# College Math Teaching

## June 7, 2016

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

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)

## February 5, 2016

### More fun with selective sums of divergent series

Just a reminder: if $\sum_{k=1}^{\infty} a_k$ is a series and $c_1, c_2, ...c_n ,,$ is some sequence consisting of 0’s and 1’s then a selective sum of the series is $\sum_{k=1}^{\infty} c_k a_k$. The selective sum concept is discussed in the MAA book Real Infinite Series (MAA Textbooks) by Bonar and Khoury (2006) and I was introduced to the concept by Ferdinands’s article Selective Sums of an Infinite Series in the June 2015 edition of Mathematics Magazine (Vol. 88, 179-185).

There is much of interest there, especially if one considers convergent series or alternating series.

This post will be about divergent series of positive terms for which $lim_{n \rightarrow \infty} a_n = 0$ and $a_{n+1} < a_n$ for all $n$.

The first fun result is this one: any selected $x > 0$ is a selective sum of such a series. The proof of this isn’t that bad. Since $lim_{n \rightarrow \infty} a_n = 0$ we can find a smallest $n$ such that $a_n \leq x$. Clearly if $a_n = x$ we are done: our selective sum has $c_n = 1$ and the rest of the $c_k = 0$.

If not, set $n_1 = n$ and note that because the series diverges, there is a largest $m_1$ so that $\sum_{k=n_1}^{m_1} a_k \leq x$. Now if $\sum_{k=n_1}^{m_1} a_k = x$ we are done, else let $\epsilon_1 = x - \sum_{k=n_1}^{m_1} a_k$ and note $\epsilon_1 < a_{m_1+1}$. Now because the $a_k$ tend to zero, there is some first $n_2$ so that $a_{n_2} \leq \epsilon_1$. If this is equality then the required sum is $a_{n_2} + \sum_{k=n_1}^{m_1} a_k$, else we can find the largest $m_2$ so that $\sum_{k=n_1}^{m_1} a_k + \sum_{k=n_2}^{m_2} a_k \leq x$

This procedure can be continued indefinitely. So if we label $\sum_{k=n_j}^{m_{j}} a_k = s_j$ we see that $s_1 + s_2 + ...s_{n} = t_{n}$ form an increasing, bounded sequence which converges to the least upper bound of its range, and it isn’t hard to see that the least upper bound is $x$ because $x-t_{n} =\epsilon_n < a_{m_n+1}$

So now that we can obtain any positive real number as the selective sum of such a series, what can we say about the set of all selective sums for which almost all of the $c_k = 0$ (that is, all but a finite number of the $c_k$ are zero).

Answer: the set of all such selective sums are dense in the real line, and this isn’t that hard to see, given our above construction. Let $(a,b)$ be any open interval in the real line and let $a < x < b$. Then one can find some $N$ such that for all $n > N$ we have $x - a_n > a$. Now consider our construction and choose $m$ large enough such that $x - t_m > x - a_n > a$. Then the $t_m$ represents the finite selected sum that lies in the interval $(a,b)$.

We can be even more specific if we now look at a specific series, such as the harmonic series $\sum_{k=1}^{\infty} \frac{1}{k}$. We know that the set of finite selected sums forms a dense subset of the real line. But it turns out that the set of select sums is the rationals. I’ll give a slightly different proof than one finds in Bonar and Khoury.

First we prove that every rational in $(0,1]$ is a finite select sum. Clearly 1 is a finite select sum. Otherwise: Given $\frac{p}{q}$ we can find the minimum $n$ so that $\frac{1}{n} \leq \frac{p}{q} < \frac{1}{n-1}$. If $\frac{p}{q} = \frac{1}{n}$ we are done. Otherwise: the strict inequality shows that $pn-p < q$ which means $pn-q < p$. Then note $\frac{p}{q} - \frac{1}{n} = \frac{pn-q}{qn}$ and this fraction has a strictly smaller numerator than $p$. So we can repeat our process with this new rational number. And this process must eventually terminate because the numerators generated from this process form a strictly decreasing sequence of positive integers. The process can only terminate when the new faction has a numerator of 1. Hence the original fraction is some sum of fractions with numerator 1.

Now if the rational number $r$ in question is greater than one, one finds $n_1$ so that $\sum^{n_1}_{k=1} \frac{1}{k} \leq r$ but $\sum^{n_1+1}_{k=1} \frac{1}{k} > r$. Then write $r-\sum^{n_1+1}_{k=1} \frac{1}{k}$ and note that its magnitude is less than $\frac{1}{n_1+1}$. We then use the procedure for numbers in $(0,1)$ noting that our starting point excludes the previously used terms of the harmonic series.

