Why we can always choose $n_{k+1}>n_k$?
$begingroup$
Let $E$ is a complex inner product space and $(x_n)_{n},(y_n)_{n}subseteq E$ are unit sequences.
If we suppose that it is not true that there is a constant $c<1$ such that
$$|langle x_n|y_nrangle|leq c <1 ,$$
for all $n$ sufficiently large.
Therefore, for every positive integer $k$, the number $c_k=1-1/k$ does not satisfy
$$|langle x_n|y_nrangle|leq 1-1/k ,tag{*}$$
for all $n$ sufficiently large, which means that it is not true that there exists an $N$ such that for all $n>N$ the inequality $(*)$ is valid.
Therefore, there exists some $n_k$ such that
$$1geq |langle x_{n_k}|y_{n_k}rangle|> 1-1/k, $$
where the inequality $1geq |langle x_{n_k}|y_{n_k}rangle|$ follows from the fact that $x_n,y_n$ are unit vectors.
Hence
$$lim_{ktoinfty}|langle x_{n_k}|y_{n_k}rangle|=1.$$
Why we can always choose $n_{k+1}>n_k$?
real-analysis
asked Jan 9 at 15:25
SchülerSchüler
1,4481421
$endgroup$
add a comment |
$begingroup$
Let $E$ is a complex inner product space and $(x_n)_{n},(y_n)_{n}subseteq E$ are unit sequences.
If we suppose that it is not true that there is a constant $c<1$ such that
$$|langle x_n|y_nrangle|leq c <1 ,$$
for all $n$ sufficiently large.
Therefore, for every positive integer $k$, the number $c_k=1-1/k$ does not satisfy
$$|langle x_n|y_nrangle|leq 1-1/k ,tag{*}$$
for all $n$ sufficiently large, which means that it is not true that there exists an $N$ such that for all $n>N$ the inequality $(*)$ is valid.
Therefore, there exists some $n_k$ such that
$$1geq |langle x_{n_k}|y_{n_k}rangle|> 1-1/k, $$
where the inequality $1geq |langle x_{n_k}|y_{n_k}rangle|$ follows from the fact that $x_n,y_n$ are unit vectors.
Hence
$$lim_{ktoinfty}|langle x_{n_k}|y_{n_k}rangle|=1.$$
Why we can always choose $n_{k+1}>n_k$?
real-analysis
asked Jan 9 at 15:25
SchülerSchüler
1,4481421
$endgroup$
add a comment |
$begingroup$
Let $E$ is a complex inner product space and $(x_n)_{n},(y_n)_{n}subseteq E$ are unit sequences.
If we suppose that it is not true that there is a constant $c<1$ such that
$$|langle x_n|y_nrangle|leq c <1 ,$$
for all $n$ sufficiently large.
Therefore, for every positive integer $k$, the number $c_k=1-1/k$ does not satisfy
$$|langle x_n|y_nrangle|leq 1-1/k ,tag{*}$$
for all $n$ sufficiently large, which means that it is not true that there exists an $N$ such that for all $n>N$ the inequality $(*)$ is valid.
Therefore, there exists some $n_k$ such that
$$1geq |langle x_{n_k}|y_{n_k}rangle|> 1-1/k, $$
where the inequality $1geq |langle x_{n_k}|y_{n_k}rangle|$ follows from the fact that $x_n,y_n$ are unit vectors.
Hence
$$lim_{ktoinfty}|langle x_{n_k}|y_{n_k}rangle|=1.$$
Why we can always choose $n_{k+1}>n_k$?
real-analysis
asked Jan 9 at 15:25
SchülerSchüler
1,4481421
$endgroup$
Let $E$ is a complex inner product space and $(x_n)_{n},(y_n)_{n}subseteq E$ are unit sequences.
If we suppose that it is not true that there is a constant $c<1$ such that
$$|langle x_n|y_nrangle|leq c <1 ,$$
for all $n$ sufficiently large.
