# The Unapologetic Mathematician

## Some Banach Spaces

To complete what we were saying about the $L^p$ spaces, we need to show that they’re complete. As it turns out, we can adapt the proof that mean convergence is complete, but we will take a somewhat different approach. It suffices to show that for any sequence of functions $\{u_n\}$ in $L^p$ so that the series of $L^p$-norms converges

$\displaystyle\sum\limits_{n=1}^\infty\lVert u_n\rVert_p<\infty$

the series of functions converges to some function $f\in L^p$.

For finite $p$, Minkowski’s inequality allows us to conclude that

$\displaystyle\int\left(\sum\limits_{n=1}^\infty\lvert u_n\rvert\right)^p\,d\mu=\left(\sum\limits_{n=1}^\infty\lVert u_n\rVert_p\right)^p<\infty$

The monotone convergence theorem now tells us that the limiting function

$\displaystyle f=\sum\limits_{n=1}^\infty u_n$

is defined a.e., and that $f\in L^p$. The dominated convergence theorem can now verify that the partial sums of the series are $L^p$-convergent to $f$:

$\displaystyle\int\left\lvert f-\sum\limits_{k=1}^n u_k\right\rvert^p\,d\mu\leq\int\left(\sum\limits_{l=n+1}^\infty\lvert u_l\rvert\right)^p\to0$

In the case $p=\infty$, we can write $c_n=\lVert u_n\rVert_\infty$. Then $\lvert u_n(x)\rvert except on some set $E_n$ of measure zero. The union of all the $E_n$ must also be negligible, and so we can throw it all out and just have $\lvert u_n(x)\rvert. Now the series of the $c_n$ converges by assumption, and thus the series of the $u_n$ must converge to some function bounded by the sum of the $c_n$ (except on the union of the $E_n$).

August 31, 2010

## The Supremum Metric

We can actually extend what we’ve been doing with Hölder’s inequality and Minkowski’s inequality a little further. Given a metric space $(X,\mathcal{S},\mu)$, we’ve already discussed the idea of an “essentially bounded” function — one for which there is some real constant $c$ so that $f(x)\leq c$ for almost all $x\in X$. We will write $L^\infty(X)$ for the collection of essentially bounded functions on the measure space. It should be clear that these form a vector space.

We also discussed the “essential supremum” $\text{ess sup}(\lvert f\rvert)$ of an essentially bounded function. We’ll now write this as $\lVert f\rVert_\infty$, suggesting that it’s a norm. And it’s clear that $\lVert cf\rVert_\infty=\lvert c\rvert\lVert f\rVert_\infty$, and that $\lVert f\rVert_\infty=0$ if and only if $f=0$ almost everywhere. Verifying the triangle identity is exactly Minkowski’s inequality.

And, indeed, we know that $\lvert f(x)\rvert\leq\lVert f\rVert_\infty$ and $\lvert g(x)\rvert\leq\lVert g\rVert_\infty$ a.e., so $\lvert f(x)+g(x)\rvert\leq\lvert f(x)\rvert+\lvert g(x)\rvert\leq\lVert f\rVert_\infty+\lVert g\rVert_\infty$ a.e., so whatever the least such essential upper bound is smaller still. That is, $\lVert f+g\rVert_\infty=\lVert f\rVert_\infty+\lVert g\rVert_\infty$.

Now for Hölder’s inequality. For this purpose we consider $\frac{1}{\infty}=0$, and thus $\frac{1}{1}+\frac{1}{\infty}=1$, which means that $1$ and $\infty$ are Hölder-conjugates. Thus our assertion is that if $f$ is integrable and $g$ is essentially bounded, then $fg$ is integrable and $\lVert fg\rVert_1\leq\lVert f\rVert_1\lVert g\rVert_\infty$. Indeed, we know that $\lvert g(x)\rvert\leq\lVert g\rVert_\infty$, and so $\lvert f(x)g(x)\rvert=\lvert f(x)\rvert\lvert g(x)\rvert\leq \lvert f(x)\rvert\lVert g\rVert_\infty$ — both inequalities holding almost everywhere. From this, we conclude that

\displaystyle\begin{aligned}\lVert fg\rVert_1&=\int\lvert fg\rvert\,d\mu\\&\leq\int\lvert f\rvert\lVert g\rVert_\infty\,d\mu\\&=\int\lvert f\rvert\,d\mu\lVert g\rVert_\infty\\&=\lVert f\rVert_1\lVert g\rVert_\infty\end{aligned}

as we asserted. From now on, we’ll allow $p=\infty$ (and $q=1$) whenever we’re talking about a Hölder-conjugate pair or $L^p$-space.