There is more we can do, but I’ll stop here for now.

## January 20, 2016

### Congratulations to the Central Missouri State Mathematics Department

Filed under: advanced mathematics, editorial, number theory — Tags: — collegemathteaching @ 10:43 pm

The largest known prime has been discovered by mathematicians at Central Missouri State University.

For what it is worth, it is: $2^{74,207,281} -1$.

Now if you want to be depressed, go to the Smithsonian Facebook page and read the comment. The Dunning-Kruger effect is real. Let’s just say that in our era, our phones are smarter than our people. 🙂

## July 28, 2015

### J. H. Conway, Terry Tao and avoiding work

Filed under: advanced mathematics, algebra, media — Tags: , , , , , — collegemathteaching @ 7:48 pm

The mainstream media recently had some excellent articles on two mathematical giants:

John Conway and Terrance Tao. I’ve never met Terry Tao though I do read (or try to follow) his blog.

I did meet John Conway when he visited the University of Texas. He is a friend of my dissertation advisor and gave some talks on knot diagram colorings.

I had a private conversation with him at a party, and he gave me some ideas which resulted in three papers for me! Here is one of them.

Yes, I am avoiding studying a book on the theory of interest; I am teaching that course this fall and need to get ahead of the game.

Unfortunately, when I don’t teach, my use of time becomes undisciplined.

## June 25, 2015

### Workshop in Geometric Topology: TCU 2015 morning session 1

Filed under: advanced mathematics, conference, editorial, topology — Tags: — collegemathteaching @ 3:59 pm

I’ll be blunt: I’ve been teaching at a 11-12 hour load (mostly 11; one time I had a 9 hour load; 3 courses) since fall, 1991. Though I’ve published, most of what I’ve done has been extremely “bare handed”; it is tough to learn the most advanced techniques (which is a full time job in and of itself)

So, at math conferences, I get to see how much further behind I’ve fallen.

But these things help in the following way:

1. They are an excellent change of pace from the usual routine of teaching calculus.
2. I do learn things, even if it is “looking up” a definition or two; for example I looked up the definition of “pure braid group” in between the 20 minute talks.
3. I have to review my own stuff to see if I am indeed making progress; I don’t want to say something idiotic in front of some very smart, informed people.

But yes, the talks have been given by smart, (mostly) young, energetic people who have been studying the topic that they are talking about very intensely for a long time; frequently it is tough to hang in to the second half of the 20 minute talks. But I can see WHAT is being studied, what tools are being used and, as I said before, find stuff to look up.

The final talk: didn’t understand much beyond the general gist but it was well organized, well presented..exactly what you get when you have a brilliant energetic young researcher working full time in mathematical research.

On one hand, I envy his talent. On the other hand, I am glad that we have some smart humans among us; they benefit all of us.

The trip here The plane was about 2.5 hours late getting in, then there was a long ride to the car rental place and a 35 minute drive to campus, then finding my way around in the dark. So no morning run; I might do a gentle “after the talks” focused walk (5K-ish?).

I talk at 9 am tomorrow and I want to make it worth their while.

## June 19, 2015

### Scientific American article about finite simple groups

Filed under: advanced mathematics, algebra, mathematician — Tags: , — collegemathteaching @ 2:42 pm

For those of you who are a bit rusty: a finite group is a group that has a finite number of elements. A simple group is one that has no proper non-trivial normal subgroups (that is, only the identity and the whole group are normal subgroups).

It is a theorem that if $G$ is a finite simple group then $G$ falls into one of the following categories:

1. Cyclic (of prime order, of course)
2. Alternating (and not isomorphic to $A_4$ of course)
3. A member of a subclass of Lie Groups
4. One of 26 other groups that don’t fall into 1, 2 or 3.

Scientific American has a nice article about this theorem and the effort to get it written down and understood; the problem is that the proof of such a theorem is far from simple; it spans literally hundreds of research articles and would take thousands of pages to be complete. And, those who have an understanding of this result are aging and won’t be with us forever.

Here is a link to the preview of the article; if you don’t subscribe to SA it is probably in your library.

## April 10, 2015

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

Filed under: advanced mathematics, analysis, point set topology, sequences — Tags: , , , — 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.

