Prove that the limit superior of $(a_n)_n=m^infty$ and $(a_n)_n=m'^infty$ are equivilant

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I came across the following exercise while self-studying Terrence Tao's book Analysis I:

Exercise 6.4.2 Let $(a_n)_n=m^infty$ be an arbitrary sequence of real numbers and let $m'ge m$ and $kge 0$ be integers. Show that $limsup_ntoinfty a_n$ is equivalent to

1. The limit superior of of $(a_n)_n=m'^infty$.


2. The limit superior of $(a_n+k)_n=m^infty$.

My Attempt: Let us first prove 1. Suppose that the limit superior of $(a_n)_n=m^infty$ is equal to some real number $cinBbb R$, i.e. $inf(a_N^+)_N=m^infty = c$ where $(a_N^+)_N=m^infty$ denotes the sequence of all $a_N=sup(a_n)_n=N^infty$. If $M'ge Mge m$ then $a_M'^+le a_M^+$, as if $a_M^+le a_M'^+$ then $$forall nge M: a_nle a_M^+le a_M'^+,$$ and $a_M'^+$ cannot be the lowest upper bound of $(a_n)_n=M'^infty$. Thus, $(a_N^+)_N=m^infty$ is decreasing. To prove that $c$ is a lower bound of $(a^+_N)_N=m'^infty$, suppose that there is some $a_n'<c$ where $n'le m'$. Then $$exists n'ge m'forall nge m:a_n'<cle a_nimplies exists n'ge m': a_n'<a_n',$$ a contradiction. Conversely, suppose that $inf(a^+_N)_N=m'^infty = c$. Then once again, $(a^+_N)_N=m'^infty$ is decreasing and as such, $$forall n'ge m'ge nge m:a_nge a_n'ge c.$$ Hence, $inf(a_N^+)_N=m^infty = c$. For 2, note that $(a_n+k)_n=m^infty$ is precisely the same sequence as $(a_n)_n=m+k^infty$, thus part 1 is applicable here since $mle m+k$.

Is this reasoning correct, or have I done something wrong? Thanks in advance.

Edit: I realize now that it was unnecessary to prove that $(a^+_N)_N=m^infty$ in the first bit of my proof of 1, even though it is applicable to the second part. Granted, one could cope with this by using a contradiction like $$exists n'ge m'forall nge m:a_n'<cle a_nimplies exists n'ge m': a_n'<a_n'+1,$$
instead (this should imply that $(a^+_N)_N=m^infty$ is not decreasing).

real-analysis self-learning limsup-and-liminf

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edited Jul 24 at 21:25

asked Jul 24 at 13:50

Crosby

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I came across the following exercise while self-studying Terrence Tao's book Analysis I:

Exercise 6.4.2 Let $(a_n)_n=m^infty$ be an arbitrary sequence of real numbers and let $m'ge m$ and $kge 0$ be integers. Show that $limsup_ntoinfty a_n$ is equivalent to

1. The limit superior of of $(a_n)_n=m'^infty$.


2. The limit superior of $(a_n+k)_n=m^infty$.

My Attempt: Let us first prove 1. Suppose that the limit superior of $(a_n)_n=m^infty$ is equal to some real number $cinBbb R$, i.e. $inf(a_N^+)_N=m^infty = c$ where $(a_N^+)_N=m^infty$ denotes the sequence of all $a_N=sup(a_n)_n=N^infty$. If $M'ge Mge m$ then $a_M'^+le a_M^+$, as if $a_M^+le a_M'^+$ then $$forall nge M: a_nle a_M^+le a_M'^+,$$ and $a_M'^+$ cannot be the lowest upper bound of $(a_n)_n=M'^infty$. Thus, $(a_N^+)_N=m^infty$ is decreasing. To prove that $c$ is a lower bound of $(a^+_N)_N=m'^infty$, suppose that there is some $a_n'<c$ where $n'le m'$. Then $$exists n'ge m'forall nge m:a_n'<cle a_nimplies exists n'ge m': a_n'<a_n',$$ a contradiction. Conversely, suppose that $inf(a^+_N)_N=m'^infty = c$. Then once again, $(a^+_N)_N=m'^infty$ is decreasing and as such, $$forall n'ge m'ge nge m:a_nge a_n'ge c.$$ Hence, $inf(a_N^+)_N=m^infty = c$. For 2, note that $(a_n+k)_n=m^infty$ is precisely the same sequence as $(a_n)_n=m+k^infty$, thus part 1 is applicable here since $mle m+k$.

Is this reasoning correct, or have I done something wrong? Thanks in advance.

