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Prove the following metric space is complete
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I've been struggling with this question for many, hours.
Even reading the professor's proof outline, and understanding parts of it, I'm struggling with completely reproducing the proof on my own (pun unintended). His outline is in a different language, so I am translating and filling in all the technical details. My goal in this exchange is proof verification.
Set $X= lbrace (a_n)_n=1^infty mid forall n in mathbbN
:a_n in mathbbR rbrace $, the set of all infinite real sequences.
Define a metric $D$ on $X$:
$D((a_n)_n=1^infty,(b_n)_n=1^infty) := undersetnsup lbracebar d(a_n,b_n)/nrbrace$
Where $bar d(a,b):=min lbrace |a-b|,1rbrace $, for any $a,b inmathbbR$.
Prove that $(X,D)$ is a complete metric space.
I know that I need to show that every Cauchy sequence in $(X,D)$ converges in $(X,D)$.
So, given a Cauchy sequence, $bar x_n = lbrace x_n^i rbrace _i=1
^infty$, I imagine setting up an infinite square matrix, where the $n^th$ column is the $n^th$ sequence in $bar x_n$ and the $i^th$ row is a sequence of all the $i^th$ elements in each sequence.
- First, I want to show that every row sequence $lbrace x_n^i rbrace _n=1^infty $, is Cauchy. I'm not sure if I did this correctly.
Let $i in mathbbN$ be arbitrary. Since $bar x_n$ is Cauchy, for every $epsilon > 0$, there exists $Nin mathbbN$ such that for any $m>n>N$:
$D(bar x_n,bar x_m)= undersetisup lbracebar d(x_n^i , x_m^i)/irbrace = undersetisup lbrace min(|x_n^i-x_n^i|,1)/irbrace<epsilon /i$.
It follows that $forall m>n>N: |x^i_n - x^i_m|/i<epsilon /i$.
$implies |x^i_n - x^i_m|<epsilon$, proving that $lbrace x_n^i rbrace _n=1^infty $, is Cauchy in $mathbbR$.
Since $(mathbbR,|cdot|)$ is complete, $forall i in mathbbN, lbrace x_n^i rbrace _n=1^infty oversetn rightarrow inftylongrightarrow L_i < infty$.
$ $
- Next, define $L= lbrace L_i rbrace _i=1^infty$. Showing that $ undersetn rightarrow inftylim bar x_n=L$ completes the proof.
Let $epsilon >0$ be arbitrary. Set $K > epsilon^-1.$ Since $forall i in mathbbN : undersetn rightarrow inftylim x_n^i=L_i $, there exists $N$ such that $forall n>N$:
$forall 1leq i leq K : |x_n^i-L_i| < i epsilon implies |x_n^i-L_i|/ i < epsilon implies underset 1leq i leq Ksup lbrace bar d(x_n^i , L_i)/irbrace < epsilon$
$forall i > K : i > epsilon^-1 implies 1/i < epsilon
implies underseti>Ksup lbrace bar d(x_n^i , L_i)/irbrace < epsilon$
Finally implying $D(lbrace x_n^i rbrace _i=1^infty, lbrace L_i rbrace _i=1^infty) < epsilon implies D(bar x_n, L) < epsilon$
Q.E.D
Wow, this took a very long time to write.
ANY feedback would be appreciated.
real-analysis proof-verification metric-spaces complete-spaces
asked Jul 24 at 16:51
ikoikoia
1399
add a comment |Â
up vote
4
down vote
favorite
I've been struggling with this question for many, hours.
Even reading the professor's proof outline, and understanding parts of it, I'm struggling with completely reproducing the proof on my own (pun unintended). His outline is in a different language, so I am translating and filling in all the technical details. My goal in this exchange is proof verification.
Set $X= lbrace (a_n)_n=1^infty mid forall n in mathbbN
:a_n in mathbbR rbrace $, the set of all infinite real sequences.
