Thu-24-04-2025 18:03
profe: Natalia Accomazzo Scotti
status:
tags: Espacios métricos
Def. : {\color{Cyan} \text{Def. :} } Def. :
Una sucesi o ˊ n ( x n ) n ∈ N ⊆ M es de Cauchy si ∀ ε > 0 ∃ n 0 ∈ N ∣ s i n , m ≥ n 0 ⟹ d ( x n , x m ) < ε \begin{array}{l}
\text{Una sucesión $( x_{n} )_{n \in \mathbb{N}}\subseteq M$ es de Cauchy si $\forall \varepsilon>0 \:\exists\:n_{0} \in \mathbb{N}\bigm|$}\\
si\:n,m\geq n_{0}\implies d(x_{n},x_{m})<\varepsilon
\end{array} Una sucesi o ˊ n ( x n ) n ∈ N ⊆ M es de Cauchy si ∀ ε > 0 ∃ n 0 ∈ N s i n , m ≥ n 0 ⟹ d ( x n , x m ) < ε E j e m p l o s : Ejemplos: E j e m pl os :
x n − 1 n x_{n}- \frac{1}{n} x n − n 1 ( R , ∣ ⋅ ∣ ) (\mathbb{R},|\cdot|) ( R , ∣ ⋅ ∣ ) ( M , δ ) . ( x n ) n ∈ N ⊆ M (M,\delta).( x_{n} )_{n \in \mathbb{N}}\subseteq M ( M , δ ) . ( x n ) n ∈ N ⊆ M
δ ( x n , x m ) < ε → S i ε < 1 → x n = x m ∀ n , m ≥ n 0 \delta(x_{n},x_{m})<\varepsilon\to Si\:\varepsilon<1\to x_{n}=x_{m}\quad \forall n,m\geq n_{0} δ ( x n , x m ) < ε → S i ε < 1 → x n = x m ∀ n , m ≥ n 0 Luego, las sucesiones de Cauchy son las eventualmente constantes.
En ( C [ 0 , 1 ] , d 1 ) (C[0,1],d_{1}) ( C [ 0 , 1 ] , d 1 )
d 1 ( f , g ) = ∫ 0 1 ∣ f ( x ) − g ( x ) ∣ d x d_{1}(f,g)=\int _{0}^1 |f(x)-g(x)| \, dx d 1 ( f , g ) = ∫ 0 1 ∣ f ( x ) − g ( x ) ∣ d x f n ( x ) = { 0 x ∈ [ 0 , 1 2 − 1 n ] l i n e a l x ∈ [ 1 2 − 1 n , 1 2 ] 1 x ∈ [ 1 2 , 1 ] f_{n}(x)=\begin{cases}
0 & x \in\left[ 0, \frac{1}{2}- \frac{1}{n} \right] \\
lineal & x \in\left[ \frac{1}{2}- \frac{1}{n}, \frac{1}{2} \right] \\
1 & x \in \left[ \frac{1}{2},1 \right]
\end{cases} f n ( x ) = ⎩ ⎨ ⎧ 0 l in e a l 1 x ∈ [ 0 , 2 1 − n 1 ] x ∈ [ 2 1 − n 1 , 2 1 ] x ∈ [ 2 1 , 1 ] con n ≥ 3 n\geq 3 n ≥ 3
( f n ) n ∈ N ( f_{n} )_{n \in \mathbb{N}} ( f n ) n ∈ N
d 1 ( f n , f m ) = ? d_{1}(f_{n},f_{m})=? d 1 ( f n , f m ) = ?
Las áreas se van achicando.
