An Explicit Example of a Sequence Which has Convergent Subnet but no Convergent Subsequence

I get confused by my (faked) intuition that all subnets of a sequence should also be a subsequence. This turns out to be wrong, and I try to seek for easy example.

The following is the easiest one:

Example. Let $\mathcal I$ be the set of vectors $(n_1,n_2,\dots)$, where $n_i\in \N$ for each $i$ and $n_1<n_2<\cdots$. The cardinality of $\mathcal I$ is easily seen to be $|\R|$. Now our desired sequence will be a sequence of functions defined on $\mathcal I$ as follows:

Let $i=(n_1,n_2,\dots)\in \mathcal I$, we define \[
f_n(i)=\begin{cases}
(-1)^k,&\text{if }n=n_k,\\
0,&\text{otherwise}.
\end{cases}
\] It is easy to see that given an $i=(n_1,n_2,\dots)\in \mathcal I$, $\{f_{n_k}(i)\}=\{(-1)^k\}$ diverges.

With a little abuse of notation, we define $f_n = (f_n(i))_{\mathcal I}$. Now $f_1,f_2,\dots \in [-1,1]^{\mathcal I}$. By Tychonoff's Theorem $f_1,f_2,\dots \in [-1,1]^{\mathcal I}$ is compact w.r.t. the product topology, therefore $\{f_n\}$, being a net, must have a convergent subnet by a standard exercise on nets.

We claim that $\{f_n\}$ has no convergent subsequence. Suppose it does, then there is $\{n_k\}$ such that $\{f_{n_k}\}$ converges w.r.t. product topology. By definition, it has coordinatewise convergence: For every $i\in \mathcal I$, $f_{n_k}(i)$ converges. This is a contradiction if we choose $i=(n_1,n_2,\dots)$.$\qed$

Polar Decomposition of Matrices with an Application in Deriving SVD (to be added in my linear algebra notes)

In this post I want to record a standard result in linear algebra that I haven't paid attention before---the polar decomposition. An immediate consequence that I come up with is the SVD decompsition theorem, which I prove as two consequences, one is for square matrices and one is for general matrices.

In what follows we say that a matrix $A$ is positive if $x\mapsto x\cdot Ax$ is always nonnegative, and strictly positive if $x\mapsto x\cdot Ax$ forms an inner product.

Digression. When we say that an operator $T:H\to H$ on a complex Hilbert space is positive: $\inner{Tx,x}\ge 0$ for every $x$, it is necessarily (i.e., can be proved to be) self-adjoint! The key is that we can always recover the value $\inner{Tx,y}$ by just knowing what is $\inner{Tx,x}$ for every $x\in H$, while this may not happen in real Hilbert spaces.

Theorem (Polar Decomposition). Let $A$ be an $n\times n$ matrix over $\F=\R$ or $\C$, then there is a unitary (orthogonal when $\F=\R$) matrix $U$ and a positive matrix $P$ such that \[
A=UP,
\] where $P=\sqrt{A^*A}$, the unique positive square root of the positive matrix $A^*A$.

PhD Qualifying Exam for Real Analysis, Spring 2013-14 by Dr Li

I just solved all of them except for 4. As a usual practice answer of qualifying exam will not be unveiled (even to PG students in UST), let me record my solution below for future use.

The following is the set of problems a few days ago:

Problem 1. Let $F$ be a Lebesgue nonmeasurable subset of $[0,1]$. Prove that there is $c\in (0,1)$ such that whenever $E\subseteq[0,1]$ is Lebesgue measurable and $m(E)\ge c$, then $F\cap E$ is Lebesgue nonmeasurable.

Problem 2. Let $r_1,r_2,r_3,\dots$ be a sequence containing each rational number in $[0,1]$ exactly once. Let $f:[0,1]\setminus \{r_1,r_2,r_3\dots\}\to \R$ be defined by $\dis f(x)=\sum_{k=1}^\infty \frac{1}{k^2|x-r_k|^{1/2}}$. Prove that $f(x)<\infty$ a.e. on $[0,1]\setminus \{r_1,r_2,r_3\dots\}$.

Problem 3. Let $S$ be a vector subspace of $L^2[0,1]$ with $\|f\|=(\int_{[0,1]}|f|^2\,dm )^{1/2}$ and suppose that there is a constant $K$ such that for every $f\in S$ and every $x\in [0,1]$, $|f(x)|\leq K\|f\|$. Show that the dimension of $S$ is finite.

My Remark. The converse is also true, and this can be generalized to every $L^p(K)$, where $p\in (1,\infty)$ and $K$ is compact.