Define a metric $D$ on $X$:
$D((a_n)_n=1^infty,(b_n)_n=1^infty) := undersetnsup lbracebar d(a_n,b_n)/nrbrace$
Where $bar d(a,b):=min lbrace |a-b|,1rbrace $, for any $a,b inmathbbR$.
Prove that $(X,D)$ is a complete metric space.
I know that I need to show that every Cauchy sequence in $(X,D)$ converges in $(X,D)$.
So, given a Cauchy sequence, $bar x_n = lbrace x_n^i rbrace _i=1
^infty$, I imagine setting up an infinite square matrix, where the $n^th$ column is the $n^th$ sequence in $bar x_n$ and the $i^th$ row is a sequence of all the $i^th$ elements in each sequence.
- First, I want to show that every row sequence $lbrace x_n^i rbrace _n=1^infty $, is Cauchy. I'm not sure if I did this correctly.
Let $i in mathbbN$ be arbitrary. Since $bar x_n$ is Cauchy, for every $epsilon > 0$, there exists $Nin mathbbN$ such that for any $m>n>N$:
$D(bar x_n,bar x_m)= undersetisup lbracebar d(x_n^i , x_m^i)/irbrace = undersetisup lbrace min(|x_n^i-x_n^i|,1)/irbrace<epsilon /i$.
It follows that $forall m>n>N: |x^i_n - x^i_m|/i<epsilon /i$.
$implies |x^i_n - x^i_m|<epsilon$, proving that $lbrace x_n^i rbrace _n=1^infty $, is Cauchy in $mathbbR$.
Since $(mathbbR,|cdot|)$ is complete, $forall i in mathbbN, lbrace x_n^i rbrace _n=1^infty oversetn rightarrow inftylongrightarrow L_i < infty$.
$ $
- Next, define $L= lbrace L_i rbrace _i=1^infty$. Showing that $ undersetn rightarrow inftylim bar x_n=L$ completes the proof.
Let $epsilon >0$ be arbitrary. Set $K > epsilon^-1.$ Since $forall i in mathbbN : undersetn rightarrow inftylim x_n^i=L_i $, there exists $N$ such that $forall n>N$:
$forall 1leq i leq K : |x_n^i-L_i| < i epsilon implies |x_n^i-L_i|/ i < epsilon implies underset 1leq i leq Ksup lbrace bar d(x_n^i , L_i)/irbrace < epsilon$
$forall i > K : i > epsilon^-1 implies 1/i < epsilon
implies underseti>Ksup lbrace bar d(x_n^i , L_i)/irbrace < epsilon$
Finally implying $D(lbrace x_n^i rbrace _i=1^infty, lbrace L_i rbrace _i=1^infty) < epsilon implies D(bar x_n, L) < epsilon$
Q.E.D
Wow, this took a very long time to write.
ANY feedback would be appreciated.
real-analysis proof-verification metric-spaces complete-spaces
asked Jul 24 at 16:51
ikoikoia
1399
Your proof is okay, just remember not to use $i$ as an index right after you have fixed it as in "Let $iinmathbbN$ be arbitrary...".
– SEBASTIAN VARGAS LOAIZA
Jul 26 at 22:56
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I've been struggling with this question for many, hours.
Even reading the professor's proof outline, and understanding parts of it, I'm struggling with completely reproducing the proof on my own (pun unintended). His outline is in a different language, so I am translating and filling in all the technical details. My goal in this exchange is proof verification.
Set $X= lbrace (a_n)_n=1^infty mid forall n in mathbbN
:a_n in mathbbR rbrace $, the set of all infinite real sequences.
Define a metric $D$ on $X$:
$D((a_n)_n=1^infty,(b_n)_n=1^infty) := undersetnsup lbracebar d(a_n,b_n)/nrbrace$
Where $bar d(a,b):=min lbrace |a-b|,1rbrace $, for any $a,b inmathbbR$.
Prove that $(X,D)$ is a complete metric space.
I know that I need to show that every Cauchy sequence in $(X,D)$ converges in $(X,D)$.