Prop. : {\color{Orange} \text{Prop. :} } Prop. :
Sea ( x n ) n ∈ N ⊆ M y x n → x ⟹ ( x n ) es de Cauchy. \begin{array}{l}
\text{Sea $( x_{n} )_{n \in \mathbb{N}}\subseteq M$ y $x_{n}\to x\implies(x_{n})$ es de Cauchy.}
\end{array} Sea ( x n ) n ∈ N ⊆ M y x n → x ⟹ ( x n ) es de Cauchy. Dem: {\color{Orange} \text{Dem:} } Dem: ε > 0 \varepsilon>0 ε > 0
d ( x n , x m ) ≤ d ( x n , x ) ⏟ < ε 2 + d ( x , x m ) ⏟ < ε 2 < ε d(x_{n},x_{m})\leq \underbrace{ d(x_{n},x) }_{ < \frac{\varepsilon}{2} }+\underbrace{ d(x,x_{m}) }_{ < \frac{\varepsilon}{2} }<\varepsilon d ( x n , x m ) ≤ < 2 ε d ( x n , x ) + < 2 ε d ( x , x m ) < ε Sea n 0 ∈ N ∣ n ≥ n 0 ⟹ d ( x n , x ) < ε 2 n_{0}\in \mathbb{N}\bigm|n\geq n_{0}\implies d(x_{n},x)< \frac{\varepsilon}{2} n 0 ∈ N n ≥ n 0 ⟹ d ( x n , x ) < 2 ε n , m ≥ n 0 : n,m\geq n_{0}: n , m ≥ n 0 :
d ( x n , x m ) < ε d(x_{n},x_{m})<\varepsilon d ( x n , x m ) < ε □ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \square □ Prop. : {\color{Orange} \text{Prop. :} } Prop. :
Si ( x n ) n ∈ N ⊆ M es de Cauchy entonces ( x n ) n ∈ N es acotada. \begin{array}{l}
\text{Si $( x_{n} )_{n \in \mathbb{N}}\subseteq M$ es de Cauchy entonces $( x_{n} )_{n \in \mathbb{N}}$ es acotada.}
\end{array} Si ( x n ) n ∈ N ⊆ M es de Cauchy entonces ( x n ) n ∈ N es acotada. Dem: {\color{Orange} \text{Dem:} } Dem: ε = 1 \varepsilon=1 ε = 1 n 0 ∈ N ∣ n_{0}\in \mathbb{N}\bigm| n 0 ∈ N n , m ≥ n 0 , d ( x n , x m ) < 1 n,m\geq n_{0},d(x_{n},x_{m})<1 n , m ≥ n 0 , d ( x n , x m ) < 1 x n ∈ B ( x n 0 , 1 ) ∀ n ≥ n 0 x_{n}\in B(x_{n_{0}},1)\quad\forall n\geq n_{0} x n ∈ B ( x n 0 , 1 ) ∀ n ≥ n 0
R = m a x { 1 , d ( x 1 , x n 0 ) , d ( x 2 , x n 0 ) , … , d ( x n 0 − 1 , x n 0 ) } R=max\{ 1,d(x_{1},x_{n_{0}}),d(x_{2},x_{n_{0}}),\dots,d(x_{n_{0}-1},x_{n_{0}}) \} R = ma x { 1 , d ( x 1 , x n 0 ) , d ( x 2 , x n 0 ) , … , d ( x n 0 − 1 , x n 0 )} Luego, si r > R ⟹ x n ∈ B ( x n 0 , r ) ∀ n ⟹ ( x n ) n ∈ N es acotada r>R\implies x_{n}\in B(x_{n_{0}},r)\quad\forall n\implies ( x_{n} )_{n \in \mathbb{N}}\text{ es acotada} r > R ⟹ x n ∈ B ( x n 0 , r ) ∀ n ⟹ ( x n ) n ∈ N es acotada
Prop.: {\color{Orange}\text{Prop.:}} Prop.:
Sea ( x n ) n ∈ N una sucesi o ˊ n de Cauchy en el espacio m e ˊ trico ( M , d ) , y supongamos que existe una sub‑sucesi o ˊ n ( x n j ) j ∈ N tal que x n j → x . Entonces x n → x . \begin{array}{l}
\text{Sea $(x_n)_{n\in\mathbb{N}}$ una sucesión de Cauchy en el espacio métrico $(M,d)$,}\\
\text{y supongamos que existe una sub‑sucesión $(x_{n_j})_{j\in\mathbb{N}}$ tal que }x_{n_j}\to x.\\
\text{Entonces }x_n\to x.
\end{array} Sea ( x n ) n ∈ N una sucesi o ˊ n de Cauchy en el espacio m e ˊ trico ( M , d ) , y supongamos que existe una sub‑sucesi o ˊ n ( x n j ) j ∈ N tal que x n j → x . Entonces x n → x . Dem.: {\color{Orange}\text{Dem.:}} Dem.:
Dado ε > 0 \varepsilon>0 ε > 0 ( x n ) (x_n) ( x n ) N 1 ∈ N N_1\in\mathbb{N} N 1 ∈ N
n , m ≥ N 1 ⟹ d ( x n , x m ) < ε 2 . n,m\ge N_1\;\Longrightarrow\;d(x_n,x_m)<\tfrac{\varepsilon}{2}. n , m ≥ N 1 ⟹ d ( x n , x m ) < 2 ε . Por otra parte, al converger la sub‑sucesión, existe J ∈ N J\in\mathbb{N} J ∈ N
d ( x n J , x ) < ε 2 y adem a ˊ s n J ≥ N 1 . d(x_{n_J},x)<\tfrac{\varepsilon}{2}
\quad\text{y además}\quad
n_J\ge N_1. d ( x n J , x ) < 2 ε y adem a ˊ s n J ≥ N 1 . Ahora, para cualquier n ≥ N 1 n\ge N_1 n ≥ N 1
d ( x n , x ) ≤ d ( x n , x n J ) + d ( x n J , x ) < ε 2 + ε 2 = ε . \begin{aligned}
d(x_n,x)
&\le d(x_n,x_{n_J})+d(x_{n_J},x)
< \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2}
= \varepsilon.
