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

Saturday, March 26, 2011

烏籠...

今日上 math 323 tutorial, 而且今日功課 dead line, tutorial 完左照舊又係同人吹吹 topology d 野, 跟住發覺我將所有功課裏面有關喺 topological space $  (X,\mathcal{T})$ 上,  collection of limit point of $  A$ 嘅 definition 寫左做 \[
x\in A' \iff \forall U\in N_x\triangleq \{V\in \mathcal{T}:x\in V\},( U-\{x\})\cap A\neq \emptyset.
\] 咁我呢 D 乖學生黎, 有錯梗係即改, 但改好哂之後就喺到諗, $  ( U-\{x\})\cap A$ 同 $  U\cap (A-\{x\})$ 好似冇分別喎......=.=.
張貼者： 於
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