Therefore, for every positive integer $k$, the number $c_k=1-1/k$ does not satisfy
$$|langle x_n|y_nrangle|leq 1-1/k ,tag{*}$$
for all $n$ sufficiently large, which means that it is not true that there exists an $N$ such that for all $n>N$ the inequality $(*)$ is valid.
Therefore, there exists some $n_k$ such that
$$1geq |langle x_{n_k}|y_{n_k}rangle|> 1-1/k, $$
where the inequality $1geq |langle x_{n_k}|y_{n_k}rangle|$ follows from the fact that $x_n,y_n$ are unit vectors.
Hence
$$lim_{ktoinfty}|langle x_{n_k}|y_{n_k}rangle|=1.$$
Why we can always choose $n_{k+1}>n_k$?
real-analysis
real-analysis
asked Jan 9 at 15:25
SchülerSchüler
1,4481421
asked Jan 9 at 15:25
SchülerSchüler
1,4481421
asked Jan 9 at 15:25
SchülerSchüler
1,4481421
asked Jan 9 at 15:25
SchülerSchüler
1,4481421
asked Jan 9 at 15:25
SchülerSchüler
1,4481421
1,4481421
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The contrapositive of the statement "There exists $NinBbb N$ such that
$$
|langle x_n|y_nrangle|leq 1-1/k
$$
holds for all $nge N$" is
"For each $Nin Bbb N$, we can find $m > N$ such that
$$
|langle x_m|y_mrangle| > 1-1/k
$$
holds."
Now, we just take $N=n_k$ and take $n_{k+1}$ to be $m$ from the above statement.
answered Jan 9 at 15:39
BigbearZzzBigbearZzz
8,49221652
$endgroup$
$begingroup$
Thank you for answer. Please how can I write correctely the sentence ''there is a constant $theta<1$ such that $|langle x_n|y_nrangle|leq c <1 $ for all $n$ sufficiently large.'' ? It is true to write ''there is a constant $theta<1$ and $Nin mathbb{N}$ such that for all $ngeq N$ we have $$|langle x_n|y_nrangle|leq c <1. $$ ?
$endgroup$
– Schüler
Jan 9 at 15:55
$begingroup$
@Schüler You are correct.
$endgroup$
– BigbearZzz
Jan 9 at 16:12
add a comment |
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1 Answer
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active
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1 Answer
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oldest
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$begingroup$
The contrapositive of the statement "There exists $NinBbb N$ such that
$$
|langle x_n|y_nrangle|leq 1-1/k
$$
holds for all $nge N$" is
"For each $Nin Bbb N$, we can find $m > N$ such that
$$
|langle x_m|y_mrangle| > 1-1/k
$$
holds."
Now, we just take $N=n_k$ and take $n_{k+1}$ to be $m$ from the above statement.
answered Jan 9 at 15:39
BigbearZzzBigbearZzz
8,49221652
$endgroup$
$begingroup$
Thank you for answer. Please how can I write correctely the sentence ''there is a constant $theta<1$ such that $|langle x_n|y_nrangle|leq c <1 $ for all $n$ sufficiently large.'' ? It is true to write ''there is a constant $theta<1$ and $Nin mathbb{N}$ such that for all $ngeq N$ we have $$|langle x_n|y_nrangle|leq c <1. $$ ?
$endgroup$
– Schüler
Jan 9 at 15:55
$begingroup$
@Schüler You are correct.
$endgroup$
– BigbearZzz
Jan 9 at 16:12
add a comment |
$begingroup$
The contrapositive of the statement "There exists $NinBbb N$ such that
$$
|langle x_n|y_nrangle|leq 1-1/k
$$
holds for all $nge N$" is
"For each $Nin Bbb N$, we can find $m > N$ such that
$$
|langle x_m|y_mrangle| > 1-1/k
$$
holds."
Now, we just take $N=n_k$ and take $n_{k+1}$ to be $m$ from the above statement.
answered Jan 9 at 15:39
BigbearZzzBigbearZzz
8,49221652
$endgroup$
$begingroup$
Thank you for answer. Please how can I write correctely the sentence ''there is a constant $theta<1$ such that $|langle x_n|y_nrangle|leq c <1 $ for all $n$ sufficiently large.'' ? It is true to write ''there is a constant $theta<1$ and $Nin mathbb{N}$ such that for all $ngeq N$ we have $$|langle x_n|y_nrangle|leq c <1. $$ ?