\end{aligned} d ( x n , x ) ≤ d ( x n , x n J ) + d ( x n J , x ) < 2 ε + 2 ε = ε . Por lo tanto x n → x x_n\to x x n → x
Q.E.D. □ \quad \quad \quad \quad \quad \quad \quad \quad \boxed{\text{Q.E.D.}} \quad \square Q.E.D. □ Def. : {\color{Cyan} \text{Def. :} } Def. :
Sea ( M , d ) un espacio m e ˊ trico. M es completo si toda sucesi o ˊ n de Cauchy es convergente \begin{array}{l}
\text{Sea $(M,d)$ un espacio métrico. $M$ es completo si toda sucesión de Cauchy es convergente}
\end{array} Sea ( M , d ) un espacio m e ˊ trico. M es completo si toda sucesi o ˊ n de Cauchy es convergente (si tiene límite en M M M
E j e m p l o s : Ejemplos: E j e m pl os :
( Q , ∣ ⋅ ∣ ) (\mathbb{Q},|\cdot|) ( Q , ∣ ⋅ ∣ )
( q n ) n ∈ N ⊆ Q ∣ q n → 2 ( q_{n} )_{n \in \mathbb{N}}\subseteq \mathbb{Q}\bigm| q_{n}\to \sqrt{ 2 } ( q n ) n ∈ N ⊆ Q q n → 2 ⟹ ( q n ) \implies(q_{n}) ⟹ ( q n ) Q \mathbb{Q} Q
( M , δ ) (M,\delta) ( M , δ ) M = { 1 n : n ∈ N } M=\left\{ \frac{1}{n}:n \in \mathbb{N} \right\} M = { n 1 : n ∈ N } d 1 ( x , y ) = ∣ x − y ∣ d_{1}(x,y)=|x-y| d 1 ( x , y ) = ∣ x − y ∣ x n = 1 n , ( x n ) n ∈ N ⊆ M x_{n}= \frac{1}{n},\quad( x_{n} )_{n \in \mathbb{N}}\subseteq M x n = n 1 , ( x n ) n ∈ N ⊆ M M M M ⟹ ( M , d 1 ) \implies(M,d_{1}) ⟹ ( M , d 1 )
( M , δ ) si es completo (M,\delta)\text{ si es completo} ( M , δ ) si es completo tarea: δ ∼ d 1 \delta \sim d_{1} δ ∼ d 1
O b s : Obs: O b s : M M M d 1 d_{1} d 1 d 2 d_{2} d 2 ( M , d 1 ) (M,d_{1}) ( M , d 1 ) ⟹ ( M , d 2 ) \cancel{ \implies }(M,d_{2}) ⟹ ( M , d 2 )
Ejercicio:
M M M d 1 d_{1} d 1 d 2 d_{2} d 2 ∃ C 1 , C 2 > 0 ∣ \:\exists\:C_{1},C_{2}>0\bigm| ∃ C 1 , C 2 > 0
C 1 ⋅ d 1 ( x , y ) ≤ d 2 ( x , y ) ≤ C 2 ⋅ d 1 ( x , y ) ∀ x , y ∈ M C_{1}\cdot d_{1}(x,y)\leq d_{2}(x,y)\leq C_{2}\cdot d_{1}(x,y)\quad \forall x,y \in M C 1 ⋅ d 1 ( x , y ) ≤ d 2 ( x , y ) ≤ C 2 ⋅ d 1 ( x , y ) ∀ x , y ∈ M Si ( M , d 1 ) (M,d_{1}) ( M , d 1 ) ⟹ ( M , d 2 ) \implies(M,d_{2}) ⟹ ( M , d 2 )
Teorema : {\color{violet} \text{Teorema :} } Teorema :
( R , ∣ ⋅ ∣ ) es completo. \begin{array}{l}
\text{$(\mathbb{R},|\cdot|)$ es completo.}
\end{array} ( R , ∣ ⋅ ∣ ) es completo. Dem: {\color{violet} \text{Dem:} } Dem: ( x n ) n ∈ N ⊆ R ( x_{n} )_{n \in \mathbb{N}}\subseteq \mathbb{R} ( x n ) n ∈ N ⊆ R
