Merge branch 'main' of https://github.com/Trance-0/NoteNextra
This commit is contained in:
@@ -3,6 +3,8 @@
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FROM node:18-alpine AS base
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ENV NODE_OPTIONS="--max-old-space-size=8192"
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# 1. Install dependencies only when needed
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FROM base AS deps
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# Check https://github.com/nodejs/docker-node/tree/b4117f9333da4138b03a546ec926ef50a31506c3#nodealpine to understand why libc6-compat might be needed.
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3
Jenkinsfile
vendored
3
Jenkinsfile
vendored
@@ -1,7 +1,8 @@
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pipeline {
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environment {
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registry = "trance0/NoteNextra"
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registry = "trance0/notenextra"
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version = "1.0"
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NODE_OPTIONS = "--max-old-space-size=8192"
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}
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agent any
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@@ -3,7 +3,7 @@ services:
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build:
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context: ./
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dockerfile: ./Dockerfile
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image: trance0/notenextra:v1.1.5
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image: trance0/notenextra:v1.1.7
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restart: on-failure:5
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ports:
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- 13000:3000
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19918
package-lock.json
generated
19918
package-lock.json
generated
File diff suppressed because it is too large
Load Diff
@@ -1,11 +1,12 @@
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{
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"scripts": {
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"dev": "next",
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"build": "export NODE_OPTIONS=--max-old-space-size=8192 && next build",
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"build": "next build",
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"start": "next start"
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},
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"dependencies": {
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"@cloudflare/next-on-pages": "^1.13.7",
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"@napi-rs/simple-git": "^0.1.19",
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"@vercel/analytics": "^1.4.1",
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"@vercel/speed-insights": "^1.1.0",
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"next": "^15.0.3",
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@@ -16,4 +16,5 @@ export default {
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CSE332S_L11: "Object-Oriented Programming Lab (Lecture 11)",
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CSE332S_L12: "Object-Oriented Programming Lab (Lecture 12)",
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CSE332S_L13: "Object-Oriented Programming Lab (Lecture 13)",
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CSE332S_L14: "Object-Oriented Programming Lab (Lecture 14)",
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}
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77
pages/CSE559A/CSE559A_L14.md
Normal file
77
pages/CSE559A/CSE559A_L14.md
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# CSE559A Lecture 14
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## Neural Network Training
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## Object Detection
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AP (Average Precision)
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### Benchmarks
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#### PASCAL VOC Challenge
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20 Challenge classes.
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CNN increases the accuracy of object detection.
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#### COCO dataset
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Common objects in context.
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Semantic segmentation. Every pixel is classified to tags.
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Instance segmentation. Every pixel is classified and grouped into instances.
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### Object detection: outline
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Proposal generation
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Object recognition
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#### R-CNN
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Proposal generation
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Use CNN to extract features from proposals.
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with SVM to classify proposals.
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Use selective search to generate proposals.
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Use AlexNet finetuned on PASCAL VOC to extract features.
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Pros:
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- Much more accurate than previous approaches
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- Andy deep architecture can immediately be "plugged in"
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Cons:
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- Not a single end-to-end trainable system
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- Fine-tune network with softmax classifier (log loss)
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- Train post-hoc linear SVMs (hinge loss)
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- Train post-hoc bounding box regressors (least squares)
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- Training is slow 2000CNN passes for each image
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- Inference (detection) was slow
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#### Fast R-CNN
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Proposal generation
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Use CNN to extract features from proposals.
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##### ROI pooling and ROI alignment
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ROI pooling:
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- Pooling is applied to the feature map.
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- Pooling is applied to the proposal.
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ROI alignment:
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- Align the proposal to the feature map.
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- Align the proposal to the feature map.
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Use bounding box regression to refine the proposal.
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@@ -16,4 +16,5 @@ export default {
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CSE559A_L11: "Computer Vision (Lecture 11)",
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CSE559A_L12: "Computer Vision (Lecture 12)",
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CSE559A_L13: "Computer Vision (Lecture 13)",
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CSE559A_L14: "Computer Vision (Lecture 14)",
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}
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@@ -1 +1,74 @@
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# Lecture 21
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# Math4121 Lecture 21
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## Rolling from last lecture
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### Convergence of integrals
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#### Arzela-Osgood Theorem
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Let $\{f_n\}$ be a sequence of function, $f(x)=\lim_{n\to\infty}f_n(x)$ for every $x\in [0,1]$, if $f\in \mathscr{R}[0,1]$, and $\exists B>0$ such that $|f_n(x)|\leq B \forall x\in [0,1]$. (uniformly bounded and integrable)
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$$
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\lim_{n\to\infty}\int_0^1 f_n(x) dx = \int_0^1 f(x) dx
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$$
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If we let $\Gamma_{\alpha}$ be the set of intervals where $f_n$ is not continuous,
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$$
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\Gamma_{\alpha} = \{x\in [0,1] : \textup{ for any }m\in \mathbb{N}, \delta > 0, \exists n\geq m, y\in (x-\delta, x+\delta) \text{ s.t. } |f_n(y)-f(y)|>\alpha\}
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$$
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Fact: $\Gamma_{\alpha}$ is closed and nowhere dense.