So, given a Cauchy sequence, $bar x_n = lbrace x_n^i rbrace _i=1
^infty$, I imagine setting up an infinite square matrix, where the $n^th$ column is the $n^th$ sequence in $bar x_n$ and the $i^th$ row is a sequence of all the $i^th$ elements in each sequence.
- First, I want to show that every row sequence $lbrace x_n^i rbrace _n=1^infty $, is Cauchy. I'm not sure if I did this correctly.
Let $i in mathbbN$ be arbitrary. Since $bar x_n$ is Cauchy, for every $epsilon > 0$, there exists $Nin mathbbN$ such that for any $m>n>N$:
$D(bar x_n,bar x_m)= undersetisup lbracebar d(x_n^i , x_m^i)/irbrace = undersetisup lbrace min(|x_n^i-x_n^i|,1)/irbrace<epsilon /i$.
It follows that $forall m>n>N: |x^i_n - x^i_m|/i<epsilon /i$.
$implies |x^i_n - x^i_m|<epsilon$, proving that $lbrace x_n^i rbrace _n=1^infty $, is Cauchy in $mathbbR$.
Since $(mathbbR,|cdot|)$ is complete, $forall i in mathbbN, lbrace x_n^i rbrace _n=1^infty oversetn rightarrow inftylongrightarrow L_i < infty$.
$ $
- Next, define $L= lbrace L_i rbrace _i=1^infty$. Showing that $ undersetn rightarrow inftylim bar x_n=L$ completes the proof.
Let $epsilon >0$ be arbitrary. Set $K > epsilon^-1.$ Since $forall i in mathbbN : undersetn rightarrow inftylim x_n^i=L_i $, there exists $N$ such that $forall n>N$:
$forall 1leq i leq K : |x_n^i-L_i| < i epsilon implies |x_n^i-L_i|/ i < epsilon implies underset 1leq i leq Ksup lbrace bar d(x_n^i , L_i)/irbrace < epsilon$
$forall i > K : i > epsilon^-1 implies 1/i < epsilon
implies underseti>Ksup lbrace bar d(x_n^i , L_i)/irbrace < epsilon$
Finally implying $D(lbrace x_n^i rbrace _i=1^infty, lbrace L_i rbrace _i=1^infty) < epsilon implies D(bar x_n, L) < epsilon$
Q.E.D
Wow, this took a very long time to write.
ANY feedback would be appreciated.
real-analysis proof-verification metric-spaces complete-spaces
asked Jul 24 at 16:51
ikoikoia
1399
I've been struggling with this question for many, hours.
Even reading the professor's proof outline, and understanding parts of it, I'm struggling with completely reproducing the proof on my own (pun unintended). His outline is in a different language, so I am translating and filling in all the technical details. My goal in this exchange is proof verification.
Set $X= lbrace (a_n)_n=1^infty mid forall n in mathbbN
:a_n in mathbbR rbrace $, the set of all infinite real sequences.
Define a metric $D$ on $X$:
$D((a_n)_n=1^infty,(b_n)_n=1^infty) := undersetnsup lbracebar d(a_n,b_n)/nrbrace$
Where $bar d(a,b):=min lbrace |a-b|,1rbrace $, for any $a,b inmathbbR$.
Prove that $(X,D)$ is a complete metric space.
I know that I need to show that every Cauchy sequence in $(X,D)$ converges in $(X,D)$.
So, given a Cauchy sequence, $bar x_n = lbrace x_n^i rbrace _i=1
^infty$, I imagine setting up an infinite square matrix, where the $n^th$ column is the $n^th$ sequence in $bar x_n$ and the $i^th$ row is a sequence of all the $i^th$ elements in each sequence.
- First, I want to show that every row sequence $lbrace x_n^i rbrace _n=1^infty $, is Cauchy. I'm not sure if I did this correctly.