某人真係好鬼污糟

話說我地 office 有一個起碼讀左 7 年 PhD 嘅怪人 (人稱包哥)，出哂名污糟邋遢，成日唔沖涼 (貌似最近有人見佢沖...，但有咩用，肯定佢唔係成日洗衫)。

有一次食飯前我地三五成群去 PG office 最近嘅某一個廁所放水，喺果到我地目擊到一單好神奇嘅事：包兄痾完尿之後，竟然會好勤力咁洗手！

事後傾返，原來我地都不約而同地心諗一直錯怪左佢。

但，因為我係最後一個放水嘅關係，我睇到嘅係另一回事！

包兄洗完手之後竟然又再去痾多鑊。因為我地痾嘅尿兜相近，我唔知點解夜神月上身咁一路痾一路忍笑十幾秒。最後佢痾完後立即揚長而去。__ ！原來都係唔洗手果 d __ 街黎。

散散心

同一班好耐冇見嘅師兄行山。其中一個喺 BU 做緊 research assistant professor，話如果有困難嘅話可以去佢果到萬五蚊咁做兩年 RA。非常感激呢番好意，但呢類 RA 工對我前途應該冇太大幫助 ...。暫時盡力去搵中大嘅工，留喺果到學習喺科大比較難學到嘅野。

漫長艱辛嘅等 offer 期結束，要搵工 lu

最近都係冇 offer，但又難怪，自己寫 personal statement 求求其其 (幾乎間間學校用同一份 ...)，報學校又報得少，toefl 同 gre math 又考得唔好，冇得怨，因為須要再改進嘅地方實在太多，只好下年再黎同認真少少。

期望可以搵到一份可以留喺大學到繼續讀書嘅工。經 email 同一個較熟嘅中大 professor 傾過後，有機會去 CU。期望去到喇...，因為好想喺果到跟真係做緊 analysis 嘅 professor 學野，聽一 d 真正自己想聽嘅 seminar。

當初寧願等多年都唔應該入科大，喺科大幾乎冇野學到 (當然我仍然好感激 kin li 幫我 R 個 mphil 位返黎)。

開始要備戰下學期考 toefl 了。Peter 送左我一大堆 download 返黎嘅電映，我要開始俾心機煲哂佢同學多 d 生字。

Connected Components

In the past I learnt connected components in a quite high generality. Special case is of particular importance, the following are the few used implicitly in complex analysis on $\C$.

Fact 1. If $U$ is open in a locally connected topological space $X$, then each of its connected components is open. In particular, if $K$ is closed, $\C\setminus K$ is a countable union of open connected components.

Proof. Suppose $C\subseteq X$ is a connected component, let $x\in C$, then there is a connected neighborhood $V$, $x\in V$. Since $V\cap C\neq \emptyset$, and since $V,C\subseteq X$, it must happen that $V\subseteq C$, as desired.$\qed$

Fact 2. If $D$ is open connected, then $D$ is a component of $\C\setminus \partial D$.

Proof. Let $D\subseteq U\subseteq \C\setminus \partial D$ with $U$ connected, we show that $D=U$. Since $U=D\cup (U\setminus D)$. As $U\cap \partial D=\emptyset$, thus we have \[U=D\cup (U\setminus \ol{D}).\] Since $U$ is connected, and both $D$ and $U\setminus \ol D$ are open, one of them must be empty, and since $D\neq \emptyset$, we have $U=D$, as desired.$\qed$

Fact 3. If $X$ is locally path connected, then connected subsets are path-connected.

Proof. Let $U\subseteq X$ be connected, fix $x\in U$, then the set $U_x$ of points that can be connected with $x$ by a path in $U$ is both open and closed. Since $x$ and $x$ can be connected by a path in $U$, $U_x\neq \emptyset$, thus $U_x=U$, as desired.$\qed$

Fact 4. If $C$ is a connected component of a locally connected topological subspace $X$ of some $Y$, then $\partial C\subseteq \partial X$. A typical example of $X$ is open subset $\C\setminus K$ of $\C$. This is a particularly important example in potential theory as it pops up quite frequently and naturally when using maximum principle for subharmonic functions (since $\partial C\subseteq \partial (\C\setminus K) =\partial K$).

Proof 1. To show $\ol C\setminus C^\circ \subseteq \ol X\setminus X^\circ$, we pick $x\in \ol C$. We now show that if $x\in X^\circ$, then $x\in C^\circ$. Suppose $x\in X^\circ$, then there is a connected neighborhood $V\subseteq X$ of $x$. Now $V\cap C\neq \emptyset$, thus $V\subseteq C$, so $x\in C^\circ$.

Now we have shown that $\ol C\setminus C^\circ \subseteq \ol C\setminus X^\circ$, thus $\partial C\subseteq \partial X$.$\qed$

Proof 2 (When $C$ is just a component of $\mathbb P\setminus K$ with $K$ compact). This is a very common scenario. Take $x\in \partial C$. For the sake of contradiction, suppose that $x\not\in \partial(\mathbb P \setminus K)=\partial K$, then there is a path connected open $V$, with $x\in V$, such that $V\cap \partial(\mathbb P\setminus K)=\emptyset$. Now $V\cap C\neq \emptyset$ (since $x\in \ol C$) and as $C$ must be open, $V\cup C$ is a connected open set.

We expect $V\cap K=\emptyset$, if not, then there is $v\in V\cap K$, joining this point with a $c\in C\subseteq \mathbb P\setminus K$ using a path in $V\cup C$, we get a point in \[
(V\cup C)\cap (\partial K) = C\cap \partial K\subseteq C\cap K=\emptyset,
\] a contradiction. We conclude that $V\subseteq \mathbb P\setminus K$.

Therefore $V\cup C$ is an open connected subset of $\mathbb P\setminus K$, and as $C$ is a component of $\mathbb P\setminus K$, we have $C\cup V=C$, so $V\subseteq C$, a contradiction since $V$ is a neighborhood of $x\in \partial C$.$\qed$