August 30, 2010 Posted by | Analysis, Measure Theory | 1 Comment

## What’s Happening?

You might be expecting another “Sunday Sample” here today, as there have been for the last 187 weeks, but not today. And you’ll find that they don’t even exist here anymore!

What happened? They’ve all been moved (along with this week’s installment) to drmathochist.wordpress.com, which will be my place for more personal posts. I’ll also relax the filter there, so Dr. Friedman can quit complaining about self-censorship (and I can post Cee Lo Green’s new track!)

At the same time, I’ve just opened up The Unapologetic Programmer for my more computer-programming related posts. It’ll be much less organized than here, I’m sure, but I know there’s a bunch of people into that sort of thing here.

## Minkowski’s Inequality

We continue our project to show that the $L^p$ spaces are actually Banach spaces with Minkowski’s inequality. This will allow us to conclude that $L^p$ is a normed vector space. It states that if $f$ and $g$ are both in $L^p$, then their sum $f+g$ is in $L^p$, and we have the inequality

$\displaystyle\lVert f+g\rVert_p\leq\lVert f\rVert_p+\lVert g\rVert_p$

We start by considering Hölder’s inequality in a toy space I’ll whip up right now. Take two isolated points, and let each one have measure $1$; the whole space of both points has measure $2$. A function is just an assignment of a pair of real values $(a_1,a_2)$, and integration just means adding them together. Hölder’s inequality for this space tells us that

$\displaystyle\lvert a_1b_1+a_2b_2\rvert\leq\left(\lvert a_1\rvert^p+\lvert a_2\rvert^p\right)^\frac{1}{p}\left(\lvert b_1\rvert^q+\lvert b_2\rvert^q\right)^\frac{1}{q}$

where $p$ and $q$ are Hölder-conjugate to each other. We can set $a_1=\lvert f\rvert^p$, $a_2=\lvert g\rvert^p$, and $b_1=b_2=\lvert f+g\rvert^{p-1}$ and use this inequality to find

\displaystyle\begin{aligned}\lvert f+g\rvert^p&=\lvert f+g\rvert\,\lvert f+g\rvert^{p-1}\\&\leq(\lvert f\rvert+\lvert g\rvert)\lvert f+g\rvert^{p-1}\\&=\lvert f\rvert\,\lvert f+g\rvert^{p-1}+\lvert g\rvert\,\lvert f+g\rvert^{p-1}\\&\leq\left(\lvert f\rvert^p+\lvert g\rvert^p\right)^\frac{1}{p}\left(2\lvert f+g\rvert^{q(p-1)}\right)^\frac{1}{q}\\&\leq\left(\lvert f\rvert^p+\lvert g\rvert^p\right)^\frac{1}{p}2^\frac{1}{q}\lvert f+g\rvert^{p-1}\end{aligned}

Dividing out $\lvert f+g\rvert^{p-1}$ and raising both sides to the $p$th power, we conclude that $\lvert f+g\rvert^p\leq 2^\frac{p}{q}\left(\lvert f\rvert^p+\lvert g\rvert^p\right)$. Thus if both $\lvert f\rvert^p$ and $\lvert g\rvert^p$ are integrable, then so is $\lvert f+g\rvert^p$. Thus $f+g$ must be in $L^p$.

Now we calculate

\displaystyle\begin{aligned}\lVert f+g\rVert_p^p&=\int\lvert f+g\rvert^p\,d\mu\\&\leq\int\lvert f\rvert\,\lvert f+g\rvert^{p-1}\,d\mu+\int\lvert g\rvert\,\lvert f+g\rvert^{p-1}\,d\mu\\&\leq\left(\int\lvert f\rvert^p\,d\mu\right)^\frac{1}{p}\left(\int\lvert f+g\rvert^{q(p-1)}\,d\mu\right)^\frac{1}{q}+\left(\int\lvert g\rvert^p\,d\mu\right)^\frac{1}{p}\left(\int\lvert f+g\rvert^{q(p-1)}\,d\mu\right)^\frac{1}{q}\\&\leq\left(\int\lvert f\rvert^p\,d\mu\right)^\frac{1}{p}\left(\left(\int\lvert f+g\rvert^p\,d\mu\right)^\frac{1}{p}\right)^\frac{p}{q}+\left(\int\lvert g\rvert^p\,d\mu\right)^\frac{1}{p}\left(\left(\int\lvert f+g\rvert^p\,d\mu\right)^\frac{1}{p}\right)^\frac{p}{q}\\&=\left(\lVert f\rVert_p+\lVert g\rVert_p\right)\left(\lVert f+g\rVert_p\right)^\frac{p}{q}\end{aligned}

