# College Math Teaching

## 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.

## 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.