$endgroup$
– Schüler
Jan 9 at 15:55
$begingroup$
@Schüler You are correct.
$endgroup$
– BigbearZzz
Jan 9 at 16:12
add a comment |
$begingroup$
The contrapositive of the statement "There exists $NinBbb N$ such that
$$
|langle x_n|y_nrangle|leq 1-1/k
$$
holds for all $nge N$" is
"For each $Nin Bbb N$, we can find $m > N$ such that
$$
|langle x_m|y_mrangle| > 1-1/k
$$
holds."
Now, we just take $N=n_k$ and take $n_{k+1}$ to be $m$ from the above statement.
answered Jan 9 at 15:39
BigbearZzzBigbearZzz
8,49221652
$endgroup$
The contrapositive of the statement "There exists $NinBbb N$ such that
$$
|langle x_n|y_nrangle|leq 1-1/k
$$
holds for all $nge N$" is
"For each $Nin Bbb N$, we can find $m > N$ such that
$$
|langle x_m|y_mrangle| > 1-1/k
$$
holds."
Now, we just take $N=n_k$ and take $n_{k+1}$ to be $m$ from the above statement.
answered Jan 9 at 15:39
BigbearZzzBigbearZzz
8,49221652
answered Jan 9 at 15:39
BigbearZzzBigbearZzz
8,49221652
answered Jan 9 at 15:39
BigbearZzzBigbearZzz
8,49221652
answered Jan 9 at 15:39
BigbearZzzBigbearZzz
8,49221652
8,49221652
$begingroup$
Thank you for answer. Please how can I write correctely the sentence ''there is a constant $theta<1$ such that $|langle x_n|y_nrangle|leq c <1 $ for all $n$ sufficiently large.'' ? It is true to write ''there is a constant $theta<1$ and $Nin mathbb{N}$ such that for all $ngeq N$ we have $$|langle x_n|y_nrangle|leq c <1. $$ ?
$endgroup$
– Schüler
Jan 9 at 15:55
$begingroup$
@Schüler You are correct.
$endgroup$
– BigbearZzz
Jan 9 at 16:12
add a comment |
$begingroup$
Thank you for answer. Please how can I write correctely the sentence ''there is a constant $theta<1$ such that $|langle x_n|y_nrangle|leq c <1 $ for all $n$ sufficiently large.'' ? It is true to write ''there is a constant $theta<1$ and $Nin mathbb{N}$ such that for all $ngeq N$ we have $$|langle x_n|y_nrangle|leq c <1. $$ ?
$endgroup$
– Schüler
Jan 9 at 15:55
$begingroup$
@Schüler You are correct.
$endgroup$
– BigbearZzz
Jan 9 at 16:12
$begingroup$
Thank you for answer. Please how can I write correctely the sentence ''there is a constant $theta<1$ such that $|langle x_n|y_nrangle|leq c <1 $ for all $n$ sufficiently large.'' ? It is true to write ''there is a constant $theta<1$ and $Nin mathbb{N}$ such that for all $ngeq N$ we have $$|langle x_n|y_nrangle|leq c <1. $$ ?
$endgroup$
– Schüler
Jan 9 at 15:55
$begingroup$
Thank you for answer. Please how can I write correctely the sentence ''there is a constant $theta<1$ such that $|langle x_n|y_nrangle|leq c <1 $ for all $n$ sufficiently large.'' ? It is true to write ''there is a constant $theta<1$ and $Nin mathbb{N}$ such that for all $ngeq N$ we have $$|langle x_n|y_nrangle|leq c <1. $$ ?
$endgroup$
– Schüler
Jan 9 at 15:55
$begingroup$
@Schüler You are correct.
$endgroup$
– BigbearZzz
Jan 9 at 16:12
$begingroup$
@Schüler You are correct.
$endgroup$
– BigbearZzz
Jan 9 at 16:12
add a comment |
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