\( \newcommand{\N}{\mathbb{N}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\P}{\mathcal P} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathbb{F}} \newcommand{\E}{\mathcal E} \newcommand{\brac}[1]{\left(#1\right)} \newcommand{\abs}[1]{\left|#1\right|} \newcommand{\matrixx}[1]{\begin{bmatrix}#1\end {bmatrix}} \newcommand{\vmatrixx}[1]{\begin{vmatrix} #1\end{vmatrix}} \newcommand{\lims}{\mathop{\overline{\lim}}} \newcommand{\limi}{\mathop{\underline{\lim}}} \newcommand{\limn}{\lim_{n\to\infty}} \newcommand{\limsn}{\lims_{n\to\infty}} \newcommand{\limin}{\limi_{n\to\infty}} \newcommand{\nul}{\mathop{\mathrm{Nul}}} \newcommand{\col}{\mathop{\mathrm{Col}}} \newcommand{\rank}{\mathop{\mathrm{Rank}}} \newcommand{\dis}{\displaystyle} \newcommand{\spann}{\mathop{\mathrm{span}}} \newcommand{\range}{\mathop{\mathrm{range}}} \newcommand{\inner}[1]{\langle #1 \rangle} \newcommand{\innerr}[1]{\left\langle #1 \right \rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\toto}{\rightrightarrows} \newcommand{\upto}{\nearrow} \newcommand{\downto}{\searrow} \newcommand{\qed}{\quad \blacksquare} \newcommand{\tr}{\mathop{\mathrm{tr}}} \newcommand{\bm}{\boldsymbol} \newcommand{\cupp}{\bigcup} \newcommand{\capp}{\bigcap} \newcommand{\sqcupp}{\bigsqcup} \newcommand{\re}{\mathop{\mathrm{Re}}} \newcommand{\im}{\mathop{\mathrm{Im}}} \newcommand{\comma}{\text{,}} \newcommand{\foot}{\text{。}} \)

Thursday, April 24, 2014

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。

Wednesday, April 16, 2014

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。

Wednesday, April 2, 2014

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 $x\in \partial C$.$\qed$

Tuesday, April 1, 2014

反思

臨就黎畢業,PhD application 唔敢抱樂觀態度,再諗返,究竟係我喺科大學到學得到 d 乜?我可以好肯定咁同自己講,幾乎冇野學到!我真正感覺到自己知識有正增長嘅時期係 year 1,喺中大讀書嘅時期。

作為一個做分析嘅人,喺科大讀書只可以話係一埸悲劇。仲 active 喺 research in analysis 嘅 professor 得返一個,十個 seminar 九個同 analysis 無關。反觀人地中大個 IMS 不斷有 mini course,seminar 同 analysis 有關,我真係羨慕到不得了,但又唔能夠下下都走過去中大聽。

如果我年半前係跟一個 active 喺 research 嘅 professor,話唔定我已有一年真真正正嘅 research experience,已經有 paper 喺手,已經學到好多有對我有意義,有趣同深奧嘅知識,可惜喺科大嘅第二年學到 (及被迫去學) 嘅野對 analysis 及對我做緊嘅 complex analysis research 一 d 幫助都冇。

有志做 analysis 嘅人,唔應該入科大。入得科大,唔好諗住做 analysis。真係喺科大揀左做 analysis 嘅話,只好願上天保祐。