A proof that there exists an infinite collection of sets whose cartesian product is nonempty.

I stumbled upon this succinct proof while discussing topology with some friends.

Suppose that, for any infinite collection of sets $A$, the cartesian product, i.e. the collection of all choice functions on $A$, is empty. I claim this implies Tychonoff’s theorem, which states that the product of any collection of compact topological spaces is a compact topological space.

To see this, we note that Tychonoff’s theorem holds easily for any finite collection of topological spaces. Now, for any infinite collection of compact topological spaces, the product of all of these is empty by assumption, hence compact.

Thus, Tychonoff’s theorem is shown to be true. This implies the axiom of choice, which contradicts our hypothesis.

By Modus Tollens, we thus conclude that there exists some infinite collection of sets, the cartesian product of all of which is nonempty.

Using the Seifert-van Kampen theorem to calculate the fundamental group of the torus, aided by illustrations from mspaint

One of my favorite theorems is the Seifert-van Kampen theorem. It’s a very handy result in algebraic topology which allows us to calculate the fundamental group of complicated spaces by breaking them down into simpler spaces. The version of the theorem I’ll be using here can be stated as follows:

$\textbf{Theorem (Seifert-van Kampen):}$ Let $X$ be a path connected topological space, and let $U_1, U_2$ be subsets of $X$ such that $U_1, U_2$ are both open and path connected, $U_1 \cap U_2$ is path connected, and $U_1 \cup U_2 = X$. If for some $x_0 \in U_1 \cap U_2$ we have

$\pi_1(U_1, x_0) \cong \langle a_1, \dots, a_n \mid R_1 \rangle$

$\pi_1(U_2, x_0) \cong \langle b_1, \dots, b_m \mid R_2 \rangle$

$\pi_1(U_1 \cap U_2, x_0) \cong \langle c_1, \dots, c_j \mid R_3 \rangle$

where $R_1, R_2, R_3$ are a set of relations, then we can calculate the fundamental group of $U_1 \cup U_2 = X$ in the following way. If $I_1:U_1\cap U_2\to U_1$ and $I_2:U_1\cap U_2 \to U_2$ are the inclusion maps, then we have

$\pi_1(U_1 \cup U_2, x_0) \cong \langle a_1, \dots a_n, b_1, \dots, b_m \mid R_1, R_2, R_4\rangle$

where $R_4 = \{I_1(c_k) = I_2(c_k) \mid k\in \{1,\dots, j\}\}$

$\textbf{Important corollary:}$ If $U_1 \cap U_2$ is simply connected, then the fundamental group of $X$ is the free product of the fundamental groups of $U_1$ and $U_2$. This follows by noting that $I_i(C_k) = 1$ for all $i \in \{1,2\}, k \in \{1,\dots, j\}$, and hence $R_4$ consists only of the trivial relation.

Side note: from now on, because $X$ is path connected, the choice of base point does not affect the isomorphism class (this is a classic result). As such, I will be simplifying notation by referring to the fundamental group of $X$ as simply $\pi_1(X)$.

We will use this to calculate $\pi_1(T)$, there $T$ is the torus, which is realized as the following identification space:

We will let $U_1$ be a small disk in the center, and we will let $U_2$ be the complement of $U_1$, plus a little bit, so the intersection will be an open annulus. Here is an illustration:

We can see that $U_1$ is simply connected, so $\pi_1(U_1)$ is trivial. Further, because the intersection of $U_1$ and $U_2$ is an annulus, it has fundamental group isomorphic to $\mathbb{Z}\cong\langle a\mid \emptyset \rangle$, where $a$ is a clockwise loop which encircles $U_1$. We next need to find the fundamental group of $U_1$, which is the punctured torus.

First, we take the “hole” at the center of the square and “stretch” it out:

$\to$

Now, as in the statement of the Seifert-van Kampen theorem above, we will let $U_1$ be defined as follows:

Similarly, here is $U_2$

So, here is $U_1 \cap U_2$

We see that $U_1$ and $U_2$ are both annuli, and hence have fundamental group $\mathbb{Z}$. Further, their intersection is in an X shape (as in, shaped like an uppercase X), so is simply connected. Thus, the fundamental group of the punctured torus is the free product of $\mathbb{Z}$ with itself, which is isomorphic to the free group on two generators.