Dividing out by $\left(\lVert f+g\rVert_p\right)^\frac{p}{q}$ we find that

$\displaystyle\lVert f+g\rVert_p=\left(\lVert f+g\rVert_p\right)^\frac{p}{p}=\left(\lVert f+g\rVert_p\right)^{p\left(1-\frac{1}{q}\right)}=\left(\lVert f+g\rVert_p\right)^{p-\frac{p}{q}}\leq\lVert f\rVert_p+\lVert g\rVert_p$

This lets us conclude that $L^2$ is a vector space. But we can also verify the triangle identity now. Indeed, if $f$, $g$, and $h$ are all in $L^p$, then Minkowski’s inequality shows us that

$\displaystyle\rho_p(f,g)=\lVert f-g\rVert_p\leq\lVert f-h\rVert_p+\lVert h-g\rVert_p=\rho_p(f,h)+\rho_p(h,g)$

which is exactly the triangle inequality we want. Thus $\lVert\cdot\rVert_p$ is a norm, and $L^p$ is a normed vector space.

August 27, 2010 Posted by | Analysis, Measure Theory | 2 Comments

## Hölder’s Inequality

We’ve seen the space of integrable functions on a measure space $X$, which we called $L^1$ or $L^1(\mu)$. We’ve seen that this gives us a complete normed vector space — a Banach space. This is what we’d like to generalize.

Given a real number $p>1$, we define the space $L^p$ or $L^p(\mu)$ to be the collection of all measurable functions $f$ for which $\lvert f\rvert^p$ is integrable. As in the case of $L^1$, we identify two functions if they’re equal $\mu$-almost everywhere.

It will turn out that these are Banach spaces. We define the $L^p$ norm

$\displaystyle\lVert f\rVert_p=\left(\int\lvert f\rvert^p\,d\mu\right)^{\frac{1}{p}}$

and we write $\rho_p(f,g)=\lVert f-g\rVert_p$ to define a metric. This is clearly non-negative, and we see that $\rho_p(f,g)=0$ if and only if $f=g$ $\mu$-a.e., just as before. It’s also clear that $\lVert cf\rVert_p=\lvert c\rvert\lVert f\rVert_p$. What we need to work to check is the triangle inequality. It’s also not quite so apparent a problem, but we actually don’t know yet that this is a vector space at all! That is, how do we know that $\lvert f-g\rvert^p$ is integrable if $\lvert f\rvert^p$ and $\lvert g\rvert^p$ are?

As a first step in this direction, we prove Hölder’s inequality: if $p$ and $q$ are real numbers greater than $1$ such that $\frac{1}{p}+\frac{1}{q}=1$, and if $f\in L^p$ and $g\in L^q$, then the product $fg\in L^1$ and $\lVert fg\rVert_1\leq\lVert f\rVert_p\lVert g\rVert_q$. To see this, we will use the function $\phi$ defined for all positive real numbers by

$\displaystyle\phi(t)=\frac{t^p}{p}+\frac{t^{-q}}{q}$

Differentiating, we see that $\phi'(t)=t^{p-1}-t^{-q-1}=t^{-q-1}(t^{p+q}-1)$, so the only (positive) critical point of $\phi$ is $1$. Since the limit as $t$ approaches $0$ and $\infty$ are both positive infinite, $t=1$ must be a local minimum. That is

$\displaystyle\frac{t^p}{p}+\frac{t^{-q}}{q}=\phi(t)\geq\phi(1)=\frac{1}{p}+\frac{1}{q}=1$

