Divergent real sequence for which $forall epsilon >0 exists a in mathbb{R}: |x_n-a|<epsilon$ for almost...
$begingroup$
Looking at the definition of convergence:
$exists a in mathbb{R}forall epsilon >0: |x_n-a|<epsilon$ for almost all n
I switched the order of the first two quantifiers. Is there a sequence for which
$forall epsilon >0 exists b in mathbb{R}: |x_n-b|<epsilon$ for almost all n
holds but it's not convergent?
real-analysis sequences-and-series convergence
edited Dec 18 '18 at 22:47
Bernard
123k741116
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
$endgroup$
add a comment |
$begingroup$
Looking at the definition of convergence:
$exists a in mathbb{R}forall epsilon >0: |x_n-a|<epsilon$ for almost all n
I switched the order of the first two quantifiers. Is there a sequence for which
$forall epsilon >0 exists b in mathbb{R}: |x_n-b|<epsilon$ for almost all n
holds but it's not convergent?
real-analysis sequences-and-series convergence
edited Dec 18 '18 at 22:47
Bernard
123k741116
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
$endgroup$
add a comment |
$begingroup$
Looking at the definition of convergence:
$exists a in mathbb{R}forall epsilon >0: |x_n-a|<epsilon$ for almost all n
I switched the order of the first two quantifiers. Is there a sequence for which
$forall epsilon >0 exists b in mathbb{R}: |x_n-b|<epsilon$ for almost all n
holds but it's not convergent?
real-analysis sequences-and-series convergence
edited Dec 18 '18 at 22:47
Bernard
123k741116
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
$endgroup$
Looking at the definition of convergence:
$exists a in mathbb{R}forall epsilon >0: |x_n-a|<epsilon$ for almost all n
I switched the order of the first two quantifiers. Is there a sequence for which
$forall epsilon >0 exists b in mathbb{R}: |x_n-b|<epsilon$ for almost all n
holds but it's not convergent?
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
edited Dec 18 '18 at 22:47
Bernard
123k741116
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
edited Dec 18 '18 at 22:47
Bernard
123k741116
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
edited Dec 18 '18 at 22:47
Bernard
123k741116
edited Dec 18 '18 at 22:47
Bernard
123k741116
edited Dec 18 '18 at 22:47
Bernard
123k741116
123k741116
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
583
add a comment |
add a comment |
1 Answer
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oldest
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Any sequence satisfying this condition is Cauchy, and therefore convergent to some finite limit by the completeness of $mathbb{R}$. To see this, suppose $epsilon > 0$. Then there exists $b$ such that $|x_n - b| < frac{epsilon}{2}$ for almost all $n$. In order words, for such a $b$, there exists $N$ such that $|x_n - b| < frac{epsilon}{2}$ whenever $n ge N$. But then whenever $m, n ge N$, we have
$$|x_m - x_n| = |(x_m - b) - (x_n - b)| le |x_m - b| + |x_n - b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon.$$
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
9,1191721
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Any sequence satisfying this condition is Cauchy, and therefore convergent to some finite limit by the completeness of $mathbb{R}$. To see this, suppose $epsilon > 0$. Then there exists $b$ such that $|x_n - b| < frac{epsilon}{2}$ for almost all $n$. In order words, for such a $b$, there exists $N$ such that $|x_n - b| < frac{epsilon}{2}$ whenever $n ge N$. But then whenever $m, n ge N$, we have
$$|x_m - x_n| = |(x_m - b) - (x_n - b)| le |x_m - b| + |x_n - b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon.$$
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
9,1191721
$endgroup$
add a comment |
$begingroup$
Any sequence satisfying this condition is Cauchy, and therefore convergent to some finite limit by the completeness of $mathbb{R}$. To see this, suppose $epsilon > 0$. Then there exists $b$ such that $|x_n - b| < frac{epsilon}{2}$ for almost all $n$. In order words, for such a $b$, there exists $N$ such that $|x_n - b| < frac{epsilon}{2}$ whenever $n ge N$. But then whenever $m, n ge N$, we have
$$|x_m - x_n| = |(x_m - b) - (x_n - b)| le |x_m - b| + |x_n - b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon.$$
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
9,1191721
$endgroup$
add a comment |
$begingroup$
Any sequence satisfying this condition is Cauchy, and therefore convergent to some finite limit by the completeness of $mathbb{R}$. To see this, suppose $epsilon > 0$. Then there exists $b$ such that $|x_n - b| < frac{epsilon}{2}$ for almost all $n$. In order words, for such a $b$, there exists $N$ such that $|x_n - b| < frac{epsilon}{2}$ whenever $n ge N$. But then whenever $m, n ge N$, we have
$$|x_m - x_n| = |(x_m - b) - (x_n - b)| le |x_m - b| + |x_n - b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon.$$
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
9,1191721
$endgroup$
Any sequence satisfying this condition is Cauchy, and therefore convergent to some finite limit by the completeness of $mathbb{R}$. To see this, suppose $epsilon > 0$. Then there exists $b$ such that $|x_n - b| < frac{epsilon}{2}$ for almost all $n$. In order words, for such a $b$, there exists $N$ such that $|x_n - b| < frac{epsilon}{2}$ whenever $n ge N$. But then whenever $m, n ge N$, we have
$$|x_m - x_n| = |(x_m - b) - (x_n - b)| le |x_m - b| + |x_n - b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon.$$
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
9,1191721
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
9,1191721
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
9,1191721
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
9,1191721
9,1191721
add a comment |
add a comment |
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