Let $i in mathbbN$ be arbitrary. Since $bar x_n$ is Cauchy, for every $epsilon > 0$, there exists $Nin mathbbN$ such that for any $m>n>N$:
$D(bar x_n,bar x_m)= undersetisup lbracebar d(x_n^i , x_m^i)/irbrace = undersetisup lbrace min(|x_n^i-x_n^i|,1)/irbrace<epsilon /i$.
It follows that $forall m>n>N: |x^i_n - x^i_m|/i<epsilon /i$.
$implies |x^i_n - x^i_m|<epsilon$, proving that $lbrace x_n^i rbrace _n=1^infty $, is Cauchy in $mathbbR$.
Since $(mathbbR,|cdot|)$ is complete, $forall i in mathbbN, lbrace x_n^i rbrace _n=1^infty oversetn rightarrow inftylongrightarrow L_i < infty$.
$ $
- Next, define $L= lbrace L_i rbrace _i=1^infty$. Showing that $ undersetn rightarrow inftylim bar x_n=L$ completes the proof.
Let $epsilon >0$ be arbitrary. Set $K > epsilon^-1.$ Since $forall i in mathbbN : undersetn rightarrow inftylim x_n^i=L_i $, there exists $N$ such that $forall n>N$:
$forall 1leq i leq K : |x_n^i-L_i| < i epsilon implies |x_n^i-L_i|/ i < epsilon implies underset 1leq i leq Ksup lbrace bar d(x_n^i , L_i)/irbrace < epsilon$
$forall i > K : i > epsilon^-1 implies 1/i < epsilon
implies underseti>Ksup lbrace bar d(x_n^i , L_i)/irbrace < epsilon$
Finally implying $D(lbrace x_n^i rbrace _i=1^infty, lbrace L_i rbrace _i=1^infty) < epsilon implies D(bar x_n, L) < epsilon$
Q.E.D
Wow, this took a very long time to write.
ANY feedback would be appreciated.
real-analysis proof-verification metric-spaces complete-spaces
asked Jul 24 at 16:51
ikoikoia
1399
edited Jul 25 at 9:41
asked Jul 24 at 16:51
ikoikoia
1399
asked Jul 24 at 16:51
ikoikoia
1399
asked Jul 24 at 16:51
ikoikoia
1399
1399
Your proof is okay, just remember not to use $i$ as an index right after you have fixed it as in "Let $iinmathbbN$ be arbitrary...".
– SEBASTIAN VARGAS LOAIZA
Jul 26 at 22:56
add a comment |Â
Your proof is okay, just remember not to use $i$ as an index right after you have fixed it as in "Let $iinmathbbN$ be arbitrary...".
– SEBASTIAN VARGAS LOAIZA
Jul 26 at 22:56
Your proof is okay, just remember not to use $i$ as an index right after you have fixed it as in "Let $iinmathbbN$ be arbitrary...".
– SEBASTIAN VARGAS LOAIZA
Jul 26 at 22:56
Your proof is okay, just remember not to use $i$ as an index right after you have fixed it as in "Let $iinmathbbN$ be arbitrary...".
– SEBASTIAN VARGAS LOAIZA
Jul 26 at 22:56
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Here's an outline:
- Let $bigl(x(n)bigr)_ninmathbb N$ be a Cauchy sequence of elements of your space. In particular, for each $ninmathbb N$, $x(n)$ is a sequence $bigl(x(n)_kbigr)_kinmathbb N$ of real numbers.
- For each $kinmathbb N$, prove that the sequence $bigl(x(n)_kbigr)_ninmathbb N$ is a Cauchy sequence of real numbers. Let $x_k$ be its limit.
- Prove that $lim_ntoinftyx(n)=(x_k)_kinmathbb N$.
answered Jul 24 at 17:00
José Carlos Santos
113k1697176
Great. I have the same setup as you outlined. I imagine the Cauchy sequence as an infinite matrix. Every column is an infinite series. In step 2 I need to prove every row in itself is a Cauchy sequence. This results in every row converging to xk. The “last†column composed of these limits is the limit sequence in my space. I was hoping for more details on steps 2 and 3 since these are the challenging steps. Maybe if you could explain what D and d intuitively mean, I could do it on my own.
