Mixing
Complete Metric Space if and only if
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Let $(X,d)$ be a metric space. Let $mathcal{C}$ be the set of all collections ${O_i}_{i=1}^infty$ of non-empty closed subsets such that begin{align*}
&(a) O_{n+1}subset O_n forall n \
&(b) limoperatorname{diam} (O_n) = 0 as n to infty
end{align*}
Prove that $X$ is complete if and only if $forall C in mathcal{C}$ begin{align*}
bigcap_{Ain C} A not= emptyset
end{align*}
For the $if$ part: For every $n$, choose $x_n in O_n$. Then since $O_{n+1}subset O_n$, the set ${x_n,x_{n+1},x_{n+2},cdots}subset O_n$. Since $limoperatorname{diam}(O_n) = 0$, for any $epsilon > 0$ choose a natural number $N$ so that $operatorname{diam}(O_n)<epsilon$ for $ngeq N$. This means that for any $n,m geq N$, $|x_n-x_m| leqoperatorname{diam}({x_{N},x_{N+1},dots}) leqoperatorname{diam}(O_N) < epsilon$. So ${x_n}$ is a Cauchy sequence. By completeness of $X$, ${x_n}$ converges to a point, $a$. By $O_n$ being closed it must contain ${x_n,x_{n+1},dots}$. Thus, for any given $Dinmathcal{C}$ we get
begin{align*}
ain bigcap_{A in D} A.
end{align*}
hence non-empty.
Also, I am not sure how to start the other direction. I think you need to look at the tails of each sequence
metric-spaces
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up vote
1
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Let $(X,d)$ be a metric space. Let $mathcal{C}$ be the set of all collections ${O_i}_{i=1}^infty$ of non-empty closed subsets such that begin{align*}
&(a) O_{n+1}subset O_n forall n \
&(b) limoperatorname{diam} (O_n) = 0 as n to infty
end{align*}
Prove that $X$ is complete if and only if $forall C in mathcal{C}$ begin{align*}
bigcap_{Ain C} A not= emptyset
end{align*}
For the $if$ part: For every $n$, choose $x_n in O_n$. Then since $O_{n+1}subset O_n$, the set ${x_n,x_{n+1},x_{n+2},cdots}subset O_n$. Since $limoperatorname{diam}(O_n) = 0$, for any $epsilon > 0$ choose a natural number $N$ so that $operatorname{diam}(O_n)<epsilon$ for $ngeq N$. This means that for any $n,m geq N$, $|x_n-x_m| leqoperatorname{diam}({x_{N},x_{N+1},dots}) leqoperatorname{diam}(O_N) < epsilon$. So ${x_n}$ is a Cauchy sequence. By completeness of $X$, ${x_n}$ converges to a point, $a$. By $O_n$ being closed it must contain ${x_n,x_{n+1},dots}$. Thus, for any given $Dinmathcal{C}$ we get
begin{align*}
ain bigcap_{A in D} A.
end{align*}
hence non-empty.
Also, I am not sure how to start the other direction. I think you need to look at the tails of each sequence
metric-spaces
edited yesterday
José Carlos Santos
140k18111203
asked yesterday
Issacg628496
82
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Issacg628496 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $(X,d)$ be a metric space. Let $mathcal{C}$ be the set of all collections ${O_i}_{i=1}^infty$ of non-empty closed subsets such that begin{align*}
&(a) O_{n+1}subset O_n forall n \
&(b) limoperatorname{diam} (O_n) = 0 as n to infty
end{align*}
Prove that $X$ is complete if and only if $forall C in mathcal{C}$ begin{align*}
bigcap_{Ain C} A not= emptyset
end{align*}
For the $if$ part: For every $n$, choose $x_n in O_n$. Then since $O_{n+1}subset O_n$, the set ${x_n,x_{n+1},x_{n+2},cdots}subset O_n$. Since $limoperatorname{diam}(O_n) = 0$, for any $epsilon > 0$ choose a natural number $N$ so that $operatorname{diam}(O_n)<epsilon$ for $ngeq N$. This means that for any $n,m geq N$, $|x_n-x_m| leqoperatorname{diam}({x_{N},x_{N+1},dots}) leqoperatorname{diam}(O_N) < epsilon$. So ${x_n}$ is a Cauchy sequence. By completeness of $X$, ${x_n}$ converges to a point, $a$. By $O_n$ being closed it must contain ${x_n,x_{n+1},dots}$. Thus, for any given $Dinmathcal{C}$ we get
begin{align*}
ain bigcap_{A in D} A.
end{align*}
hence non-empty.
