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

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: Let be a path connected

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. For all , . For real

Every set has a minimal superset

I credit Mikey and Atal with this proof. For any set which is a subset of a (lets say much larger) set , a minimal superset of is a set such that , and if , then for any superset of . I claim that any set has a minimal superset. We see this by

Showing two easy spaces are not homeomorphic.

This one is my own doing. I will be showing the following: is not homeomorphic to First, we begin with a definition and a lemma. If is any topological space, we define the of as the topological space We have the following lemma: if are two homeomorphic topological spaces, then their one point compactifications are

Addendum to the previous post

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: If are two homeomorphic

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 contains a unique prime number without using the fact that is a PID, hence the mostly in the title. We begin

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