For any two real numbers $a$ and $b$, we can consider the value

$\displaystyle t=\frac{a^\frac{1}{q}}{b^\frac{1}{p}}$

and it follows that

\displaystyle\begin{aligned}1&\leq\phi(t)\\&=\frac{t^p}{p}+\frac{t^{-q}}{q}\\&=\frac{\frac{a^\frac{p}{q}}{b}}{p}+\frac{\frac{a^{-1}}{b^{-\frac{q}{p}}}}{q}\\&=\frac{a^\frac{p}{q}}{bp}+\frac{b^\frac{q}{p}}{aq}\\&=\frac{a^{p-1}}{bp}+\frac{b^{q-1}}{aq}\end{aligned}

and thus

$\displaystyle ab\leq\frac{a^p}{p}+\frac{b^q}{q}$

which is clearly also true even if we allow $a$ or $b$ to be zero. This is known as “Young’s inequality”.

Okay, so now we can turn to the theorem itself. If either $\lVert f\rVert_p=0$ or $\lVert g\rVert_q=0$, the inequality clearly holds. Otherwise, we define

$\displaystyle a=\frac{\lvert f\rvert}{\lVert f\rVert_p}\qquad b=\frac{\lvert g\rvert}{\lVert g\rVert_q}$

we can plug these into the above inequality to find

\displaystyle\begin{aligned}\frac{\lvert fg\rvert}{\lVert f\rVert_p\lVert g\rVert_q}&=\frac{\lvert f\rvert}{\lVert f\rVert_p}\frac{\lvert g\rvert}{\lVert g\rVert_q}\\&\leq\frac{\lvert f\rvert^p}{p\lVert f\rVert_p^p}+\frac{\lvert g\rvert^q}{q\lVert g\rVert_q^q}\\&=\frac{1}{p}\frac{\lvert f\rvert^p}{\int\lvert f\rvert^p\,d\mu}+\frac{1}{q}\frac{\lvert g\rvert^q}{\int\lvert g\rvert^q\,d\mu}\end{aligned}

Since the measurability of $f$ and $g$ implies that of $\lvert fg\rvert$, and the right hand side of this inequality is integrable, we conclude that $\lvert fg\rvert$ is integrable. If we integrate, we find

$\displaystyle\frac{\lVert fg\rVert_1}{\lVert f\rVert_p\lVert g\rVert_q}\leq\frac{1}{p}\frac{\int\lvert f\rvert^p\,d\mu}{\int\lvert f\rvert^p\,d\mu}+\frac{1}{q}\frac{\int\lvert g\rvert^q\,d\mu}{\int\lvert g\rvert^q\,d\mu}=\frac{1}{p}+\frac{1}{q}=1$

and Hölder’s inequality follows.

The condition relating $p$ and $q$ is very common in this discussion, so we will say that such a pair of real numbers are “Hölder conjugates” of each other. Given $p$, the Hölder conjugate $q$ is uniquely defined by $q=\frac{p}{p-1}$, which is a strictly decreasing function sending $(1,\infty)$ to itself (with order reversed, of course). The fact that this function has a (unique) fixed point at $2$ will be important. In particular, we will see that this norm is associated with an inner product on $L^2$, and that Hölder’s inequality actually implies the Cauchy-Schwarz inequality!

August 26, 2010 Posted by | Analysis, Measure Theory | 6 Comments

## The Measure Algebra of the Unit Interval

Let $Y$ be the unit interval $[0,1]$, let $\mathcal{T}$ be the class of Borel sets on $Y$, and let $\nu$ be Lebesgue measure. If $\{\mathcal{Q}_n\}$ is a sequence of partitions of the maximal element $Y$ of the measure algebra $(\mathcal{T},\nu)$ into intervals, and if the limit of the sequence of norms $\lvert\mathcal{Q}_n\rvert$ is zero, then $\{\mathcal{Q}_n\}$ is dense.

If $\epsilon$ is a positive number, then we can find some $n$ so that $\lvert\mathcal{Q}_n\rvert<\frac{\epsilon}{2}$. If $E$ is a subinterval of $Y$, then let $E_1\in\mathcal{Q}_n$ be the unique interval containing the left endpoint of $E$. If $E_1$ also contains the right endpoint, then we can stop. Otherwise, let $E_2$ be the next interval of $\mathcal{Q}_n$ to the right of $E_1$, and keep going (at most a finite number of steps) until we get to an interval $E_k$ containing the right endpoint of $E$. The union $U$ of all the $\{E_i\}$ can overshoot $E$ by at most $\frac{\epsilon}{2}$ on the left and the same amount on the right, and so $\rho(E,U)<\epsilon$. Thus any interval can be approximated arbitrarily closely by some partition in the sequence. The general result follows, because we can always find a finite collection of intervals whose union is arbitrarily close to any given Borel set.

