of $F$ . We can call such points $x_{ \{a_i \} }$

Note: in the subspace topology, the $F^k_{(a_1, a_2, ...a_k)}$ are open sets, as well as being closed.

Finding a homeomorphism from $F$ to $C$ .
Let $f: F \rightarrow C$ be defined by $f( x_{ \{a_i \} } ) = \{a_i \}$ . This is a bijection. To show continuity: consider the open set $U = y_1 \times y_2 ....\times y_m \times \Pi^{\infty}_{i=m} D_i$ . Under $f$ this pulls back to the open set (in the subspace topology) $F^{m+1}_{(y1, y2, ...y_m, 0 ) } \cup F^{m+1}_{(y1, y2, ...y_m, 2)}$ hence $f$ is continuous. Because $F$ is compact and $C$ is Hausdorff, $f$ is a homeomorphism.

This ends part I.

We have shown that the Cantor sets defined geometrically and defined via “deleted intervals” are homeomorphic to $C$ . What we have not shown is the following:

Let $X$ be a compact Hausdorff space which is dense in itself (every point is a limit point) and totally disconnected (components are one point sets). Then $X$ is homeomorphic to $C$ . That will be part II.

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February 11, 2015

Teaching advanced stuff to undergraduates: taking stuff for granted….

Filed under: advanced mathematics, pedagogy, point set topology — Tags: connected set, homeomorphism — collegemathteaching @ 5:34 pm

I am teaching undergraduate topology this semester; you can see some stuff that I’ve posted for undergraduates here.

I took the “start with metric spaces and topology of $R, R^2, R^3..$ approach and am going slower than I’d like. But it takes some time to absorb the stuff.

So, we are finally gotten up to homeomorphisms (still in basic metric spaces) and so I figured that we were finally ready to show something like: $[0,1]$ is not homeomorphic to $S^1$ (the unit circle). Fine: suppose a homeomorphism exists and decompose $[0,1] =U \cup V\ \{x \}$ where $x \notin U, x\notin V, U \cap V = \emptyset$

Not a problem so far…so now pull back the disjoint open sets $U, V$ to the unit circle minus one point…and….then…I ….realized….that…I have not proven that the interval is a connected set; in fact I haven’t even defined “connected set”. &^%$#. Now, that isn’t that hard to do, but it does take time and one has to do some setting up.

And that is one problem with teaching this course: you have to do so much elementary stuff to even begin to prove something elementary.

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January 17, 2015

Convergence of functions and nets (from advanced calculus)

Filed under: advanced mathematics, infinity, point set topology, topology — Tags: directed sets, function space, nets, pointwise convergence, topological convergence — collegemathteaching @ 4:56 am

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.

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January 16, 2015

Power sets, Function spaces and puzzling notation

Filed under: advanced mathematics, analysis, point set topology, sequences, sequences of functions, topology — Tags: function space, point set topology, power sets, set theory — collegemathteaching @ 2:20 am

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.

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January 12, 2015

Thoughts prior to next semester

Filed under: point set topology, topology — Tags: teaching topology — 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.

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October 25, 2013

Why the sequence cos(n) diverges

Filed under: analysis, calculus, editorial, elementary number theory, number theory, point set topology, sequences — Tags: hard calculus problems — collegemathteaching @ 12:59 am

We are in the sequences section of our Freshman calculus class. One of the homework problems was to find whether the sequence $a_n = cos(\frac{n}{2})$ converged or diverged. This sequence diverges, but it isn’t easy for a freshman to see.

I’ll discuss this problem and how one might go about explaining it to a motivated student. To make things a bit simpler, I’ll discuss the sequence $a_n = cos(n)$ instead.

Of course $cos(x)$ is periodic with a fundamental region $[0, 2\pi]$ so we will work with that region. Now we notice the following:

$n (mod 2 \pi)$ is a group with the usual operation of addition.

By $n (mod 2 \pi)$ , I mean the set $n + k*2\pi$ where $k \in \{..-2, -1, 0, 1, 2, 3,...\}$ ; one can think of the analogue of modular arithmetic, or one might see the elements of the group $\{ r| r \in [0, 2 \pi), r = n - k 2\pi \}$ .

Of course, to get additive inverses, we need to include the negative integers, but ultimately that won’t matter. Example: $1, 2, 3, 4, 5, 6$ are just equal to themselves $mod 2 \pi.$ $7 = 7 - 2\pi (mod 2\pi), 13 = 13 - 4 \pi (mod 2\pi)$ , etc. So, I’ll denote the representative of $n (mod 2\pi)$ by $[n]$ .

