160 lines
4.4 KiB
Markdown
160 lines
4.4 KiB
Markdown
# Lecture 6
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## Review
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Let $A$ and $B$ be subset of $\mathbb{R}$, and consider the function $f:A \to B$ defined by $f(x)=cos(x)$. Find choices for the domain $A$ and co-domain $B$ which make $f$...
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(a) neither injective nor surjective.
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$A=\mathbb{R},B=\mathbb{R}$
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(b) surjective but not injective.
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$A=[0,\frac{3\pi}{2}],B=[-1,1]$
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(c) injective but not surjective.
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$A=[0,\pi],B=[-1,1.1]$
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(d) bijective.
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$A=(0,\pi),B=(-1,1)$
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> injective: y don't repeat, surjective: there exists x for each y
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## New Materials (Chapter 2: Basic topology (4171). Finite, countable, and uncountable sets)
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### Functions
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#### Definition 2.1/2.2
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"$f:A\to B$" means "$f$ is a function from $A$ to $B$"
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$A$:"domain", and $B$: "co-domain".
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If $S\subset A$, the **range** of $S$ under $f$ is $f(S)=\{f(x):s\in S\}$
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The "**image**" of $f$ is $f(A)$.
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If $T\subset B$, **the inverse image** (pre-image) of $T$ under $f$ is
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$$
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f^{-1}(T)=\{x\in A: f(x)\in T\}
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$$
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* $f$ is injective or one-to-one if $\forall x_1,x_2\in A$ such that $f(x_1)=f(x_2)$, then $x_1=x_2$. ($f(x_1)=f(x_2)\implies x_1=x_2 \equiv x_1\neq x_2\implies f(x_1)\neq f(x_2)$)
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* $f$ is surjective or onto if $\forall y\in B, \exists x\in A$ such that $f(x)=y$. ($f(A)=B$)
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* $f$ is bijective if it's both injective and surjective.
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#### Definition 2.3
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If $\exists$ bijection $f:A\to B$, we say:
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* $A$ and $B$ can be put into 1-1 correspondence
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* $A$ and $B$ b oth have the same cardinality
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* $A$ and $B$ are equivalent $A\sim B$
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### Cardinality
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#### Definition 2.4
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(a) $A$ is finite if $A\neq \phi$ or $\exists n\in \mathbb{N}$ such that $A\sim \{1,2,...,n\}$
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(b) $A$ is **infinite** if it's not finite
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(c) $A$ is **countable** if $A\sim \mathbb{N}$
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(d) $A$ is **uncountable** if $A$ is neither finite nor countable
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(e) $A$ is **at most countable** if it's finite or countable
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> Note in some other books call (c) **countable infinite**, and (e) for **countable**
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#### Definition 2.7
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A **sequence** in $A$ is a function $f:\mathbb{N}\to\mathbb{A}$
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Note: By conversion, instead of $f(1),...f(n)$, we usually write $x_1,x_2,...,x_3$ and we say $\{x_n\}_{n=1}^{\infty}$ is a sequence.
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#### Theorem 2.8
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Every infinite subset of countable set $A$ is countable.
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Ideas of proof: if $A$ is countable, so we can list its element in a sequence. and we iterate $E\subset A$ to create a new order function by deleting element $\notin E$
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#### Definition 2.9 (arbitrary unions and intersections)
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Let $A$ be a set (called the "index set"). For each $\alpha\in A$, let $E_{\alpha}$ be a set.
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Union: $\bigcup_{\alpha \in A}E_{\alpha}=\{x:\exists \alpha \in A$ such that $x\in E_{\alpha}\}$
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Intersection: $\bigcap_{\alpha \in A}E_{\alpha}=\{x:\exists \alpha \in A$ such that $x\in E_{\alpha}\}$
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Special notation for special cases:
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$\bigcup^{n}_{m=1}E_m$ and $E_1\cup E_2\cup ...\cup E_n$ are by definition $\bigcup_{\alpha \in \{1,..,n\}}E_{\alpha}$
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and $\bigcup^{\infty}_{m=1}E_m$ and $E_1\cup E_2\cup ...\cup E_n$ are by definition $\bigcup_{\alpha \in \mathbb{N}}E_{\alpha}$
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> Note: Despite the $\infty$ symbol this def makes no references to limits, different from infinite sums.
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### Countability
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#### Theorem 2.12
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"A countable union of countable sets is countable".
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Let $\{E_n\},n=1,2,3,...$ be a sequence of countable sets, and put
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$$
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S=\bigcup^{\infty}_{n=1} E_n
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$$
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The $S$ is countable.
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Proof by infinite grid, and form a new sequence and remove duplicates.
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#### Corollary
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An at most countable union of at most countable sets is at most countable.
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#### Theorem 2.13
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$A$ is countable, $n\in \mathbb{N}$,
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$\implies A^n=\{(a_{1},...,a_{n}):a_1\in A, a_n\in A\}$, is countable.
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<details>
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<summary>Proof</summary>
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Induct on $n$,
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Base case $n=1$,
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True by assumptions
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Induction step: suppose $A^{n-1}$ is countable. Note $A^n=\{(b,a):b\in A^{n-1},a\in A\}=\bigcup_{b\in A^{n-1}\{(b,a),a\in A\}}$.
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Since $b$ is fixed, so this is in 1-1 correspondence with $A$, so it's countable by Theorem 2.12.
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</details>
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#### Theorem 2.14
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Let $A$ be the set of all sequences for 0s and 1s. Then $A$ is uncountable.
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<details>
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<summary>Proof</summary>
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Let $E\subset A$ be a countable subset. We'll show $A\backslash E\neq \phi$ (i.e.$\exists t\in A$ such that $t\notin E$)
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$E$ is countable so we can list it's elements $S_1,S_2,S_3,...$.
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Then we define a new sequence $t$ which differs from $S_1$'s first bit and $S_2$'s second bit,...
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This is called Cantor's diagonal argument.
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</details>
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