Now, say that every separable, non-atomic, normalized measure algebra $(\mathcal{S},\mu)$ is isomorphic to the measure ring $(\mathcal{T},\nu)$.

Since the metric space $\mathfrak{S}(\mu)$ has a diameter of $1$ — no two sets can differ by more than $X$, and $\mu(X)=1$ — we can find a dense sequence $\{E_k\}$ in the space. For each $n$ we can consider sets of the form

$\displaystyle\bigcap\limits_{i=1}^n A_i$

where either $A_i=E_i$ or $A_i=X\setminus E_i$. The collection of all such sets for a given $n$ is a partition $\mathcal{P}_n$. It should be clear that the sequence $\{\mathcal{P}_n\}$ is decreasing, and the fact that $\{E_n\}$ is dense implies that the sequence of partitions is also dense. We thus conclude that $\lim\limits_{n\to\infty}\lvert\mathcal{P}_n\rvert=0$.

The first partition $\mathcal{P}_1$ has two elements $E_1$ and $X\setminus E_1$. We can define $t(E_1)=[0,\mu(E_1)]$ and $t(X\setminus E_1)=(\mu(E_1),1]$, so that $\nu(t(E_1))=\mu(E_1)$ and $\nu(t(X\setminus E_1))=\mu(X\setminus E_1)$. This gives us a partition $\mathcal{Q}_1$ of $Y$ that reflects the structure of $\mathcal{P}_1$. Similarly, we can carve up each of these intervals so that the resulting partition $\mathcal{Q}_2$ reflects the structure of $\mathcal{P}_2$. And we can continue, each time subdividing $\mathcal{Q}_n$ into smaller intervals so that the next partition $\mathcal{Q}_{n+1}$ reflects the structure of $\mathcal{P}_{n+1}$. Since this correspondence preserves measures, it follows that $\lim\limits_{n\to\infty}\lvert\mathcal{Q}_n\rvert=0$, and the above result then shows that the sequence $\{\mathcal{Q}_n\}$ is dense.

Now, we can extend $t$ from partition elements occurring in $\{\mathcal{P}_n\}$ to finite unions of such elements by sending such a finite union to the finite union of corresponding elements of $\{\mathcal{Q}_n\}$. This gives an isometry from a dense subset of $\mathfrak{S}(\mu)$ to a dense subset of $\mathfrak{T}(\nu)$. Thus we can uniquely extend it to an isometry $\bar{t}:\mathfrak{S}(\mu)\to\mathfrak{T}(\nu)$. Since $t$ preserves unions and differences, and since these operations are uniformly continuous, it follows that $\bar{t}$ gives us an isomorphism of measure algebras.

August 25, 2010

## Partitions in Measure Algebras

Let $(\mathcal{S},\mu)$ be a totally finite measure algebra, and write $X$ for the maximal element. Without loss of generality, we can assume that $\mu$ is normalized so that $\mu(X)=1$.

We define a “partition” $\mathcal{P}$ of an element $E\subseteq\mathcal{S}$ to be a finite set of “disjoint” elements of $\mathcal{S}$ whose “union” is $E$. Remember, of course, that the elements of $\mathcal{S}$ are not (necessarily) sets, so the set language is suggestive, but not necessarily literal. That is, if $\mathcal{P}=\{E_1,\dots,E_k\}$ then $E_i\cap E_j=\emptyset$ and

$\displaystyle E=\bigcup\limits_{i=1}^kE_i$

The “norm” $\lvert\mathcal{P}\rvert$ of a partition $\mathcal{P}$ is the maximum of the numbers $\{\mu(E_i)\}$. If $\mathcal{P}=\{E_1,\dots,E_k\}$ is a partition of $E$ and if $F\subseteq E$ is any element of $\mathcal{S}$ below $E$, then $\mathcal{P}\cap F=\{E_1\cap F,\dots,E_k\cap F\}$ is a partition of $F$.

