Mixing
Baby Rudin Theorem 3.7 Clarification
$begingroup$
$bf 3.7 $ Theorem $ $ The subsequential limits of a sequence ${p_n}$ in a metric space $X$ form a closed subset of $X$.
Proof $ $ Let $E^*$ be the set of all subsequential limits of ${p_n}$ and let $q$ be a limit point of $E^*$. We have to show that $qin E^*$.
$qquad$ Choose $n_1$ so that $p_{n_1}neq q$. (If no such $n_1$ exists, then $E^*$ has only one point, and there is nothing to prove.) Put $delta=d(q,p_{n_1})$. Suppose $n_1,...,n_{i-1}$ are chosen. Since $q$ is a limit point of $E^*$, there is an $xin E^*$ with $d(x,q)<2^{-i}delta$. Since $xin E^*$, there is an $n_i>n_{i-1}$ such that $d(x,p_{n_i})<2^{-i}delta$. Thus $$d(q,p_{n_i})leq 2^{1-i}delta$$ for $i=1,2,3,...$. This says that ${p_{n_i}}$ converges to $q$. Hence $qin E^*$.
Can someone explain how Rudin is selecting the $n_i$?
real-analysis
edited Aug 9 '16 at 10:01
user153330
381733
asked May 13 '15 at 7:06
AlainAlain
6151513
$endgroup$
add a comment |
$begingroup$
$bf 3.7 $ Theorem $ $ The subsequential limits of a sequence ${p_n}$ in a metric space $X$ form a closed subset of $X$.
Proof $ $ Let $E^*$ be the set of all subsequential limits of ${p_n}$ and let $q$ be a limit point of $E^*$. We have to show that $qin E^*$.
$qquad$ Choose $n_1$ so that $p_{n_1}neq q$. (If no such $n_1$ exists, then $E^*$ has only one point, and there is nothing to prove.) Put $delta=d(q,p_{n_1})$. Suppose $n_1,...,n_{i-1}$ are chosen. Since $q$ is a limit point of $E^*$, there is an $xin E^*$ with $d(x,q)<2^{-i}delta$. Since $xin E^*$, there is an $n_i>n_{i-1}$ such that $d(x,p_{n_i})<2^{-i}delta$. Thus $$d(q,p_{n_i})leq 2^{1-i}delta$$ for $i=1,2,3,...$. This says that ${p_{n_i}}$ converges to $q$. Hence $qin E^*$.
Can someone explain how Rudin is selecting the $n_i$?
real-analysis
edited Aug 9 '16 at 10:01
user153330
381733
asked May 13 '15 at 7:06
AlainAlain
6151513
$endgroup$
add a comment |
$begingroup$
$bf 3.7 $ Theorem $ $ The subsequential limits of a sequence ${p_n}$ in a metric space $X$ form a closed subset of $X$.
Proof $ $ Let $E^*$ be the set of all subsequential limits of ${p_n}$ and let $q$ be a limit point of $E^*$. We have to show that $qin E^*$.
$qquad$ Choose $n_1$ so that $p_{n_1}neq q$. (If no such $n_1$ exists, then $E^*$ has only one point, and there is nothing to prove.) Put $delta=d(q,p_{n_1})$. Suppose $n_1,...,n_{i-1}$ are chosen. Since $q$ is a limit point of $E^*$, there is an $xin E^*$ with $d(x,q)<2^{-i}delta$. Since $xin E^*$, there is an $n_i>n_{i-1}$ such that $d(x,p_{n_i})<2^{-i}delta$. Thus $$d(q,p_{n_i})leq 2^{1-i}delta$$ for $i=1,2,3,...$. This says that ${p_{n_i}}$ converges to $q$. Hence $qin E^*$.
Can someone explain how Rudin is selecting the $n_i$?
real-analysis
edited Aug 9 '16 at 10:01
user153330
381733
asked May 13 '15 at 7:06
AlainAlain
6151513
$endgroup$
$bf 3.7 $ Theorem $ $ The subsequential limits of a sequence ${p_n}$ in a metric space $X$ form a closed subset of $X$.
Proof $ $ Let $E^*$ be the set of all subsequential limits of ${p_n}$ and let $q$ be a limit point of $E^*$. We have to show that $qin E^*$.