– ikoikoia
Jul 24 at 17:13
The distance $overline d(a,b)$ is the usual distance if the numbers $a$ and $b$ are close to each other and $1$ otherwise. As far as limits (or being a Cauchy sequence) are concerned, this is the same thing as the usual didtane, but it cann never be greater than $1$. And the distance $D$ between two sequences $(a_n)_ninmathbb N$ and $(b_n)_ninmathbb N$ is small if and only if the distance between the first terms of both sequences is small.
– José Carlos Santos
Jul 24 at 17:19
Would the question be very different if d was defined to always be 1?
– ikoikoia
Jul 24 at 17:29
@ikoikoia It sure would. For instance, neither $overline d$ nor $D$ would be metrics.
– José Carlos Santos
Jul 24 at 17:35
How about the other way around? Why is the 1 necessary?
– ikoikoia
Jul 24 at 17:53
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Here's an outline:
- Let $bigl(x(n)bigr)_ninmathbb N$ be a Cauchy sequence of elements of your space. In particular, for each $ninmathbb N$, $x(n)$ is a sequence $bigl(x(n)_kbigr)_kinmathbb N$ of real numbers.
- For each $kinmathbb N$, prove that the sequence $bigl(x(n)_kbigr)_ninmathbb N$ is a Cauchy sequence of real numbers. Let $x_k$ be its limit.
- Prove that $lim_ntoinftyx(n)=(x_k)_kinmathbb N$.
answered Jul 24 at 17:00
José Carlos Santos
113k1697176
Great. I have the same setup as you outlined. I imagine the Cauchy sequence as an infinite matrix. Every column is an infinite series. In step 2 I need to prove every row in itself is a Cauchy sequence. This results in every row converging to xk. The “last†column composed of these limits is the limit sequence in my space. I was hoping for more details on steps 2 and 3 since these are the challenging steps. Maybe if you could explain what D and d intuitively mean, I could do it on my own.
– ikoikoia
Jul 24 at 17:13
The distance $overline d(a,b)$ is the usual distance if the numbers $a$ and $b$ are close to each other and $1$ otherwise. As far as limits (or being a Cauchy sequence) are concerned, this is the same thing as the usual didtane, but it cann never be greater than $1$. And the distance $D$ between two sequences $(a_n)_ninmathbb N$ and $(b_n)_ninmathbb N$ is small if and only if the distance between the first terms of both sequences is small.
– José Carlos Santos
Jul 24 at 17:19
Would the question be very different if d was defined to always be 1?
– ikoikoia
Jul 24 at 17:29
@ikoikoia It sure would. For instance, neither $overline d$ nor $D$ would be metrics.
– José Carlos Santos
Jul 24 at 17:35
How about the other way around? Why is the 1 necessary?
– ikoikoia
Jul 24 at 17:53
 |Â
show 1 more comment
up vote
1
down vote
Here's an outline:
- Let $bigl(x(n)bigr)_ninmathbb N$ be a Cauchy sequence of elements of your space. In particular, for each $ninmathbb N$, $x(n)$ is a sequence $bigl(x(n)_kbigr)_kinmathbb N$ of real numbers.
- For each $kinmathbb N$, prove that the sequence $bigl(x(n)_kbigr)_ninmathbb N$ is a Cauchy sequence of real numbers. Let $x_k$ be its limit.
- Prove that $lim_ntoinftyx(n)=(x_k)_kinmathbb N$.
answered Jul 24 at 17:00
José Carlos Santos
113k1697176
Great. I have the same setup as you outlined. I imagine the Cauchy sequence as an infinite matrix. Every column is an infinite series. In step 2 I need to prove every row in itself is a Cauchy sequence. This results in every row converging to xk. The “last†column composed of these limits is the limit sequence in my space. I was hoping for more details on steps 2 and 3 since these are the challenging steps. Maybe if you could explain what D and d intuitively mean, I could do it on my own.