## March 24, 2015

### Google Doodle

Filed under: advanced mathematics, algebra, famous mathematicians — Tags: , — collegemathteaching @ 2:52 am

For the prize: what is the significance of the two chains, and which one directly applies to the human subject of this doodle?

The human subject of the doodle is Emmy Noether.

## March 16, 2015

### Compact Spaces and Tychonoff’s Theorem II

Filed under: advanced mathematics, topology — Tags: , , — collegemathteaching @ 6:10 pm

Ok, now lets prove the following: If $X, Y$ are compact spaces, then $X \times Y$ is compact (in the usual product topology). Note: this effectively proves that the finite product of compact spaces is compact. One might call this a “junior” Tychonoff Theorem.

Proof. We will prove this theorem a couple of times; the first proof is the more elementary but less elegant proof. It can NOT be easily extended to show that the arbitrary product of compact spaces is compact (which is the full Tychonoff Theorem).

We will show that an open covering of $X \times Y$ by basis elements of the form $U \times V$, $U$ open in $X$ and $V$ open in $Y$ has a finite subcover.

So let $\mathscr{U}$ be an open cover of $X \times Y$. Now fix $x_{\beta} \in X$ and consider the subset $x_{\beta} \times Y$. This subset is homeomorphic to $Y$ and is therefore compact; therefore there is a finite subcollection of $\mathscr{U}$ which overs $x_{\beta} \times Y$, say $\cup^{\beta, k}_{i=1} U_{\beta, i} \times V_{\beta, i}$ Note that each $U_{\beta, i}$ is an open set in $X$ which contains $x_{\beta}$ and there are only a finite number of these. Hence $\cap^{\beta k}_{i=1} U_{\beta i} = U_{\beta}$ is also an open set which contains $x_{\beta}$. Also know that $U_{\beta} \times Y \subset \cup^{\beta, k}_{i=1} U_{\beta, i} \times V_{\beta, i}$

We can do this for each $x_{\beta} \in X$ and so obtain an open cover of $X$ by $\cup_{x_{\beta} \in X} U_{\beta}$ and because $X$ is compact, a finite subcollection of these covers $X$. Call these $U_1, U_2, U_3....U_m$. For each one of these, we have $U_j \times Y \subset \cup^{j, k}_{i=1} U_{j, i} \times V_{j, i}$.

So, our finite subcover of $X \times Y$ is $\cup^m_{j=1}\cup^{j, k}_{i=1} U_{j, i} \times V_{j, i}$.

Now while this proof is elementary, it doesn’t extend to the arbitrary infinite product case.

So, to set up such an extension, we’ll give some “equivalent” definitions of compactness. Note: at some point, we’ll use some elementary cardinal arithmetic.

## March 15, 2015

### Compact spaces and Tychonoff’s Theorem I

Note to the reader: when I first started this post, I thought it would be a few easy paragraphs. The size began to swell, so I am taking this post in several bite sized parts. This is part I.

Pretty much everyone knows what a compact space is. But not everyone is up on equivalent definitions and on how to prove that the arbitrary product of compact spaces is compact.
The genesis of this blog post is David Wright’s Proceedings of the American Mathematical Society paper (1994) on Tychonoff’s Theorem.

Since I am also writing for my undergraduate topology class, I’ll keep things elementary where possible and perhaps put in more detail than a professional mathematician would have patience for.

I should start by saying why this topic is dear to me: my research area is knot theory; in particular I studied embeddings of the circle into the 3-dimensional sphere $S^3$, which can be thought of as the “compactification” of $R^3$; basically one starts with $R^3$ and then adds a point $\infty$ and declares that the neighborhoods of this new point will be generated by sets of the following form: $\{ (x,y,z) | x^2 + y^2 + z^2 > M^2 \}$ The reason we do this: we often study the complement of the embedded circle, and it is desirable to have a compact set as the complement.

I’ve also studied (in detail) certain classes of embeddings of the real line into non-compact manifolds; to make this study a bit more focused, I insist that such embeddings be “proper” in that the inverse image of a compact set be compact. Hence compactness comes up again, even when the objects of study are not compact.

So, what do we mean by “compact”?