$qquad$ Choose $n_1$ so that $p_{n_1}neq q$. (If no such $n_1$ exists, then $E^*$ has only one point, and there is nothing to prove.) Put $delta=d(q,p_{n_1})$. Suppose $n_1,...,n_{i-1}$ are chosen. Since $q$ is a limit point of $E^*$, there is an $xin E^*$ with $d(x,q)<2^{-i}delta$. Since $xin E^*$, there is an $n_i>n_{i-1}$ such that $d(x,p_{n_i})<2^{-i}delta$. Thus $$d(q,p_{n_i})leq 2^{1-i}delta$$ for $i=1,2,3,...$. This says that ${p_{n_i}}$ converges to $q$. Hence $qin E^*$.
Can someone explain how Rudin is selecting the $n_i$?
real-analysis
real-analysis
edited Aug 9 '16 at 10:01
user153330
381733
asked May 13 '15 at 7:06
AlainAlain
6151513
edited Aug 9 '16 at 10:01
user153330
381733
asked May 13 '15 at 7:06
AlainAlain
6151513
edited Aug 9 '16 at 10:01
user153330
381733
edited Aug 9 '16 at 10:01
user153330
381733
edited Aug 9 '16 at 10:01
user153330
381733
381733
asked May 13 '15 at 7:06
AlainAlain
6151513
asked May 13 '15 at 7:06
AlainAlain
6151513
asked May 13 '15 at 7:06
AlainAlain
6151513
6151513
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The sequence is choosen inductively. First he picks $n_1$ so that $p_{n_1}neq q$ and set $delta = d(p_{n_1}, q)$. Then he consider $delta/2$. As $q$ is a limit point, there is $n_2$ (which can be chosen so that $n_2 > n_1$) so that
$$d(p_{n_2}, q) < delta /2$$
Inductively, he choose $n_k>n_{k-1}> cdots n_2 >n_1$ so that
$$d(p_{n_i}, q)< frac{delta}{2^i}.$$
$endgroup$
add a comment |
$begingroup$
If $E^* = emptyset$, then $E^*$ is a closed subset of $X$.
If $E^* = {q}$ for some $q in X$, then $E^*$ is a closed subset of $X$.
We assume that $#E^* geq 2$.
Let $q$ be a limit point of $E^*$.
If $p_n = q$ for all $n in {1,2,cdots}$, then any subsequence of ${p_n}$ converges to $q$. So $E^* = {q}$. But we assumed that $#E^* geq 2$. So there is $n_1 in {1,2,cdots}$ such that $p_{n_1} neq q$.
Because $q$ is a limit point of $E^*$, there exists $x_2 in E^*$ such that $d(x_2, q) < frac{delta}{2^2}$.
Because $x_2$ is a subsequential limit of ${p_n}$, there exists $n_2 in {1,2,cdots}$ such that $n_2 > n_1$ and $d(p_{n_2}, x_2) < frac{delta}{2^2}$.
By the triangle equality in $X$, $d(p_{n_2}, q) leq d(p_{n_2}, x_2) + d(x_2, q) < frac{delta}{2^2} + frac{delta}{2^2} = frac{delta}{2}$.
Because $q$ is a limit point of $E^*$, there exists $x_3 in E^*$ such that $d(x_3, q) < frac{delta}{2^3}$.
Because $x_3$ is a subsequential limit of ${p_n}$, there exists $n_3 in {1,2,cdots}$ such that $n_3 > n_2$ and $d(p_{n_3}, x_3) < frac{delta}{2^3}$.
By the triangle equality in $X$, $d(p_{n_3}, q) leq d(p_{n_3}, x_3) + d(x_3, q) < frac{delta}{2^3} + frac{delta}{2^3} = frac{delta}{2^2}$.