Also, I am not sure how to start the other direction. I think you need to look at the tails of each sequence
metric-spaces
edited yesterday
José Carlos Santos
140k18111203
asked yesterday
Issacg628496
82
New contributor
Issacg628496 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $(X,d)$ be a metric space. Let $mathcal{C}$ be the set of all collections ${O_i}_{i=1}^infty$ of non-empty closed subsets such that begin{align*}
&(a) O_{n+1}subset O_n forall n \
&(b) limoperatorname{diam} (O_n) = 0 as n to infty
end{align*}
Prove that $X$ is complete if and only if $forall C in mathcal{C}$ begin{align*}
bigcap_{Ain C} A not= emptyset
end{align*}
For the $if$ part: For every $n$, choose $x_n in O_n$. Then since $O_{n+1}subset O_n$, the set ${x_n,x_{n+1},x_{n+2},cdots}subset O_n$. Since $limoperatorname{diam}(O_n) = 0$, for any $epsilon > 0$ choose a natural number $N$ so that $operatorname{diam}(O_n)<epsilon$ for $ngeq N$. This means that for any $n,m geq N$, $|x_n-x_m| leqoperatorname{diam}({x_{N},x_{N+1},dots}) leqoperatorname{diam}(O_N) < epsilon$. So ${x_n}$ is a Cauchy sequence. By completeness of $X$, ${x_n}$ converges to a point, $a$. By $O_n$ being closed it must contain ${x_n,x_{n+1},dots}$. Thus, for any given $Dinmathcal{C}$ we get
begin{align*}
ain bigcap_{A in D} A.
end{align*}
hence non-empty.
Also, I am not sure how to start the other direction. I think you need to look at the tails of each sequence
metric-spaces
metric-spaces
edited yesterday
José Carlos Santos
140k18111203
asked yesterday
Issacg628496
82
New contributor
Issacg628496 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited yesterday
José Carlos Santos
140k18111203
asked yesterday
Issacg628496
82
New contributor
Issacg628496 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited yesterday
José Carlos Santos
140k18111203
edited yesterday
José Carlos Santos
140k18111203
edited yesterday
José Carlos Santos
140k18111203
140k18111203
asked yesterday
Issacg628496
82
New contributor
Issacg628496 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked yesterday
Issacg628496
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asked yesterday
Issacg628496
82
82
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1 Answer
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Consider the Cauchy sequence ${x_n}$. Define $$F_n :=overline{ {x_m vert m geq n}}$$.Note that $F_n $ is closed and satisfies the given two conditions. Let $mathcal{F} = {F_n}_{n=1}^infty$. Then $$F=bigcap_{F_nin mathcal {F} } F_n not= emptyset$$. Take $xin F $ and prove that $x $ indeed is the limit of the sequence ${x_n}$. You may have to use the fact that $d(A)= d(overline {A})$, where $d(A) $ is the diameter of the set $A $ and $overline {A}$ is the closure of $A $.
answered yesterday
Thomas Shelby
1219
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Consider the Cauchy sequence ${x_n}$. Define $$F_n :=overline{ {x_m vert m geq n}}$$.Note that $F_n $ is closed and satisfies the given two conditions. Let $mathcal{F} = {F_n}_{n=1}^infty$. Then $$F=bigcap_{F_nin mathcal {F} } F_n not= emptyset$$. Take $xin F $ and prove that $x $ indeed is the limit of the sequence ${x_n}$. You may have to use the fact that $d(A)= d(overline {A})$, where $d(A) $ is the diameter of the set $A $ and $overline {A}$ is the closure of $A $.
answered yesterday
Thomas Shelby
1219
add a comment |
up vote
0
down vote
accepted
Consider the Cauchy sequence ${x_n}$. Define $$F_n :=overline{ {x_m vert m geq n}}$$.Note that $F_n $ is closed and satisfies the given two conditions. Let $mathcal{F} = {F_n}_{n=1}^infty$. Then $$F=bigcap_{F_nin mathcal {F} } F_n not= emptyset$$. Take $xin F $ and prove that $x $ indeed is the limit of the sequence ${x_n}$. You may have to use the fact that $d(A)= d(overline {A})$, where $d(A) $ is the diameter of the set $A $ and $overline {A}$ is the closure of $A $.
answered yesterday
Thomas Shelby
1219
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Consider the Cauchy sequence ${x_n}$. Define $$F_n :=overline{ {x_m vert m geq n}}$$.Note that $F_n $ is closed and satisfies the given two conditions. Let $mathcal{F} = {F_n}_{n=1}^infty$. Then $$F=bigcap_{F_nin mathcal {F} } F_n not= emptyset$$. Take $xin F $ and prove that $x $ indeed is the limit of the sequence ${x_n}$. You may have to use the fact that $d(A)= d(overline {A})$, where $d(A) $ is the diameter of the set $A $ and $overline {A}$ is the closure of $A $.
answered yesterday
Thomas Shelby
1219
Consider the Cauchy sequence ${x_n}$. Define $$F_n :=overline{ {x_m vert m geq n}}$$.Note that $F_n $ is closed and satisfies the given two conditions. Let $mathcal{F} = {F_n}_{n=1}^infty$. Then $$F=bigcap_{F_nin mathcal {F} } F_n not= emptyset$$. Take $xin F $ and prove that $x $ indeed is the limit of the sequence ${x_n}$. You may have to use the fact that $d(A)= d(overline {A})$, where $d(A) $ is the diameter of the set $A $ and $overline {A}$ is the closure of $A $.
answered yesterday
Thomas Shelby
1219
edited yesterday
answered yesterday
Thomas Shelby
1219
answered yesterday
Thomas Shelby
1219
answered yesterday
Thomas Shelby
1219
1219
add a comment |
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