Instead of just blurting out the definition and equivalent formulations, I’ll try to give some intuition. If we are talking about a subset of a metric space, a compact subset is one that is both closed and bounded. Now that is NOT the definition of compactness, though it is true that:

Given a set $X \subset R^n$, $X$ is compact if and only if $X$ is both closed (as a topological subset) and bounded (in that it fits within a sufficiently large closed ball). In $R^1$ compact subsets can be thought of as selected finite unions and arbitrary intersections of closed intervals. In the higher dimensions, think of the finite union and arbitrary intersections of things like closed balls.

Now it is true that if $f:X \rightarrow Y$ is continuous, then if $X$ is a compact topological space, then $f(X)$ is compact (either as a space, or in the subspace topology, if $f$ is not onto.

This leads to a big theorem of calculus: the Extreme Value Theorem: if $f:R^n \rightarrow R$ is continuous over a compact subset $D \subset R^n$ then $f$ attains both a maximum and a minimum value over $D$.

Now in calculus, we rarely use the word “compact” but instead say something about $D$ be a closed, bounded subset. In the case where $n = 1$ we usually say that $D =[a,b]$, a closed interval.

So, in terms of intuition, if one is thinking about subsets of $R^n$, one can think of a compact space as a domain on which any continuous real valued function always attains both a minimum and a maximum value.

Now for the definition
We need some terminology: a collection of open sets $U_{\alpha}$ is said to be an open cover of a space $X$ if $\cup_{\beta \in I } U_{\beta} = X$ and if $A \subset X$ a collection of open sets is said to be an open cover of $A$ if $A \subset \cup_{\beta \in I } U_{\beta}$ A finite subcover is a finite subcollection of the open sets such that $\cup^k_{i=1} U_i = \cup_{\beta \in I} U_{\beta}$.

Here is an example: $(\frac{3}{4}, 1] \cup^{\infty}_{n=1} [0, \frac{n}{n+1})$ is an open cover of $[0,1]$ in the subspace topology. A finite subcover (from this collection) would be $[0, \frac{4}{5}) \cup (\frac{3}{4}, 1]$

Let $X$ be a topological space. We say that $X$ is a compact topological space if any open over of $X$ has a finite subcover. If $C \subset X$ we say that $C$ is a compact subset of $X$ if any open cover of $C$ has a finite subcover.

Prior to going through examples, I think that it is wise to mention something. One logically equivalent definition is this: A space (or a subset) is compact if every cover by open basis elements has a finite subcover. Here is why: if $X$ is compact, then ANY open cover has a finite subcover, and an open cover by basis elements is an open cover. On the other hand: if we assume the “every open cover by open basis elements has a finite subcover” condition: then if $\mathscr{U}$ is an open cover, then we can view this open cover as an open cover of the basis elements whose union is each open $U_{\beta} \in \mathscr{U}$. This open cover of basis elements has a finite subcover of basis elements..say $B_1, B_2, ....B_k$. Then for each basis element, choose a single $U_i \in \mathscr{U}$ for which $B_i \subset U_i$. That is the required open subcover.

Now, when convenient, we can assume that the open cover in question (during a proof) consists of basic open sets. That will simplify things at times.

So, what are some compact spaces and sets, and what are some basic facts?

Let’s see some compact sets, some non compact sets and see some basic facts.

1. Let $X$ be any topological space and $A \subset X$ a finite subset. Then $A$ is a compact subset. Proof: given any open cover of $A$ choose one open set per element of $A$ which contains said element.

2. Let $R$ have the usual topology. Then the integers $Z \subset R^1$ is not a compact subset; choose the open cover $\cup^{\infty}_{n = -\infty} (n - \frac{1}{4}, n+ \frac{1}{4})$ is an infinite cover with no finite subcover. In fact, ANY unbounded subset $A \subset R^n$ in the usual metric topology fails to be compact: for $a \in A$ with $d(a, 0) \geq n$ choose $B_a(\frac{1}{n})$; clearly this open cover can have no finite subcover.

3. The finite union of compact subsets is compact (easy exercise).

4. If $C \subset X$ is compact and $X$ is a Hausdorff topological space ($T_2$) then $C$ is closed. Here is why: let $x \notin C$ and for every $c \in C$ choose $U_c, V_c$ open where $x \in U_c, c \in V_c$. Now $\cup_{c \in C}V_c$ is an open set which contains $C$ and has a finite subcover $\cup_{i=1}^k V_i$ Note that each $U_i$ is an open set which contains $x$ and now we have only a finite number of these. Hence $x \in \cap^k_{i=1} U_i$ which is disjoint from $\cup_{i=1}^k V_i$ which contains $C$. Because $x$ was an arbitrary point in $X -C$, $X-C$ is open which means that $C$ is closed. Note: this proof, with one minor addition, shows that a compact Hausdorff space is regular ($T_3$) we need only show that a closed subset of a compact Hausdorff space is compact. That is easy enough to do: let $\mathscr{U}$ be an open cover for $C$; then the collection $\mathscr{U} \cup (X-C)$ is an open cover for $X$, which has a finite subcover. Let that be $\cup^k_{i=1} U_i \cup (X-C)$ where each $U_i \in \mathscr{U}$. Now since $X-C$ does not cover $C, \cup^k_{i=1} U_i$ does.