$cdots$
answered Jan 18 at 4:30
tchappy hatchappy ha
763412
$endgroup$
add a comment |
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2 Answers
2
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2 Answers
2
active
oldest
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active
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$begingroup$
The sequence is choosen inductively. First he picks $n_1$ so that $p_{n_1}neq q$ and set $delta = d(p_{n_1}, q)$. Then he consider $delta/2$. As $q$ is a limit point, there is $n_2$ (which can be chosen so that $n_2 > n_1$) so that
$$d(p_{n_2}, q) < delta /2$$
Inductively, he choose $n_k>n_{k-1}> cdots n_2 >n_1$ so that
$$d(p_{n_i}, q)< frac{delta}{2^i}.$$
$endgroup$
add a comment |
$begingroup$
The sequence is choosen inductively. First he picks $n_1$ so that $p_{n_1}neq q$ and set $delta = d(p_{n_1}, q)$. Then he consider $delta/2$. As $q$ is a limit point, there is $n_2$ (which can be chosen so that $n_2 > n_1$) so that
$$d(p_{n_2}, q) < delta /2$$
Inductively, he choose $n_k>n_{k-1}> cdots n_2 >n_1$ so that
$$d(p_{n_i}, q)< frac{delta}{2^i}.$$
$endgroup$
add a comment |
$begingroup$
The sequence is choosen inductively. First he picks $n_1$ so that $p_{n_1}neq q$ and set $delta = d(p_{n_1}, q)$. Then he consider $delta/2$. As $q$ is a limit point, there is $n_2$ (which can be chosen so that $n_2 > n_1$) so that
$$d(p_{n_2}, q) < delta /2$$
Inductively, he choose $n_k>n_{k-1}> cdots n_2 >n_1$ so that
$$d(p_{n_i}, q)< frac{delta}{2^i}.$$
$endgroup$
The sequence is choosen inductively. First he picks $n_1$ so that $p_{n_1}neq q$ and set $delta = d(p_{n_1}, q)$. Then he consider $delta/2$. As $q$ is a limit point, there is $n_2$ (which can be chosen so that $n_2 > n_1$) so that
$$d(p_{n_2}, q) < delta /2$$
Inductively, he choose $n_k>n_{k-1}> cdots n_2 >n_1$ so that
$$d(p_{n_i}, q)< frac{delta}{2^i}.$$
answered May 13 '15 at 7:23
user99914
add a comment |
add a comment |
$begingroup$
If $E^* = emptyset$, then $E^*$ is a closed subset of $X$.
If $E^* = {q}$ for some $q in X$, then $E^*$ is a closed subset of $X$.
We assume that $#E^* geq 2$.
Let $q$ be a limit point of $E^*$.
If $p_n = q$ for all $n in {1,2,cdots}$, then any subsequence of ${p_n}$ converges to $q$. So $E^* = {q}$. But we assumed that $#E^* geq 2$. So there is $n_1 in {1,2,cdots}$ such that $p_{n_1} neq q$.
Because $q$ is a limit point of $E^*$, there exists $x_2 in E^*$ such that $d(x_2, q) < frac{delta}{2^2}$.
Because $x_2$ is a subsequential limit of ${p_n}$, there exists $n_2 in {1,2,cdots}$ such that $n_2 > n_1$ and $d(p_{n_2}, x_2) < frac{delta}{2^2}$.
By the triangle equality in $X$, $d(p_{n_2}, q) leq d(p_{n_2}, x_2) + d(x_2, q) < frac{delta}{2^2} + frac{delta}{2^2} = frac{delta}{2}$.
Because $q$ is a limit point of $E^*$, there exists $x_3 in E^*$ such that $d(x_3, q) < frac{delta}{2^3}$.
Because $x_3$ is a subsequential limit of ${p_n}$, there exists $n_3 in {1,2,cdots}$ such that $n_3 > n_2$ and $d(p_{n_3}, x_3) < frac{delta}{2^3}$.
By the triangle equality in $X$, $d(p_{n_3}, q) leq d(p_{n_3}, x_3) + d(x_3, q) < frac{delta}{2^3} + frac{delta}{2^3} = frac{delta}{2^2}$.
$cdots$
answered Jan 18 at 4:30
tchappy hatchappy ha
763412
$endgroup$
add a comment |
$begingroup$
If $E^* = emptyset$, then $E^*$ is a closed subset of $X$.
If $E^* = {q}$ for some $q in X$, then $E^*$ is a closed subset of $X$.
We assume that $#E^* geq 2$.
Let $q$ be a limit point of $E^*$.
If $p_n = q$ for all $n in {1,2,cdots}$, then any subsequence of ${p_n}$ converges to $q$. So $E^* = {q}$. But we assumed that $#E^* geq 2$. So there is $n_1 in {1,2,cdots}$ such that $p_{n_1} neq q$.
Because $q$ is a limit point of $E^*$, there exists $x_2 in E^*$ such that $d(x_2, q) < frac{delta}{2^2}$.
Because $x_2$ is a subsequential limit of ${p_n}$, there exists $n_2 in {1,2,cdots}$ such that $n_2 > n_1$ and $d(p_{n_2}, x_2) < frac{delta}{2^2}$.
By the triangle equality in $X$, $d(p_{n_2}, q) leq d(p_{n_2}, x_2) + d(x_2, q) < frac{delta}{2^2} + frac{delta}{2^2} = frac{delta}{2}$.
Because $q$ is a limit point of $E^*$, there exists $x_3 in E^*$ such that $d(x_3, q) < frac{delta}{2^3}$.