– ikoikoia
Jul 24 at 17:13
The distance $overline d(a,b)$ is the usual distance if the numbers $a$ and $b$ are close to each other and $1$ otherwise. As far as limits (or being a Cauchy sequence) are concerned, this is the same thing as the usual didtane, but it cann never be greater than $1$. And the distance $D$ between two sequences $(a_n)_ninmathbb N$ and $(b_n)_ninmathbb N$ is small if and only if the distance between the first terms of both sequences is small.
– José Carlos Santos
Jul 24 at 17:19
Would the question be very different if d was defined to always be 1?
– ikoikoia
Jul 24 at 17:29
@ikoikoia It sure would. For instance, neither $overline d$ nor $D$ would be metrics.
– José Carlos Santos
Jul 24 at 17:35
How about the other way around? Why is the 1 necessary?
– ikoikoia
Jul 24 at 17:53
 |Â
show 1 more comment
up vote
1
down vote
up vote
1
down vote
Here's an outline:
- Let $bigl(x(n)bigr)_ninmathbb N$ be a Cauchy sequence of elements of your space. In particular, for each $ninmathbb N$, $x(n)$ is a sequence $bigl(x(n)_kbigr)_kinmathbb N$ of real numbers.
- For each $kinmathbb N$, prove that the sequence $bigl(x(n)_kbigr)_ninmathbb N$ is a Cauchy sequence of real numbers. Let $x_k$ be its limit.
- Prove that $lim_ntoinftyx(n)=(x_k)_kinmathbb N$.
answered Jul 24 at 17:00
José Carlos Santos
113k1697176
Here's an outline:
- Let $bigl(x(n)bigr)_ninmathbb N$ be a Cauchy sequence of elements of your space. In particular, for each $ninmathbb N$, $x(n)$ is a sequence $bigl(x(n)_kbigr)_kinmathbb N$ of real numbers.
- For each $kinmathbb N$, prove that the sequence $bigl(x(n)_kbigr)_ninmathbb N$ is a Cauchy sequence of real numbers. Let $x_k$ be its limit.
- Prove that $lim_ntoinftyx(n)=(x_k)_kinmathbb N$.
answered Jul 24 at 17:00
José Carlos Santos
113k1697176
answered Jul 24 at 17:00
José Carlos Santos
113k1697176
answered Jul 24 at 17:00
José Carlos Santos
113k1697176
answered Jul 24 at 17:00
José Carlos Santos
113k1697176
113k1697176
Great. I have the same setup as you outlined. I imagine the Cauchy sequence as an infinite matrix. Every column is an infinite series. In step 2 I need to prove every row in itself is a Cauchy sequence. This results in every row converging to xk. The “last†column composed of these limits is the limit sequence in my space. I was hoping for more details on steps 2 and 3 since these are the challenging steps. Maybe if you could explain what D and d intuitively mean, I could do it on my own.
– ikoikoia
Jul 24 at 17:13
The distance $overline d(a,b)$ is the usual distance if the numbers $a$ and $b$ are close to each other and $1$ otherwise. As far as limits (or being a Cauchy sequence) are concerned, this is the same thing as the usual didtane, but it cann never be greater than $1$. And the distance $D$ between two sequences $(a_n)_ninmathbb N$ and $(b_n)_ninmathbb N$ is small if and only if the distance between the first terms of both sequences is small.
– José Carlos Santos
Jul 24 at 17:19
Would the question be very different if d was defined to always be 1?
– ikoikoia
Jul 24 at 17:29
@ikoikoia It sure would. For instance, neither $overline d$ nor $D$ would be metrics.
– José Carlos Santos
Jul 24 at 17:35
How about the other way around? Why is the 1 necessary?
