Completing a Metric Space, Intuitively

An incomplete metric space is very much like a golf course: it has a lot of missing points!

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What's a Transitive Group Action?

Let a group $G$ act on a set $X$. The action is said to be transitive if for any two $x,y\in X$ there is a $g\in G$ such that $g\cdot x=y$. This is equivalent to saying there is an $x\in X$ such that $\text{orb}(x)=X$, i.e. there is exactly one orbit. And all this is just the fancy way of saying that $G$ shuffles all the elements of $X$ among themselves. In other words…

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One Unspoken Rule of Measure Theory

Here's a measure theory trick: when asked to prove that a set of points in $\mathbb{R}$ (or some measure space $X$) has a certain property, try to show that the set of points which does NOT have that property has measure 0! This technique is used quite often.

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Two Ways to be Small

  In real analysis, there are two ways a measurable set $E$ can be small. Either

  • the measure of $E$ is 0, OR
  • $E$ is nowhere dense.

Intuitively, to say the measure of $E$ is $0$ means that...

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Need to Prove Your Ring is NOT a UFD?

You're given a ring $R$ and are asked to show it's not a UFD. Where do you begin? One standard trick is to apply the Rational Roots Theorem….

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Operator Norm, Intuitively

If $X$ and $Y$ are normed vector spaces, a linear map $T:X\to Y$ is said to be <b>bounded</b> if $\|T\|< \infty$ where

$$\|T\|=\sup_{\underset{x\neq 0}{x\in X}}\left\{\frac{|T(x)|}{|x|}\right\}.$$

(Note that $|T(x)|$ is the norm in $Y$ whereas $|x|$ is the norm in $X$.) One can show that this is equivalent to

$$\|T\|=\sup_{x\in X}\{|T(x)|:|x|=1\}.$$ So intuitively (at least in two dimensions), we can think of $\|T\|$ this way…

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Borel-Cantelli Lemma (Pictorially)

The Borel-Cantelli Lemma says that if $(X,\Sigma,\mu)$ is a measure space with $\mu(X)<\infty$ and if $\{E_n\}_{n=1}^\infty$ is a sequence of measurable sets such that $\sum_n\mu(E_n)<\infty$, then $$\mu\left(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k\right)=\mu\left(\limsup_{n\to\infty} En \right)=0.$$

(For the record, I didn't understand this when I first saw it (or for a long time afterwards). My only thought was, "But what does that mean? In English??") To help our intuition…

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What do Polygons and Galois Theory Have in Common?

Galois Theory is all about symmetry. So, perhaps not surprisingly, symmetries found among the roots of polynomials (via Galois theory) are closely related to symmetries of polygons in the plane (via geometry). In fact, the two are highly analogous! 

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One Unspoken Rule of Algebra

Here's an algebra tip! Whenever you're asked to prove $$A/B\cong C$$ where $A,B,C$ are groups, rings, fields, modules, etc., mostly likely the The First Isomorphism Theorem involved!

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Motivation for the Tensor Product

In general, if $F$ is a field and $V$ is a vector space over $F$, the tensor product answers the question "How can I define scalar multiplication on $V$ by some larger field which contains $F$?" (Of course this holds if we replace the word "field" by "ring" and consider the same scenario with modules.)

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Why are Noetherian Rings Special?

In short, "Noetherian-ness" is a property which generalizes "PID-ness." As Keith Conrad so nicely puts it, "The property of all ideals being singly generated is often not preserved under common ring-theoretic constructions (e.g. $\mathbb{Z}$ is a PID but $\mathbb{Z}[x]$ is not), but the property of all ideals being finitely generated does remain valid under many constructions of new rings from old rings. For example... every quadratic ring $\mathbb{Z}[\sqrt{d}]$ is Noetherian, even though many of these rings are not PIDs." (italics added)

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What is Galois Theory Anyway?

Perhaps you've heard of Évariste Galois? (Pronounced "GAL-wah.") You know, the French mathematician who died tragically in 1832 in a duel at the tender age of 20? (Supposedly over a girl! C'est romantique, n'est-ce pas?)  Well, today we're taking a bird's-eye view of his most well-known contribution to mathematics: the appropriately named Galois theory. The goal of this post is twofold...

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Baire Category & Nowhere Differentiable Functions (Part Two)

Welcome to part two of our discussion on Baire's Category Theorem. Today we'll sketch the proof that we can find a continuous function on $[0,1]$ which is nowhere differentiable.

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A Non-Measurable Set

Today we're looking at a fairly simple proof of a standard result in measure theory:

Theorem: Any measurable subset $A$ of the real line with positive measure contains a non-measurable subset.

(Remark: we used this theorem last week to prove the existence of a Lebesgue measurable set which is not a Borel set.)

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Lebesgue Measurable But Not Borel

Our goal for today is to construct a Lebesgue measurable set which is not a Borel set. In summary, we will define a homeomorphism from $[0,1]$ to $[0,2]$  which will map a (sub)set (of the Cantor set) of measure 0 to a set of measure 1. This set of measure 1 contains a non-measurable subset, say $N$. And the preimage of $N$ will be Lebesgue measurable but will not be a Borel set.

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Baire Category & Nowhere Differentiable Functions (Part One)

The Baire Category Theorem is a powerful result that relates a metric space to its underlying topology. (And sadly no, nothing to do with category theory!) Informally, the theorem says that if you can find a metric with respect to which your topological space is complete, then that space cannot be written as a countable union of nowhere dense sets. In other words, a metric can put a restriction on the topology.

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