As a side note, we can specify what the generators of this group are. A loop which starts at the bottom left of the square and travels along the left edge is not homotopic to a loop which starts at the bottom left and travels along the bottom edge.

So, back to this:

To summarize so far, we have:

$\pi_1(U_1) \cong \langle \mid e = 1 \rangle$

$\pi_1(U_2)\cong \langle a, b \mid \emptyset \rangle$

$\pi_1(U_1\cap U_2)\cong \langle c \mid \emptyset \rangle$

We know the fundamental groups of all the relevant spaces, so we are almost done. All we have to do is calculate $R_4$, and we can put all the pieces together. We see that $I_1(c) = 1$. What is $I_2(c)$?

Here is what $I_2(c)$ looks like as a loop in $U_2$. We make the looper wider, pressing it towards the boundary of the square as follows:

$\to$$\to$

The loop is clockwise, so when it reaches the edge, we can trace the loop by going up along the left edge, right along the top edge, down along the right edge, and left along the bottom edge. We noted before that if the fundamental group of the punctured torus is $\langle a, b \mid \emptyset \rangle$, then $a$ can be a loop starting at the bottom left and traversing the left edge, and $b$ can be a loop starting at the bottom left and traversing the bottom edge. Because this is a quotient space, we can see that $a$ is also the same as starting at the top right corner, going “up” to reach the bottom right corner, and traversing the right edge, and $b$ is the same as starting at the top right corner, going “right” to reach the top left corner, and traversing the top edge.

So, $I_2(c)$ is a loop traversing the edges of the square in a clockwise manner. In other words, $I_2(c)$ can be achieved by traversing $a$, followed by $b$, followed by $a^{-1}$, followed by $b^{-1}$. So, $I_2(c) = aba^{-1}b^{-1} = [a, b]$.

So, here’s the punchline:

$\pi_1(T) \cong \langle a, b, e \mid e = 1, I_1(c) = I_2(c) \rangle \cong \langle a, b \mid ab = ba \rangle$

In other words, $\pi_1(T) \cong \mathbb{Z}^2$!

A Classic Joke Proof

While this may stretch the stated point of this blog, I’m ultimately the one in control, and I’ll decide what gets posted here. And I think it’s really funny and would be fun to write up.

I am nowhere near the first person who has ever told this joke.

$\textbf{Claim:}$ For all $n > 2$, $\sqrt[n]{2}\not\in\mathbb{Q}$.

$\textbf{Proof:}$ For real numbers $a, b \in \mathbb{R}$, such that $b \neq 0$, suppose that $(\frac{a}{b})^n = 2$. We have that $a^n = 2b^n = b^n + b^n$.

By Fermat’s last theorem, we have $a, b$ are not both integers, and hence that $\frac{a}{b}$ cannot be rational (it obviously implies $a, b$ aren’t both integers, but how does it imply that they aren’t both rational? Think on this…).

Unfortunately, I have not yet read and understood Wiles’ proof of FLT, so I do not know if this is a circular argument.

Every set has a minimal superset

I credit Mikey and Atal with this proof.

For any set $A$ which is a subset of a (lets say much larger) set $K$, a minimal superset of $A$ is a set $B\subset K$ such that $A\subset B$, and if $A \subset C$, then $B\subset C$ for any superset $C$ of $A$.

I claim that any set $A$ has a minimal superset. We see this by using Zorn’s Lemma on the elements of $\mathcal{P}(K)$ which contain $A$ (I denote this collection of sets as $K_A$). The ordering will be the reverse of the typical: if $I, J \in K_A$ such that $I\subset J$, then we say $I > J$. If $S_1 < S_2 < \dots$ is a chain in $K_A$, then this chain is bounded above by $\bigcap_{i=1}^\infty S_i$. Thus, by Zorn’s Lemma, there is a maximal element $B$.

I claim this $B$ is a minimal subset. We see that $B$ must contain $A$ by definition, and further $B$ contains no other elements of $K_A$. Suppose that $C$ is in $K_A$. We have that $A\subset C$ and $A \subset C\cap B$, as both $C$ and $B$ are in $K_A$. Further, $C\cap B \subset B$. Because $C\cap B$ contains $A$ and is a subset of $B$, we are forced to conclude, by maximality of $B$, that $B = C\cap B$, meaning that $B \subset C$. Thus, $B$ is a minimal superset of $A$.