X 1 = { x n : n ∈ N } ⊆ R X 2 = { x n : n > 2 } ⊆ R X k = { x n : n > k } \begin{array}{c}
X_{1}=\{ x_{n}:n \in \mathbb{N} \}\subseteq \mathbb{R} \\
X_{2}=\{ x_{n}:n> 2 \}\subseteq \mathbb{R} \\
X_{k}=\{ x_{n}:n>k \}
\end{array} X 1 = { x n : n ∈ N } ⊆ R X 2 = { x n : n > 2 } ⊆ R X k = { x n : n > k } X 1 X_{1} X 1 R \mathbb{R} R
X 1 ⊇ X 2 ⊇ X 3 ⊇ ⋯ ⊇ X k ⊇ … X_{1}\supseteq X_{2}\supseteq X_{3}\supseteq\dots \supseteq X_{k}\supseteq\dots X 1 ⊇ X 2 ⊇ X 3 ⊇ ⋯ ⊇ X k ⊇ … ⟹ \implies ⟹ X k X_{k} X k
a 1 = i n f ( X 1 ) a 2 = i n f ( X 2 ) ⋮ a k = i n f ( X k ) \begin{array}{c}
a_{1}=inf(X_{1}) \\
a_{2}=inf(X_{2}) \\
\vdots \\
a_{k}=inf(X_{k})
\end{array} a 1 = in f ( X 1 ) a 2 = in f ( X 2 ) ⋮ a k = in f ( X k ) Como
a 1 ≤ a 2 ≤ a 3 ≤ ⋯ ≤ a k ≤ ⋯ ≤ b = s u p ( X 1 ) a_{1}\leq a_{2}\leq a_{3}\leq \dots\leq a_{k}\leq \dots \leq b=sup(X_{1}) a 1 ≤ a 2 ≤ a 3 ≤ ⋯ ≤ a k ≤ ⋯ ≤ b = s u p ( X 1 ) Luego ( a n ) n ∈ N ( a_{n} )_{n \in \mathbb{N}} ( a n ) n ∈ N
⟹ ∃ a = lim n → ∞ a n \implies \:\exists\:a=\underset{ n\to \infty }{ \lim } a_{n} ⟹ ∃ a = n → ∞ lim a n
Acá ( a n ) (a_{n}) ( a n ) X i X_{i} X i
Falta ver que a = lim n → ∞ x n a=\underset{ n\to \infty }{ \lim }x_{n} a = n → ∞ lim x n x n x_{n} x n x n j ⟶ a x_{n_{j}}\longrightarrow a x n j ⟶ a
Vamos a ver que ∃ ( x n j ) j ∈ N ∣ x n j → a \:\exists\:( x_{n_{j}} )_{j \in \mathbb{N}}\bigm|x_{n_{j}}\to a ∃ ( x n j ) j ∈ N x n j → a n 1 ∣ a 1 ≤ x n 1 ≤ a 1 + 1 n_{1}\bigm|a_{1}\leq x_{n_{1}}\leq a_{1}+1 n 1 a 1 ≤ x n 1 ≤ a 1 + 1 n 2 ∣ a n 1 ≤ x n 2 ≤ a n 1 n_{2}\bigm|a_{n_{1}}\leq x_{n_{2}}\leq a_{n_{1}} n 2 a n 1 ≤ x n 2 ≤ a n 1 1 2 \frac{1}2{} 2 1 n j ∣ a n j − 1 ⏟ → a ≤ x n j ≤ a n j − 1 + 1 j ⏟ → a n_{j}\bigm|\underbrace{ a_{n_{j-1}} }_{ \to a }\leq x_{n_{j}}\leq \underbrace{ a_{n_{j}-1}+ \frac{1}{j} }_{ \to a } n j → a a n j − 1 ≤ x n j ≤ → a a n j − 1 + j 1
Tomo límite en j → ∞ j\to \infty j → ∞
x n j → a x_{n_{j}}\to a x n j → a □ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \square □
Usé propiedades de ínfimo y teorema del chanwich.
Corolario : {\color{Red} \text{Corolario }:} Corolario :
R n con d 1 , d 2 y d ∞ es completo. \begin{array}{l}
\text{$\mathbb{R}^{n}$ con $d_{1},d_{2}$ y $d_{\infty}$ es completo.}
\end{array} R n con d 1 , d 2 y d ∞ es completo. Citas y Comentarios 