If $\mathcal{P}_1$ and $\mathcal{P}_2$ are partitions, then we write $\mathcal{P}_1\leq\mathcal{P}_2$ if each element in $\mathcal{P}_1$ is contained in an element of $\mathcal{P}_2$. We say that a sequence of partitions is “decreasing” if $\mathcal{P}_{n+1}\leq\mathcal{P}_n$ for each $n$. A sequence of partitions is “dense” if for every $E\in\mathcal{S}$ and every positive number $\epsilon$ there is some $n$ and an element $E_0\in\mathcal{S}$ so that $\rho(E,E_0)<\epsilon$, and $E_0$ is exactly the union of some elements in $\mathcal{P}_n$. That is, we can use the elements in a fine enough partition in the sequence to approximate any element of $\mathcal{S}$ as closely as we want.

Now, if $(\mathcal{S},\mu)$ is a totally finite, non-atomic measure algebra, and if $\{\mathcal{P}_n\}$ is a dense, decreasing sequence of partitions of $X$, then $\lim\limits_{n\to\infty}\lvert\mathcal{P}_n\rvert=0$. Indeed, the sequence of norms $\{\lvert\mathcal{P}_n\rvert\}$ is monotonic and bounded in the interval $[0,1]$, and so it must have a limit. We will assume that this limit is some positive number $\delta>0$, and find a contradiction.

So if $\mathcal{P}_1=\{E_1,\dots,E_k\}$ then at least one of the $E_i$ must be big enough that $\lvert\mathcal{P}_n\cap E_i\rvert\geq\delta$ for all $n$. Otherwise the sequence of norms would descend below $\delta$ and that couldn’t be the limit. Let $F_1$ be just such an element, and consider the sequence $\{\mathcal{P}\cap F_1\}$ of partitions of $F_1$. The same argument is just as true, and we find another element $F_2\subseteq F_1$ from the partition $\mathcal{P}_2$, and so on.

Now, let $F$ be the intersection of the sequence $\{F_n\}$. By assumption, each of the $F_n$ has $\mu(F_n)\geq\delta$, and so $\mu(F)\geq\delta$ as well. Since $(\mathcal{S},\mu)$ is non-atomic, $F$ can’t be an atom, and so there must be an $F_0\subseteq F$ with $0<\mu(F_0)<\mu(F)$. This element must be either contained in or disjoint from each element of each partition $\mathcal{P}_n$.

We can take $\epsilon$ smaller than either $\mu(F_0)$ or $\mu(F)-\mu(F_0)$. Now no set made up of the union of any elements of any partition $\mathcal{P}_n$ can have a distance less than $\epsilon$ from $F_0$. This shows that the sequence of partitions cannot be dense, which is the contradiction we were looking for. Thus the limit of the sequence of norms is zero.

August 24, 2010 Posted by | Analysis, Measure Theory | 1 Comment

## Stone Spaces

The Stone space functor we’ve been working with sends Boolean algebras to topological spaces. Specifically, it sends them to compact Hausdorff spaces. There’s another functor floating around, of course, though it might not be the one you expect.

The clue is in our extended result. Given a topological space $X$ we define $S(X)$ to be the Boolean algebra of all clopen subsets. This functor is contravariant — given a continuous map $f:X\to Y$, we get a homomorphism of Boolean algebras $S(f)$ sending the clopen set $Z\subseteq Y$ to its preimage $f^{-1}(Z)\subseteq X$. It’s straightforward to see that this preimage is clopen. Another surprise is that this is known as the “Stone functor”, not to be confused with the Stone space functor $S(\mathcal{B})$.

So what happens when we put these two functors together? If we start with a Boolean algebra $\mathcal{B}$ and build its Stone space $S(\mathcal{B})$, then the Stone functor applied to this space gives us a Boolean algebra $S(S(\mathcal{B}))$. This is, by construction, isomorphic to $\mathcal{B}$ itself. Thus the category $\mathbf{Bool}$ is contravariantly equivalent to some subcategory $\mathbf{Stone}$ of $\mathbf{CHaus}$. But which compact Hausdorff spaces arise as the Stone spaces of Boolean algebras?