Now if $n \ne m$ then $[n] \ne [m]$ ; for if $[n]=[m]$ then there would be integers $j, k$ so that $n + j2\pi = m +k2\pi$ which would imply that $|m-n|$ is a multiple of $\pi$ . Therefore there are an infinite number of $[n]$ in $[0, 2\pi]$ which means that the set $\{[n]\}$ has a limit point in the compact set $[0, 2\pi]$ which means that given any positive integer $m$ there is some interval of width $\frac{2\pi}{m}$ that contains two distinct $[i], [j]$ (say, $j$ greater than $i$ .)

This means that $[j-i] \in (0, \frac{2\pi}{m})$ so there is some integers $k_2, k_3,$ so that $k_2[j-i] \in (\frac{2\pi}{m}, \frac{2*2\pi}{m}), k_3[j-i] \in (\frac{2*2\pi}{m}, \frac{3*2\pi}{m})$ , etc. Therefore there is some multiple of $[j-i]$ in every interval of width $\frac{2\pi}{m}$ . But $m$ was an arbitrary positive integer; this means that the $[n]$ are dense in $[0,2\pi]$ . It follows that $cos([n]) = cos(n)$ is dense in $[-1,1]$ and hence $a_n = cos(n)$ cannot converge as a sequence.

Frankly, I think that this is a bit tough for most Freshman calculus classes (outside of, say those at MIT, Harvard, Cal Tech, etc.).

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September 22, 2013

Mathematics journal articles: terse but is it the author?

Filed under: advanced mathematics, point set topology, research — Tags: mathematical research, mathematical writing — collegemathteaching @ 11:54 pm

Via Recursivity:

It’s a sad truth, but the mathematics research literature is very tough going for beginners. By “beginners” I mean bright high-school students, or university students, or beginning graduate students, or even professional mathematicians who are trained in an area different from the article he/she is trying to read. […]

Things like this permeate the mathematical literature. Take compactness, for example. Compactness is a marvelous tool that lets you deduce — usually in a non-constructive fashion — the existence of objects (particularly infinite ones) from the existence of finite “approximations”. Formally, compactness is the property that a collection of closed sets has a nonempty intersection if every finite subcollection has a nonempty intersection; alternatively, if every open cover has a finite subcover.

Now compactness is a topological property, so to use it, you really should say explicitly what the topological space is, and what the open and closed sets are. But mathematicians rarely, if ever, do that. In fact, they usually don’t specify anything at all about the setting; they just say “by the usual compactness argument” and move on. That’s great for experts, but not so great for beginners.

I really wonder who was the very first to take this particular lazy approach to mathematical exposition.

Hmmm, often it is the reviewer, referee or editor. They accept your paper, but make you take out some details (and, to be fair, add others)

A colleague and I are thinking of starting a journal called “The Journal of Omitted Details”.

But yes, this practice makes some mathematics very difficult for the non-expert to read.

Note: the usual definition of a compact set (given some topology) is: $X$ is compact if, given any collection of open sets $U_{\alpha}$ where $X \subset \cup_{i \in \alpha} U_{\alpha}$ ,there exists a finite number of the $U_{\alpha}$ where $X \subset \cup^{k}_{i=1} U_{\alpha i}$ . That is, any open cover has a finite subcover. This is equivalent to saying that any infinite set of points in $X$ has a limit point, and in a metric space this means that $X$ is both closed and bounded.

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January 31, 2011

Cantor Sets in [0,1] and Lebesgue Measure

Filed under: advanced mathematics, analysis, calculus, cantor set, cantor space, Lebesgue Integration, Measure Theory, point set topology — collegemathteaching @ 2:34 am

I still remember one of the more moronic things I have ever written on an exam; I said “Set E has measure zero and is therefore countable….”. My professor wrote “whatever happened to the Cantor Set”, which he had told us about and had covered extensively.

It was one of those things that I had heard but really didn’t internalize into a working part of my mathematical mind; the latter is difficult to do.

So what is a Cantor Set? Actually, it depends on who you ask. 🙂 A topologist is likely to give a different answer than an analyst; I’ll discuss what is going on.

First, I’ll construct the traditional Middle Thirds Cantor Set; this is an example of a subset of $[0,1]$ which
1. Has Lebesgue outer measure zero (and is therefore measurable)
2. Is uncountable.
It also has some other interesting properties and can be generalized; as George Simmons in Introduction to Topology and Modern Analysis says on page 68:

[…] the Cantor set is a very intricate mathematical object and is just the sort of thing that mathematicians delight in.