Edit: I realize now that it was unnecessary to prove that $(a^+_N)_N=m^infty$ in the first bit of my proof of 1, even though it is applicable to the second part. Granted, one could cope with this by using a contradiction like $$exists n'ge m'forall nge m:a_n'<cle a_nimplies exists n'ge m': a_n'<a_n'+1,$$
instead (this should imply that $(a^+_N)_N=m^infty$ is not decreasing).

real-analysis self-learning limsup-and-liminf

share|cite|improve this question

edited Jul 24 at 21:25

asked Jul 24 at 13:50

Crosby

341117

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I came across the following exercise while self-studying Terrence Tao's book Analysis I:

Exercise 6.4.2 Let $(a_n)_n=m^infty$ be an arbitrary sequence of real numbers and let $m'ge m$ and $kge 0$ be integers. Show that $limsup_ntoinfty a_n$ is equivalent to

1. The limit superior of of $(a_n)_n=m'^infty$.


2. The limit superior of $(a_n+k)_n=m^infty$.

My Attempt: Let us first prove 1. Suppose that the limit superior of $(a_n)_n=m^infty$ is equal to some real number $cinBbb R$, i.e. $inf(a_N^+)_N=m^infty = c$ where $(a_N^+)_N=m^infty$ denotes the sequence of all $a_N=sup(a_n)_n=N^infty$. If $M'ge Mge m$ then $a_M'^+le a_M^+$, as if $a_M^+le a_M'^+$ then $$forall nge M: a_nle a_M^+le a_M'^+,$$ and $a_M'^+$ cannot be the lowest upper bound of $(a_n)_n=M'^infty$. Thus, $(a_N^+)_N=m^infty$ is decreasing. To prove that $c$ is a lower bound of $(a^+_N)_N=m'^infty$, suppose that there is some $a_n'<c$ where $n'le m'$. Then $$exists n'ge m'forall nge m:a_n'<cle a_nimplies exists n'ge m': a_n'<a_n',$$ a contradiction. Conversely, suppose that $inf(a^+_N)_N=m'^infty = c$. Then once again, $(a^+_N)_N=m'^infty$ is decreasing and as such, $$forall n'ge m'ge nge m:a_nge a_n'ge c.$$ Hence, $inf(a_N^+)_N=m^infty = c$. For 2, note that $(a_n+k)_n=m^infty$ is precisely the same sequence as $(a_n)_n=m+k^infty$, thus part 1 is applicable here since $mle m+k$.

Is this reasoning correct, or have I done something wrong? Thanks in advance.

Edit: I realize now that it was unnecessary to prove that $(a^+_N)_N=m^infty$ in the first bit of my proof of 1, even though it is applicable to the second part. Granted, one could cope with this by using a contradiction like $$exists n'ge m'forall nge m:a_n'<cle a_nimplies exists n'ge m': a_n'<a_n'+1,$$
instead (this should imply that $(a^+_N)_N=m^infty$ is not decreasing).

real-analysis self-learning limsup-and-liminf

share|cite|improve this question

edited Jul 24 at 21:25

asked Jul 24 at 13:50

Crosby

341117

I came across the following exercise while self-studying Terrence Tao's book Analysis I:

Exercise 6.4.2 Let $(a_n)_n=m^infty$ be an arbitrary sequence of real numbers and let $m'ge m$ and $kge 0$ be integers. Show that $limsup_ntoinfty a_n$ is equivalent to

1. The limit superior of of $(a_n)_n=m'^infty$.


2. The limit superior of $(a_n+k)_n=m^infty$.

My Attempt: Let us first prove 1. Suppose that the limit superior of $(a_n)_n=m^infty$ is equal to some real number $cinBbb R$, i.e. $inf(a_N^+)_N=m^infty = c$ where $(a_N^+)_N=m^infty$ denotes the sequence of all $a_N=sup(a_n)_n=N^infty$. If $M'ge Mge m$ then $a_M'^+le a_M^+$, as if $a_M^+le a_M'^+$ then $$forall nge M: a_nle a_M^+le a_M'^+,$$ and $a_M'^+$ cannot be the lowest upper bound of $(a_n)_n=M'^infty$. Thus, $(a_N^+)_N=m^infty$ is decreasing. To prove that $c$ is a lower bound of $(a^+_N)_N=m'^infty$, suppose that there is some $a_n'<c$ where $n'le m'$. Then $$exists n'ge m'forall nge m:a_n'<cle a_nimplies exists n'ge m': a_n'<a_n',$$ a contradiction. Conversely, suppose that $inf(a^+_N)_N=m'^infty = c$. Then once again, $(a^+_N)_N=m'^infty$ is decreasing and as such, $$forall n'ge m'ge nge m:a_nge a_n'ge c.$$ Hence, $inf(a_N^+)_N=m^infty = c$. For 2, note that $(a_n+k)_n=m^infty$ is precisely the same sequence as $(a_n)_n=m+k^infty$, thus part 1 is applicable here since $mle m+k$.

Is this reasoning correct, or have I done something wrong? Thanks in advance.

Edit: I realize now that it was unnecessary to prove that $(a^+_N)_N=m^infty$ in the first bit of my proof of 1, even though it is applicable to the second part. Granted, one could cope with this by using a contradiction like $$exists n'ge m'forall nge m:a_n'<cle a_nimplies exists n'ge m': a_n'<a_n'+1,$$
instead (this should imply that $(a^+_N)_N=m^infty$ is not decreasing).

real-analysis self-learning limsup-and-liminf

share|cite|improve this question

edited Jul 24 at 21:25

asked Jul 24 at 13:50

Crosby

341117

share|cite|improve this question

share|cite|improve this question

edited Jul 24 at 21:25

edited Jul 24 at 21:25

edited Jul 24 at 21:25

asked Jul 24 at 13:50

Crosby

341117

asked Jul 24 at 13:50

Crosby

341117

asked Jul 24 at 13:50

Crosby

341117

341117

add a comment | 

add a comment | 

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