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Proof:
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Without loss of generality, we can assume $f=0$. Given any $\alpha > 0$, $\exists N$ such that
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$$
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\left|\int_0^1 f_n(x) dx \right| < \alpha
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$$
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for all $n\geq N$.
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Consider the set $\Gamma_{\alpha/2} = \bigcup_{n=1}^{\infty} E_n$, for each $g\in \Gamma_{\alpha/2}$, we still have $\lim_{n\to\infty}f_n(g) = 0$.
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So we define
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$$
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G_i=\{g\in \Gamma_{\alpha/2} :|f_n(g)|<\frac{\alpha}{2} \text{ for all }n\geq i\}
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$$
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So $G_1\subset G_2\subset \cdots$ and $\Gamma_{\alpha/2} = \bigcup_{i=1}^{\infty} G_i$.
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By Osgood Lemma, since $\Gamma_{\alpha/2}$ is closed, $\exists K$ such that $c_e(G_K)>c_e(\Gamma_{\alpha/2})-\frac{\alpha}{4B}$.
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By definition of $c_e$, we cna find open $I_1,\ldots,I_N$ which cover $\Gamma_{\alpha/2}$ and
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$$
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\sum_{i=1}^N \ell(I_i) < c_e(\Gamma_{\alpha/2})+\frac{\alpha}{4B}
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$$
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Let $\mathcal{U}=\bigcup_{i=1}^N I_i$, and $\mathcal{C}=[0,1]\setminus \mathcal{U}$.
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Part 1: Control the integral on $\mathcal{C}$
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for each $x\in \mathcal{C}$, $x\notin \Gamma_{\alpha/2}$, so $\exists$ and open interval $I(x)$ and an integer $m(x)$ such that $|f_{m(x)}(x)|<\frac{\alpha}{2}$ and $\forall n\geq m(x), y\in I(x)$
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So $\mathcal{C}\subset \bigcup_{x\in \mathcal{C}} I(x)$, and $\mathcal{C}$ is closed and bounded, $\exists x_1,\ldots,x_J$ such that $\mathcal{C}\subset \bigcup_{j=1}^J I(x_j)$. So if $n\geq \max_{j=1,\ldots,J} m(x_j)$, and $x\in \mathcal{C}$, then $|f_n(x)|<\frac{\alpha}{2}$.
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So $\int_\mathcal{C} |f_n(x)| dx < \frac{\alpha}{2} c_e(\mathcal{C})$.
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Part 2: Control the integral on $\mathcal{U}$
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If $[x_i,x_{i+1}]\cap G_k\neq \emptyset$, then $\inf_{x\in [x_i,x_{i+1}]} |f_n(x)| < \frac{\alpha}{2}$ for all $n\geq K$. Denote such set as $P_1$.
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Otherwise, we denote such set as $P_2$.
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So $\ell(\mathcal{U})=\ell(P_1)+\ell(P_2)\geq c_e(G_K)+\ell(P_2)$.
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This implies $\ell(P_2)\leq \frac{\alpha}{4B}$.
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Continue on Friday.
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QED
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@@ -1 +1,123 @@
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# Lecture 22
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# Math 4121 Lecture 22
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## Continue on Arzela-Osgood Theorem
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Proof:
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Part 2: Control the integral on $\mathcal{U}$
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If $[x_i,x_{i+1}]\cap G_k\neq \emptyset$, then $\inf_{x\in [x_i,x_{i+1}]} |f_n(x)| < \frac{\alpha}{2}$ for all $n\geq K$. Denote such set as $P_1$.
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Otherwise, we denote such set as $P_2$.
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So $\ell(\mathcal{U})=\ell(P_1)+\ell(P_2)\geq c_e(G_K)+\ell(P_2)$.
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This implies $\ell(P_2)\leq \frac{\alpha}{4B}$ since $c_e(G_K)\leq c_e(\mathcal{U})+\frac{\alpha}{2B}$.