Because $x_3$ is a subsequential limit of ${p_n}$, there exists $n_3 in {1,2,cdots}$ such that $n_3 > n_2$ and $d(p_{n_3}, x_3) < frac{delta}{2^3}$.
By the triangle equality in $X$, $d(p_{n_3}, q) leq d(p_{n_3}, x_3) + d(x_3, q) < frac{delta}{2^3} + frac{delta}{2^3} = frac{delta}{2^2}$.
$cdots$
answered Jan 18 at 4:30
tchappy hatchappy ha
763412
$endgroup$
add a comment |
$begingroup$
If $E^* = emptyset$, then $E^*$ is a closed subset of $X$.
If $E^* = {q}$ for some $q in X$, then $E^*$ is a closed subset of $X$.
We assume that $#E^* geq 2$.
Let $q$ be a limit point of $E^*$.
If $p_n = q$ for all $n in {1,2,cdots}$, then any subsequence of ${p_n}$ converges to $q$. So $E^* = {q}$. But we assumed that $#E^* geq 2$. So there is $n_1 in {1,2,cdots}$ such that $p_{n_1} neq q$.
Because $q$ is a limit point of $E^*$, there exists $x_2 in E^*$ such that $d(x_2, q) < frac{delta}{2^2}$.
Because $x_2$ is a subsequential limit of ${p_n}$, there exists $n_2 in {1,2,cdots}$ such that $n_2 > n_1$ and $d(p_{n_2}, x_2) < frac{delta}{2^2}$.
By the triangle equality in $X$, $d(p_{n_2}, q) leq d(p_{n_2}, x_2) + d(x_2, q) < frac{delta}{2^2} + frac{delta}{2^2} = frac{delta}{2}$.
Because $q$ is a limit point of $E^*$, there exists $x_3 in E^*$ such that $d(x_3, q) < frac{delta}{2^3}$.
Because $x_3$ is a subsequential limit of ${p_n}$, there exists $n_3 in {1,2,cdots}$ such that $n_3 > n_2$ and $d(p_{n_3}, x_3) < frac{delta}{2^3}$.
By the triangle equality in $X$, $d(p_{n_3}, q) leq d(p_{n_3}, x_3) + d(x_3, q) < frac{delta}{2^3} + frac{delta}{2^3} = frac{delta}{2^2}$.
$cdots$
answered Jan 18 at 4:30
tchappy hatchappy ha
763412
$endgroup$
If $E^* = emptyset$, then $E^*$ is a closed subset of $X$.
If $E^* = {q}$ for some $q in X$, then $E^*$ is a closed subset of $X$.
We assume that $#E^* geq 2$.
Let $q$ be a limit point of $E^*$.
If $p_n = q$ for all $n in {1,2,cdots}$, then any subsequence of ${p_n}$ converges to $q$. So $E^* = {q}$. But we assumed that $#E^* geq 2$. So there is $n_1 in {1,2,cdots}$ such that $p_{n_1} neq q$.
Because $q$ is a limit point of $E^*$, there exists $x_2 in E^*$ such that $d(x_2, q) < frac{delta}{2^2}$.
Because $x_2$ is a subsequential limit of ${p_n}$, there exists $n_2 in {1,2,cdots}$ such that $n_2 > n_1$ and $d(p_{n_2}, x_2) < frac{delta}{2^2}$.
By the triangle equality in $X$, $d(p_{n_2}, q) leq d(p_{n_2}, x_2) + d(x_2, q) < frac{delta}{2^2} + frac{delta}{2^2} = frac{delta}{2}$.
Because $q$ is a limit point of $E^*$, there exists $x_3 in E^*$ such that $d(x_3, q) < frac{delta}{2^3}$.
Because $x_3$ is a subsequential limit of ${p_n}$, there exists $n_3 in {1,2,cdots}$ such that $n_3 > n_2$ and $d(p_{n_3}, x_3) < frac{delta}{2^3}$.
By the triangle equality in $X$, $d(p_{n_3}, q) leq d(p_{n_3}, x_3) + d(x_3, q) < frac{delta}{2^3} + frac{delta}{2^3} = frac{delta}{2^2}$.
$cdots$
answered Jan 18 at 4:30
tchappy hatchappy ha
763412
edited Jan 18 at 4:42
answered Jan 18 at 4:30
tchappy hatchappy ha
763412
answered Jan 18 at 4:30
tchappy hatchappy ha
763412
answered Jan 18 at 4:30
tchappy hatchappy ha
763412
763412
add a comment |
add a comment |
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