Of course, this whole proof can be greatly simplified by noting that $A$ itself will fulfill the property of being a minimal superset of $A$, so this is a proof by immediate example. Hey, the definition doesn’t forbid it!

Showing two easy spaces are not homeomorphic.

This one is my own doing. I will be showing the following:

$(0,\infty)$ is not homeomorphic to $[0, \infty)$

First, we begin with a definition and a lemma. If $X = (A, \tau)$ is any topological space, we define the $\textbf{one point compactifiation}$ of $X$ as the topological space

$(A \cup \{ \infty_X \}, \tau \cup \{B \cup \{ \infty_X \} \mid A/B \text{ is a compact closed subset of X} \})$

We have the following lemma:

$\textbf{Lemma:}$ if $X, Y$ are two homeomorphic topological spaces, then their one point compactifications are homeomorphic.

$\textbf{Proof:}$ Because $X, Y$ are homeomorphic, there is a homeomorphism $f:X\to Y$ between them. Denote the one point compactifications of $X, Y$ as $\hat{X}, \hat{Y}$, respectively. Define $\hat{f}:\hat{X}\to\hat{Y}$ such that $f(\infty_X)=\infty_Y$, and such that $\hat{f}(x) = f(x)$ for any non-infinite $x\in X$.

I claim that $\hat{f}$ is a homeomorphism. Suppose that $U\subset \hat{Y}$ is an open subset of $\hat{Y}$ which does not contain $\infty_Y$. Then, by construction, we have that $\hat{f}^{-1}(U)$ is an open subset of $\hat{X}$ which does not contain $\infty_X$. Suppose that $U\subset \hat{Y}$ is an open subset of $\hat{Y}$ which does contain $\infty_Y$. By construction, the complement of $U$ is closed and compact. So, $f^{-1}(\hat{Y}/U)$ is a compact and closed subset of $\hat{X}$ by construction, so $\hat{f}^{-1}(U)$ is the complement of a closed compact subset of $X$ and contains $\infty_X$, so is open. Similarly, $\hat{f}^{-1}$ is continuous, so we are done.

With that out of the way, it is well known that $(0, \infty)$ is homeomorphic to $\mathbb{R}$, whose one point compactification is homeomorphic to $S^1$, the unit circle. However, $[0, \infty)$ is homeomorhic to $[0, 1)$, whose one point compactification is homeomorphic to $[0, 1]$. I claim the following:

$[0, 1]$ is not homeomorphic to $S^1$

This can be seen by noting that $\pi_1([0, 1], 1)\cong\{0\}$ while $\pi_1(S^1, 1)\cong\mathbb{Z}$, and clearly these are not isomorphic. By the functoriality of $\pi_1$, we have that $[0, 1]$ is not homeomorphic to $S^1$. By the contraposition of the lemma, we thus see that $(0, \infty)$ and $[0, \infty)$ are not homeomorphic.

The previous post is very inelegant because it invokes machinery such as the one point compactification (not too absurd for a problem of this nature I suppose) and the fundamental group (which is way overkill). The best way (in my opinion) to solve this problem is to use the following lemma:

$\textbf{Lemma:}$ If $X, Y$ are two homeomorphic topological spaces, with a homeomorphism given by $f:X\to Y$, and if $S$ is any subspace of $X$, then $S$ is homeomorphic to $f(S)$ as a subspace of $Y$, with a homeomorphism given by $f|_S$

$\textbf{Proof:}$ Let $U$ be any open subset of $f(S)$. By definition, this is of the form $U' \cap f(S)$ for some $U'$ which is open in $Y$. We have the following:

$f|_S^{-1}(U) = f|_S^{-1}(U'\cap f(S)) = f^{-1}(U')\cap f^{-1}(f(S)) = f^{-1}(U')\cap S$

The right hand side is open in $S$, so $f|_S$ is continuous. Similarly, $f|_S^{-1}$ is continuous, so we are done.