Look at the other composite; starting with a topological space $X$, we find the Boolean algebra $S(X)$ of its clopen subsets, and then the Stone space $S(S(X))$ of this Boolean algebra. We also get a function $X\to S(S(X))$. For each point $x\in X$ we define the Boolean algebra homomorphism $\lambda_x:S(X)\to\mathcal{B}_0$ that sends a clopen set $C\subseteq X$ to $1$ if and only if $x\in C$. We can see that this is a continuous map by checking that the preimage of any basic set is open. Indeed, a basic set of $S(S(X))$ is $s(C)$ for some clopen set $C\subseteq X$. That is, $\{\lambda\in S(S(X))\vert\lambda(C)=1\}$. Which functions of the form $\lambda_x$ are in $s(C)$? Exactly those for which $x\in C$. Since $C$ is clopen, this preimage is open.

Two points $x_1$ and $x_2$ are sent to the same function $\lambda_{x_1}=\lambda_{x_2}$ if and only if every clopen set containing $x_1$ also contains $x_2$, and vice versa. That is, $x_1$ and $x_2$ must be in the same connected component. Indeed, if they were in different connected components, then there would be some clopen containing one but not the other. Conversely, if there is a clopen that contains one but not the other they can’t be in the same connected component. Thus this map $X\to S(S(X))$ collapses all the connected components of $X$ into points of $S(S(X))$.

If this map $X\to S(S(X))$ is a homeomorphism, then no two points of $X$ are in the same connected component. Thus each singleton $\{x\}\subseteq X$ is a connected component, and we call the space “totally disconnected”. Clearly, such a space is in the image of the Stone space functor. On the other hand, if $X=S(\mathcal{B})$, then $S(S(X))=S(S(S(\mathcal{B})))\cong S(\mathcal{B})=X$, and so this is both a necessary and a sufficient condition. Thus the “Stone spaces” form the full subcategory of $\mathbf{CHaus}$, consisting of the totally disconnected compact Hausdorff spaces. Stone’s representation theorem shows us that this category is equivalent to the dual of the category of Boolean algebras.

As a side note: I’d intended to cover the Stone-Čech compactification, but none of the references I have at hand actually cover the details. There’s a certain level below which everyone seems to simply assert certain facts and take them as given, and I can’t seem to reconstruct them myself.

August 23, 2010

## Stone’s Representation Theorem III

We conclude our coverage of Stone’s representation theorem with a version for Boolean $\sigma$-algebra. Each such algebra $\mathcal{B}$ is isomorphic to a $\sigma$-algebra of subsets of some space, modulo a $\sigma$-ideal.

We start as we did for any Boolean algebra, by using the map $s$ sending $\mathcal{B}$ to the Boolean algebra of clopen subsets of the Stone space $S(\mathcal{B})$. This algebra is, of course, our identified base. Let $\mathcal{S}$ be the $\sigma$-ring generated by this base.

When we first dealt with measure rings, we quotiented out by the ideal of negligible sets. We don’t have a measure on $S(\mathcal{B})$, so we don’t have measurable sets, but we do have something almost as good. Just as when we discussed Baire spaces, we can use nowhere-denseness as a topological stand-in for negligibility. In fact, we’ll define a “meager” set to be any countable union of nowhere-dense sets, and we’ll let $\mathcal{N}$ be the collection of all meager sets in $\mathcal{S}$. It is straightforward to verify that $\mathcal{N}$ is a $\sigma$-ideal in $\mathcal{S}$.

A quick side note: classically, meager sets were said to be “of the first category”. Any other sets were said to be “of the second category”. This is the root of the term “category” in the Baire category theorem.

Now if $\{E_n\}$ is a sequence of clopen sets, then we can find a unique preimage $s^{-1}(E_n)\in\mathcal{B}$. Since $\mathcal{B}$ is a $\sigma$-algebra, we can take the countable union of these elements, and then apply $s$ to the result. That is, we can define the set

$\displaystyle E=s\left(\bigcup\limits_{n=1}^\infty s^{-1}(E_n)\right)$

Now, of course, we can also take the countable union of the sets themselves. If $s$ were an isomorphism of $\sigma$-algebras, then this union would be exactly $E$ itself; but we aren’t usually so lucky. Instead, I say that the difference

$\displaystyle E\setminus\bigcup\limits_{n=1}^\infty E_n$

is nowhere-dense, and thus countable unions of clopen sets are clopen modulo meager sets.