– ikoikoia
Jul 24 at 17:53
 |Â
show 1 more comment
Great. I have the same setup as you outlined. I imagine the Cauchy sequence as an infinite matrix. Every column is an infinite series. In step 2 I need to prove every row in itself is a Cauchy sequence. This results in every row converging to xk. The “last†column composed of these limits is the limit sequence in my space. I was hoping for more details on steps 2 and 3 since these are the challenging steps. Maybe if you could explain what D and d intuitively mean, I could do it on my own.
– ikoikoia
Jul 24 at 17:13
The distance $overline d(a,b)$ is the usual distance if the numbers $a$ and $b$ are close to each other and $1$ otherwise. As far as limits (or being a Cauchy sequence) are concerned, this is the same thing as the usual didtane, but it cann never be greater than $1$. And the distance $D$ between two sequences $(a_n)_ninmathbb N$ and $(b_n)_ninmathbb N$ is small if and only if the distance between the first terms of both sequences is small.
– José Carlos Santos
Jul 24 at 17:19
Would the question be very different if d was defined to always be 1?
– ikoikoia
Jul 24 at 17:29
@ikoikoia It sure would. For instance, neither $overline d$ nor $D$ would be metrics.
– José Carlos Santos
Jul 24 at 17:35
How about the other way around? Why is the 1 necessary?
– ikoikoia
Jul 24 at 17:53
Great. I have the same setup as you outlined. I imagine the Cauchy sequence as an infinite matrix. Every column is an infinite series. In step 2 I need to prove every row in itself is a Cauchy sequence. This results in every row converging to xk. The “last†column composed of these limits is the limit sequence in my space. I was hoping for more details on steps 2 and 3 since these are the challenging steps. Maybe if you could explain what D and d intuitively mean, I could do it on my own.
– ikoikoia
Jul 24 at 17:13
Great. I have the same setup as you outlined. I imagine the Cauchy sequence as an infinite matrix. Every column is an infinite series. In step 2 I need to prove every row in itself is a Cauchy sequence. This results in every row converging to xk. The “last†column composed of these limits is the limit sequence in my space. I was hoping for more details on steps 2 and 3 since these are the challenging steps. Maybe if you could explain what D and d intuitively mean, I could do it on my own.
– ikoikoia
Jul 24 at 17:13
The distance $overline d(a,b)$ is the usual distance if the numbers $a$ and $b$ are close to each other and $1$ otherwise. As far as limits (or being a Cauchy sequence) are concerned, this is the same thing as the usual didtane, but it cann never be greater than $1$. And the distance $D$ between two sequences $(a_n)_ninmathbb N$ and $(b_n)_ninmathbb N$ is small if and only if the distance between the first terms of both sequences is small.
– José Carlos Santos
Jul 24 at 17:19
The distance $overline d(a,b)$ is the usual distance if the numbers $a$ and $b$ are close to each other and $1$ otherwise. As far as limits (or being a Cauchy sequence) are concerned, this is the same thing as the usual didtane, but it cann never be greater than $1$. And the distance $D$ between two sequences $(a_n)_ninmathbb N$ and $(b_n)_ninmathbb N$ is small if and only if the distance between the first terms of both sequences is small.
– José Carlos Santos
Jul 24 at 17:19
Would the question be very different if d was defined to always be 1?
– ikoikoia
Jul 24 at 17:29
Would the question be very different if d was defined to always be 1?
– ikoikoia
Jul 24 at 17:29
@ikoikoia It sure would. For instance, neither $overline d$ nor $D$ would be metrics.
– José Carlos Santos
Jul 24 at 17:35
@ikoikoia It sure would. For instance, neither $overline d$ nor $D$ would be metrics.
– José Carlos Santos
Jul 24 at 17:35
How about the other way around? Why is the 1 necessary?
– ikoikoia
Jul 24 at 17:53
How about the other way around? Why is the 1 necessary?
– ikoikoia
Jul 24 at 17:53
 |Â
show 1 more comment
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Your proof is okay, just remember not to use $i$ as an index right after you have fixed it as in "Let $iinmathbbN$ be arbitrary...".
– SEBASTIAN VARGAS LOAIZA
Jul 26 at 22:56