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Thus, for $n\geq K$,
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$$
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L(P,f_n)\leq \ell(P_1)\frac{\alpha}{2}+\ell(P_2)B
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$$
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So
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$$
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\int_\mathcal{U} |f_n(x)| dx \leq c_e(\mathcal{U})\frac{\alpha}{2}+\frac{\alpha}{2}
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$$
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All in all,
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$$
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\begin{aligned}
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\left\vert \int_\mathcal{U} f_n(x) dx\right\vert &\leq \frac{\alpha}{2}+\frac{\alpha}{2}\\
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&= \int_0^1 |f_n(x)| dx\\
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&\leq \int_\mathcal{U} |f_n(x)| dx + \int_\mathcal{C} |f_n(x)|dx\\
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&\leq c_e(\mathcal{U})\frac{\alpha}{2}+\frac{\alpha}{2}+c_e(\mathcal{C})\frac{\alpha}{2}\\
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&= \alpha
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\end{aligned}
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$$
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$\forall N\geq K$.
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QED
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### Baire Category Theorem
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Nowhere dense sets can be large, but they canot cover an open (or closed) interval.
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#### Theorem 4.7 (Baire Category Theorem)
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An open interval cannot be covered by a countable union of nowhere dense sets.
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Proof:
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Suppose $(0,1)\subset \bigcup_{n=1}^\infty S_n$ where each $S_n$ is nowhere dense. In particular, $\exists I_1$ closed interval such that $I_1\subset (0,1)$ and $I_1\cap S_1=\emptyset$.
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Now for each $k\geq 2$, $S_k$ is not dense in $I_{k-1}$ so $\exists I_k\subsetneq I_{k-1}$ such that $I_k\cap S_k=\emptyset$ for all $j\leq k$.
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By nested interval property, $\exists x\in \bigcap_{n=1}^\infty I_n$.
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Then $x\in (0,1)$ and $x\notin \bigcup_{n=1}^\infty S_n$.
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Contradiction with the assumption that $(0,1)\subset \bigcup_{n=1}^\infty S_n$.
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QED
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#### Definition First Category
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A countable union of nowhere dense sets is called a set of **first category**.
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#### Corollary 4.8
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Complement of a set of first category in $\mathbb{R}$ is dense in $\mathbb{R}$.
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Proof:
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We need to show that for every interval $I$, $\exists x\in I\cap S^c$. ($\exists x\in I$ and $x\notin S$)
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This is equivalent to the Baire Category Theorem.
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QED
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Recall a function is pointwise discontinuous if $\mathcal{C}=\{c\in [a,b]: f\text{ is continuous at } c\}$ is dense in $[a,b]$.
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$\mathcal{D}=[a,b]\setminus \mathcal{C}$ is called the set of points of discontinuity of $f$.
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#### Corollary 4.9
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$f$ is pointwise discontinuous if and only if $\mathcal{D}$ is of first category.
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Proof:
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Part 1: If $\mathcal{D}$ is of first category, then $f$ is pointwise discontinuous.
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Immediate from Corollary 4.8.
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Part 2: If $f$ is pointwise discontinuous, then $\mathcal{D}$ is of first category.
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Let $P_k=\{x\in [a,b]: w(f;x)\geq \frac{1}{k}\}$, $\mathcal{D}=\bigcup_{k=1}^\infty P_k$.
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Need to show that each $P_k$ is nowhere dense. (under the assumption that $\mathcal{C)$ is dense).
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Let $I\subseteq [a,b]$ so $\exists c\in \mathcal{C}\cap I$. So by definition of $w(f;c)$, $\exists J\subseteq I$ and $c\in J$ such that $w(f;J)\leq \frac{1}{k}$ so for all $x\in J$, $w(f;x)\leq \frac{1}{k}$. so $J\subseteq P_k=\emptyset$.
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Thus, $P_k$ is nowhere dense.
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QED
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#### Corollary 4.10
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Let $\{f_n\}$ be a sequence of pointwise discontinuous functions. The set of points at which all $f_n$ are simultaneously continuous is dense (it's also uncountable).
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Proof:
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$$
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\bigcap_{n=1}^\infty \mathcal{C}_n=\left(\bigcup_{n=1}^\infty \mathcal{D}_n\right)^c
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$$
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The complement of a set of first category is dense.
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QED
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@@ -22,15 +22,9 @@ export default {
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Math4121_L17: "Introduction to Lebesgue Integration (Lecture 17)",
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Math4121_L18: "Introduction to Lebesgue Integration (Lecture 18)",
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Math4121_L19: "Introduction to Lebesgue Integration (Lecture 19)",
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Math4121_L20: {
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||||
display: 'hidden'
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||||
},
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Math4121_L21: {
|
||||
display: 'hidden'
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||||
},
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||||
Math4121_L22: {
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||||
display: 'hidden'
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||||
},
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Math4121_L20: "Introduction to Lebesgue Integration (Lecture 20)",
|
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Math4121_L21: "Introduction to Lebesgue Integration (Lecture 21)",
|
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Math4121_L22: "Introduction to Lebesgue Integration (Lecture 22)",
|
||||
Math4121_L23: {
|
||||
display: 'hidden'
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||||
},
|
||||
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||||
Reference in New Issue
Block a user