Indeed, the countable union of the (open) sets $E_n$ is open, and so its complement is closed. The difference above is the intersection of the (closed) set $E$ with the (closed) complement of the union, and is thus closed. So if it were dense on some nonempty open set $U$ it would have to actually contain $U$. But since $U$ is open it must be the union of some collection of basic clopen sets, and we can take $U$ to be one of these sets. That is, $U\subseteq E$ and $U\cap E_n=\emptyset$ for each $n$. Since these relations only involve clopen sets, we find that $s^{-1}(U)\subseteq s^{-1}(E)$ and $s^{-1}(U)\cap s^{-1}(E_n)=\emptyset$. But there’s no way this can happen if $s^{-1}(E)$ is the union of the $s^{-1}(E_n)$!

So the map $s$ takes elements of $\mathcal{B}$ to clopen sets in $\mathcal{S}$, and then on to their equivalence classes in $\mathcal{S}/\mathcal{N}$, and this map commutes with countable unions. All that remains to show that this is an isomorphism $\mathcal{B}\cong\mathcal{S}/\mathcal{N}$ is to show that no two clopen sets in $\mathcal{S}$ represent the same equivalence class. That is, if $U_1$ and $U_2$ are distinct clopen sets, then $U_1\Delta U_2$ cannot be meager. Equivalently, no nonempty clopen set is meager.

But this is just a consequence of the Baire category theorem, here in its formulation for compact Hausdorff spaces. Indeed, if a clopen set in any Baire space — and a compact Hausdorff space is Baire — were the countable union of closed nowhere-dense sets, then its interior would be empty. But since it’s open its interior is itself, and thus is would have to be empty. Thus if $U_1$ and $U_2$ are clopen sets representing the same equivalence class modulo $\mathcal{N}$ then their (clopen) symmetric difference $U_1\Delta U_2$ is meager, and thus empty. That is, $U_1=U_2$.

August 20, 2010

## Stone’s Representation Theorem II

We can extend yesterday’s result in the case that $\mathcal{B}$ is a Boolean algebra. Now as a ring, $\mathcal{B}$ has a unit. We adjust our definition of the Stone space to define

$\displaystyle S(\mathcal{B})=\hom_\mathbf{Ring}(\mathcal{B},\mathcal{B}_0)$

That is, we insist that the ring homomorphisms preserve the identity, sending the top element of $\mathcal{B}$ to $1\in\mathcal{B}_0$. This doesn’t really change anything we said yesterday, and it all goes through as before.

What is new is that the image of $s$ — the identified base for the topology on $S(\mathcal{B})$ — consists of all the subsets of $S(\mathcal{B})$ which are clopen — both open and closed. That is, elements of $\mathcal{B}$ correspond to unions of connected components of $S(\mathcal{B})$.

First, we must show that $s(b)$ is closed, since we already know that it’s open by definition. I say that the complement $s(b)^c$ is actually the open basic set $s(b^c)$. Indeed, $b^c=b\Delta1$, and we calculate

\displaystyle\begin{aligned}s\left(b^c\right)&=s(b\Delta1)\\&=s(b)\Delta s(1)\\&=s(b)\Delta1\\&=s(b)^c\end{aligned}

Thus each set $s(b)$ is the complement of an open set, and is thus closed as well.

It also happens that our base is closed under finite unions. Indeed, we use DeMorgan’s laws and calculate

\displaystyle\begin{aligned}s(b)\cup s(b')&=\left(s(b)^c\cap s(b')^c\right)^c\\&=\left(s\left(b^c\right)\cap s\left(b'^c\right)\right)^c\\&=s\left(b^c\cap b'^c\right)^c\\&=s\left(\left(b^c\cap b'^c\right)^c\right)\\&=s(b\cup b')\end{aligned}

And from there we can extend to any finite unions we want.

Now I say that if a base of clopen sets in a compact space is closed under finite unions, then it contains every clopen set in the space. Indeed, such a clopen set can be written as a union of sets in the base since it’s open. This union gives an open covering of the set. Since the set is closed, it is compact. And so the open covering we just found has a finite subcover. That is, we can write our clopen set as a finite union of basic sets, and so it is itself in the base by assumption.

Thus in our particular case, our base of $S(\mathcal{B})$ consists of all the clopen sets in the Stone space, as we asserted!

August 19, 2010 Posted by | Analysis, Measure Theory | 2 Comments