If $limlimits_{nto infty}|x_n|= 0$ then $suplimits_{ngeq 1}|x_n|<infty$
I'm self-studying Functional analysis and in one of the proofs, I have the following conclusion which I don't quite understand.
If $limlimits_{nto infty}|x_n|= 0$ then $suplimits_{ngeq 1}|x_n|<infty$
I suppose $limlimits_{nto infty}|x_n|= 0$ implies that $|x_n|<epsilon,;;forall,ngeq N,$ for some $N$. So, taking sup over ${x_n}_{ngeq N},$ we have
begin{align} suplimits_{ngeq N}|x_n|leqepsilon<infty.end{align}
So, how come the result? Could someone explain?
real-analysis functional-analysis limits analysis
asked Nov 29 '18 at 15:30
Omojola MichealOmojola Micheal
1,591322
add a comment |
I'm self-studying Functional analysis and in one of the proofs, I have the following conclusion which I don't quite understand.
If $limlimits_{nto infty}|x_n|= 0$ then $suplimits_{ngeq 1}|x_n|<infty$
I suppose $limlimits_{nto infty}|x_n|= 0$ implies that $|x_n|<epsilon,;;forall,ngeq N,$ for some $N$. So, taking sup over ${x_n}_{ngeq N},$ we have
begin{align} suplimits_{ngeq N}|x_n|leqepsilon<infty.end{align}
So, how come the result? Could someone explain?
real-analysis functional-analysis limits analysis
asked Nov 29 '18 at 15:30
Omojola MichealOmojola Micheal
1,591322
Mike.You forgot the $1le n lt N$ elements |x_n|.Does this change anything?
– Peter Szilas
Nov 29 '18 at 15:35
@Peter Szilas: Didn't realize that! Thanks!
– Omojola Micheal
Nov 29 '18 at 15:36
Small matter:)Welcome.
– Peter Szilas
Nov 29 '18 at 15:39
add a comment |
I'm self-studying Functional analysis and in one of the proofs, I have the following conclusion which I don't quite understand.
If $limlimits_{nto infty}|x_n|= 0$ then $suplimits_{ngeq 1}|x_n|<infty$
I suppose $limlimits_{nto infty}|x_n|= 0$ implies that $|x_n|<epsilon,;;forall,ngeq N,$ for some $N$. So, taking sup over ${x_n}_{ngeq N},$ we have
begin{align} suplimits_{ngeq N}|x_n|leqepsilon<infty.end{align}
So, how come the result? Could someone explain?
real-analysis functional-analysis limits analysis
asked Nov 29 '18 at 15:30
Omojola MichealOmojola Micheal
1,591322
I'm self-studying Functional analysis and in one of the proofs, I have the following conclusion which I don't quite understand.
If $limlimits_{nto infty}|x_n|= 0$ then $suplimits_{ngeq 1}|x_n|<infty$
I suppose $limlimits_{nto infty}|x_n|= 0$ implies that $|x_n|<epsilon,;;forall,ngeq N,$ for some $N$. So, taking sup over ${x_n}_{ngeq N},$ we have
begin{align} suplimits_{ngeq N}|x_n|leqepsilon<infty.end{align}
So, how come the result? Could someone explain?
real-analysis functional-analysis limits analysis
real-analysis functional-analysis limits analysis
asked Nov 29 '18 at 15:30
Omojola MichealOmojola Micheal
1,591322
asked Nov 29 '18 at 15:30
Omojola MichealOmojola Micheal
1,591322
edited Nov 29 '18 at 15:43
Omojola Micheal
asked Nov 29 '18 at 15:30
Omojola MichealOmojola Micheal
1,591322
asked Nov 29 '18 at 15:30
Omojola MichealOmojola Micheal
1,591322
asked Nov 29 '18 at 15:30
Omojola MichealOmojola Micheal
1,591322
1,591322
Mike.You forgot the $1le n lt N$ elements |x_n|.Does this change anything?
– Peter Szilas
Nov 29 '18 at 15:35
@Peter Szilas: Didn't realize that! Thanks!
– Omojola Micheal
Nov 29 '18 at 15:36
Small matter:)Welcome.
– Peter Szilas
Nov 29 '18 at 15:39
add a comment |
Mike.You forgot the $1le n lt N$ elements |x_n|.Does this change anything?
– Peter Szilas
Nov 29 '18 at 15:35
@Peter Szilas: Didn't realize that! Thanks!