Descriptions: I’ll describe the Cantor set by describing what is NOT in it: the intervals $(1/3, 2/3), (1/9.2/9), (7/9. 8/9), (1/27, 2/27), (7/27, 8/27), (19/27, 20/27), (25/27, 26/27), ....$ . Or, I can describe it as an intersection of closed sets: $C_{1} = [0,1/3] \cup [2/3,1], C_{2} = [0, 1/9] \cup [2/9, 1/3] \cup [2/3,7/9] \cup [8/9,1],...$ and then the Cantor set $C =\cap_{i=1}^{\infty} C_{i}$ . Another way of describing the Cantor middle thirds set is the set of numbers whose base three fraction expansion does not contain the digit “1” anywhere (as these numbers lie in the removed intervals, always the middle third).

Some facts are immediate: The Cantor set is closed as its complement is the union of open sets. It is bounded and hence compact. The middle thirds Cantor set has measure zero; here is why: if we add the measures of the complement: $1/3 + 2(1/9) + 4(1/27) + ..=(1/3)\sum_{k=0}^{\infty}(2/3)^k = (1/3)(1/(1-(2/3)) = 1$

But the Cantor set is uncountable! This can be seen in many ways; there is a topological proof; we’ll present a brute force one.
Let $C$ denote the Cantor middle thirds set and let $x \in C$ . Note that each $C_{k}$ consists of $2^{k-1}$ disjoint intervals of length $(1/3)^k$ . Now we will map $x$ to a point on the sequence of 1’s and 0’s: ${f_{1}(x), f_{2}(x), f_{3} (x), ....}$ where $f_{1}(x) = 0$ if $x \leq 1/3$ and 1 if $x \geq 2/3$ . Then $f_{2}(x) = 0$ if $x$ is in the first third of the next interval and 1 if it is in the final third. We do this at every stage and note that the map from $C$ onto the sequence is onto. The map is also one to one because if $x \neq y$ then they are at least $\epsilon$ apart, and choose a non-deleted interval that is shorter than this $\epsilon$ ; that interval cannot contain both $x$ and $y$ .

We can say much more about the Cantor set(s); I’ll conclude with a few interesting tidbits:

1. It isn’t that tough to exhibit some elements of $C$ that are NOT endpoints of deleted intervals; merely choose a base three fraction sum that does not contain 1’s in its numerator. Here is one example: choose
$2/3 + 2/9 + 2/81 + 2/729 ..... = (2/3)+ (2/9) (1 + 1/9 + 1/81 +... =(2/3)+ (2/9(1/(1-(1/9))) = 2/3 +1/4 = 11/12$
2. One can alter the construction of the Cantor middle third set to obtain a Cantor set of any measure less than 1: do the same middle thirds construction but ensure that the length of each removed interval is $\delta^k$ at stage $k$ Then one can compute the length of the removed intervals:
$\delta + 2\delta^2 + 4\delta^3.....=\delta/(1-2\delta)$ . Now if $r$ is the desired measure of the complement (less than 1, of course), one solves $r = \delta/(1-2\delta)$ for $\delta$ to obtain $\delta = r/(1+2r)$ . One notes that $r = 1$ yields $\delta = 1/3$ .

The Cantor set (of any measure) has some interesting properties. For one, every point is a limit point (remember, it has the subspace topology). This isn’t hard to see; let $I$ be a neighborhood of any point $x$ of the Cantor set of width $\epsilon$ and let $k$ be such that $(1/3)^k < \epsilon/2$ . Then this non-deleted interval containing $x$ must lie in $I$ , which means that $I$ contains other points of the Cantor set. This property is called “being dense in itself”. This property, plus being compact, is enough to prove that the set is uncountable.
The Cantor set is totally disconnected; that is, the only connected components are one point sets. This is why: given any two different points in a Cantor set, there is a deleted interval between them.

Note: it might not make sense to talk of THE Cantor set as one can construct Cantor sets of different measures. But it does make sense to talk about the Cantor set in terms of topology as every compact, dense in itself, totally disconnected metric space is homeomorphic (in fact, homeomorphic to the countably infinite product of ${0, 1}$ where every factor has the discrete topology and the whole space has the product topology. Note: sometimes this topological space is referred to as the Cantor SPACE with Cantor SET being reserved for middle thirds type of construction.

Astonishingly, every compact metric space is the image of the Cantor space (and therefore of the Cantor set); that is a topic for a later time.

What does this have to do with integration?
Remember that the standard Cantor middle thirds set has measure zero; hence two functions defined on $[0,1]$ that are equal except on the Cantor set have equal integrals; we will also see that the Cantor set can cause some mischief when we start talking about the Fundamental Theorem of Calculus. 🙂
If you can’t wait, see here.

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