Now, to use the lemma: suppose that the two spaces mentioned in the previous post were homeomorphic, and let $f$ be such a homeomorphism. For simplicity, the domain of $f$ shall be $[0,\infty)$. Consider $S = (0,\infty)$ as a subset. We see that $f(S) = (0,\infty) / f(0)$. However, in this case $S$ is connected, but $f(S)$ isn’t, a contradiction.

This seemingly dumb trick can be used to prove that many different spaces are not homeomorphic. I leave the following as exercises to the reader using this.

1. Show that a figure 8 and a circle are not homeomorphic.
2. Show that an uppercase X and an uppercase T are not homeomorphic.
3. Show that $S^1$ and $\mathbb{R}$ are not homeomorphic (see the previous post) without invoking their fundamental groups.

A (mostly) purely algebraic proof of the infinitude of prime numbers

The following proof was provided/explained to me by @grassmanian on twitter. I thank him again for the support. Unfortunately, neither he nor I have found a way to show that each maximal ideal of $\mathbb{Z}$ contains a unique prime number without using the fact that $\mathbb{Z}$ is a PID, hence the mostly in the title.

We begin with a lemma:

$\textbf{Lemma:}$ If $A$ is an integral domain which has finitely many units, and finitely many maximal ideals, then it is a field.

$\textbf{Proof:}$ It is known that any finite integral domain is a field, so we shall assume that $A$ is infinite. Consider the Jacobson radical of A, denoted Jac(A). This is defined as follows:

$\text{Jac}(A) := \{x \mid 1 + xy \text{ is a unit for all }y\in A\}$

We observe the following: because $A$ is an integral domain, we have that for any nonzero $x\in A$, and for any $y_1\neq y_2 \in A$, $1 + xy_1 \neq 1 + xy_2$. $A$ is infinite, so if Jac(A) contains any nonzero elements, then $A$ will have infinitely many units. This is forbidden by hypothesis, so we are forced to conclude that $\text{Jac}(A) = \{0\}$.

Now, because $A$ is an integral domain (in particular because $A$ is commutative), we have the following equivalence of definitions:

$\text{Jac}(A) = \bigcap_{M\in\text{MSpec}(A)}M$

Where $\text{MSpec}(A)$ denotes all maximal ideals of $A$.

By hypothesis, $A$ has only finitely many maximal ideals. That is, $\text{MSpec}(A) = \{M_1, M_2, \dots, M_n\}$ for some $n\in \mathbb{N}$. Any two maximal ideals are coprime, so we may conclude the following by the Chinese Remainder Theorem for rings:

$A = A/(0) = A/(M_1\cap M_2\cap \dots \cap M_n) \cong \prod_{i=1}^n A/M_i$

Because each $M_i$ is maximal, we have that each $A/M_i$ is in fact a field, and thus $A$ is a product of fields. However, the product of two or more fields is never an integral domain. By hypothesis $A$ is an integral domain, so we are forced to conclude that there is only one such $M_i$, meaning that $A$ is indeed a field.

With that lemma out of the way, we proceed.

$\textbf{Proof of the main claim:}$ Because $\mathbb{Z}$ is an integral domain with only finitely many units ($\{-1, 1\}$), but is not a field, we may conclude that $\mathbb{Z}$ has infinitely many maximal ideals. $\mathbb{Z}$ is a principal ideal domain, so we have that each maximal ideal is of the form $(p)$ for some prime $p\in\mathbb{Z}$. Hence, there are infinitely many primes.

The CoBook

Well, this is my attempt at a blog. This is going to be a mostly for fun repository of miscellaneous math stuff, so please nobody get angry at me if I say something dumb.

The title is a reference to “The Book,” which is a collection of the most elegant proofs to all math problems that Paul Erdős claimed God was keeping from us. It is also a dumb category theory joke, wherein the prefix “co” denotes the dual of some object. In category theory, this means reversing all the arrows, hence the url.

The content of this blog will mostly be my writeups of “inelegant” solutions to problems I come up with, and occasionally I will post something more substantive. My definition of “elegant” is usually “invokes the least,” so that should tell you something about the kind of things I’ll be posting.

I hope you enjoy this, and if you don’t, I hope you forget about it quickly.

I’m doing this basically out of boredom and a desire to practice my skills at writing things up nicely and cleanly, so don’t expect too much out of me!

I’ll post daily, maybe?