So we have proved that a closed subset of a compact set is compact.

5. Let $R$ (or any infinite set) be given the finite complement topology (that is, the open sets are the empty set together with sets whose complements consist of a finite number of points). Then ANY subset is compact! Here is why: let $C$ be any set and let $\mathscr{U}$ be any open cover. Choose $U_1 \in \mathscr{U}$. Since $X -U_1$ is a finite set of points, only a finite number of them can be in $C$, say $c_1, c_2, ...c_k$. Then for each of these, select one open set in the open cover that contains the point; that is the finite subcover.

Note: this shows that non-closed sets can be compact sets, if the underlying topology is not Hausdorff.

6. If $f: X \rightarrow Y$ is continuous and onto and $X$ is compact, then so is $Y$. Proof: let $\cup_{\beta \in I} U_{\beta}$ cover $Y$ and note that $\cup_{\beta}f^{-1}(U_{\beta})$ covers $X$, hence a finite number of these open sets cover: $X = \cup^{k}_{i=1}f^{-1}(U_i)$. Therefore $\cup^k_{i=1}U_i$ covers $Y$. Note: this shows that being compact is a topological invariant; that is, if two spaces are homeomorphic, then either both spaces are compact or neither one is.

7. Ok, let’s finally prove something. Let $R^1$ have the usual topology. Then $[0, 1]$ (and therefore any closed interval) is compact. This is (part) of the famous Heine-Borel Theorem. The proof uses the least upper bound axiom of the real numbers.

Let $\mathscr{U}$ be any open cover for $[0,1]$. If no finite subcover exists, let $x$ be the least upper bound of the subset $F$ of $[0,1]$ that CAN be covered by a finite subcollection of $\mathscr{U}$. Now $x > 0$ because at least one element of $\mathscr{U}$ contains $0$ and therefore contains $[0, \delta )$ for some $\delta > 0$. Assume that $x < 1$. Now suppose $x \in F$, that is $x$ is part of the subset that can be covered by a finite subcover. Then because $x \in U_{\beta}$ for some $U_{\beta} \in \mathscr{U}$ then $(x-\delta, x + \delta) \subset U_{\beta}$ which means that $x + \delta \in F$, which means that $x$ isn’t an upper bound for $F$.

Now suppose $x \notin F$; then because $x < 1$ there is still some $U_{\beta}$ where $(x-\delta, x+ \delta) \subset U_{\beta}$. But since $x = lub(F)$ then $x - \delta \in F$ and so $[0, x- \delta ) \subset F$. So if $F$ can be covered by $\cup^k_{i=1} U_i$ then $\cup^k_{i=1} U_i \cup U_{\beta}$ is a finite subcover of $[0, x + \delta )$ which means that $x$ was not an upper bound. It follows that $x = 1$ which means that the unit interval is compact.

Now what about the closed ball in $R^n$? The traditional way is to show that the closed ball is a closed subset of a closed hypercube in $R^n$ and so if we show that the product of compact spaces is compact, we would be done. That is for later.

8. Now endow $R^1$ with the lower limit topology. That is, the open sets are generated by basis elements $[a, b)$. Note that the lower limit topology is strictly finer than the usual topology. Now in this topology: $[0,1]$ is not compact. (note: none of $(0,1), [0,1), (0, 1]$ are compact in the coarser usual topology, so there is no need to consider these). To see this, cover $[0,1]$ by $\cup ^{\infty}_{n=1} [0, \frac{n}{n+1}) \cup [1, \frac{3}{2})$ and it is easy to see that this open cover has no finite subcover. In fact, with a bit of work, one can show that every compact subset is at most countable and nowhere dense; in fact, if $A$ is compact in the lower limit topology and $a \in A$ there exists some $y_a$ where $(y_a, a) \cap A = \emptyset$.

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