– Omojola Micheal
Nov 29 '18 at 15:36
Small matter:)Welcome.
– Peter Szilas
Nov 29 '18 at 15:39
Mike.You forgot the $1le n lt N$ elements |x_n|.Does this change anything?
– Peter Szilas
Nov 29 '18 at 15:35
Mike.You forgot the $1le n lt N$ elements |x_n|.Does this change anything?
– Peter Szilas
Nov 29 '18 at 15:35
@Peter Szilas: Didn't realize that! Thanks!
– Omojola Micheal
Nov 29 '18 at 15:36
@Peter Szilas: Didn't realize that! Thanks!
– Omojola Micheal
Nov 29 '18 at 15:36
Small matter:)Welcome.
– Peter Szilas
Nov 29 '18 at 15:39
Small matter:)Welcome.
– Peter Szilas
Nov 29 '18 at 15:39
add a comment |
2 Answers
2
active
oldest
votes
Since $x_n to 0$ then exists $bar n$ such that forall $n ge bar n$
$$0le |a_n|le 1$$
now take
$$a_{max}=max{|a_n|:n=1,2,ldots,bar n}$$
and indicate with $M=max{1,a_{max}}$ then
$$0le |a_n|le M$$
answered Nov 29 '18 at 15:39
gimusigimusi
1
add a comment |
Hint: there are finitely many terms $a_n$ with $n<N.$
answered Nov 29 '18 at 15:32
szw1710szw1710
6,4101123
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Since $x_n to 0$ then exists $bar n$ such that forall $n ge bar n$
$$0le |a_n|le 1$$
now take
$$a_{max}=max{|a_n|:n=1,2,ldots,bar n}$$
and indicate with $M=max{1,a_{max}}$ then
$$0le |a_n|le M$$
answered Nov 29 '18 at 15:39
gimusigimusi
1
add a comment |
Since $x_n to 0$ then exists $bar n$ such that forall $n ge bar n$
$$0le |a_n|le 1$$
now take
$$a_{max}=max{|a_n|:n=1,2,ldots,bar n}$$
and indicate with $M=max{1,a_{max}}$ then
$$0le |a_n|le M$$
answered Nov 29 '18 at 15:39
gimusigimusi
1
add a comment |
Since $x_n to 0$ then exists $bar n$ such that forall $n ge bar n$
$$0le |a_n|le 1$$
now take
$$a_{max}=max{|a_n|:n=1,2,ldots,bar n}$$
and indicate with $M=max{1,a_{max}}$ then
$$0le |a_n|le M$$
answered Nov 29 '18 at 15:39
gimusigimusi
1
Since $x_n to 0$ then exists $bar n$ such that forall $n ge bar n$
$$0le |a_n|le 1$$
now take
$$a_{max}=max{|a_n|:n=1,2,ldots,bar n}$$
and indicate with $M=max{1,a_{max}}$ then
$$0le |a_n|le M$$
answered Nov 29 '18 at 15:39
gimusigimusi
1
answered Nov 29 '18 at 15:39
gimusigimusi
1
answered Nov 29 '18 at 15:39
gimusigimusi
1
answered Nov 29 '18 at 15:39
gimusigimusi
1
1
add a comment |
add a comment |
Hint: there are finitely many terms $a_n$ with $n<N.$
answered Nov 29 '18 at 15:32
szw1710szw1710
6,4101123
add a comment |
Hint: there are finitely many terms $a_n$ with $n<N.$
answered Nov 29 '18 at 15:32
szw1710szw1710
6,4101123
add a comment |
Hint: there are finitely many terms $a_n$ with $n<N.$
answered Nov 29 '18 at 15:32
szw1710szw1710
6,4101123
Hint: there are finitely many terms $a_n$ with $n<N.$
answered Nov 29 '18 at 15:32
szw1710szw1710
6,4101123
answered Nov 29 '18 at 15:32
szw1710szw1710
6,4101123
answered Nov 29 '18 at 15:32
szw1710szw1710
6,4101123
answered Nov 29 '18 at 15:32
szw1710szw1710
6,4101123
6,4101123
add a comment |
add a comment |
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Mike.You forgot the $1le n lt N$ elements |x_n|.Does this change anything?
– Peter Szilas
Nov 29 '18 at 15:35
@Peter Szilas: Didn't realize that! Thanks!
– Omojola Micheal
Nov 29 '18 at 15:36
Small matter:)Welcome.
– Peter Szilas
Nov 29 '18 at 15:39