Proving the statement “$l^1$ is complete.” differently.












0












$begingroup$


I want to prove $l^1$ is complete. I want to prove that any Cauchy sequence $(x_n)$ is Convergent to $(y)$ whose $y_i$ is the limit of the sequence $x_{(n,i)}$ . I want to prove this part first then I would show that $||y||$ is bounded not the other way around.



Can anyone please help me to proceed? I am really having
trouble to visualize that the Cauchy sequence $(x_n)$ converges to $(y)$.






metric-spaces complete-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    if $(x_n)$ is Cauchy, then so are the coordinate sequences $(x^{(j)}_n)$. So if your scalar field is complete, you get a candidate for the limit sequence.
    $endgroup$
    – Matematleta
    Jan 6 at 19:01












  • $begingroup$
    Yea I understood up to the limit point convergence. I have written this in my question. $(y)$ is that sequence. But I have to prove that $(x_n)$ converges to $(y)$.@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 19:04










  • $begingroup$
    Then, I am not sure what you are asking but I have posted a sketch of a proof. If it is not what you are looking for, let me know and I will be happy to delete it.
    $endgroup$
    – Matematleta
    Jan 6 at 19:53












  • $begingroup$
    Please show me how the Cauchy sequence $(x_n)$ converges to $y$.@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:01
















0












$begingroup$


I want to prove $l^1$ is complete. I want to prove that any Cauchy sequence $(x_n)$ is Convergent to $(y)$ whose $y_i$ is the limit of the sequence $x_{(n,i)}$ . I want to prove this part first then I would show that $||y||$ is bounded not the other way around.



Can anyone please help me to proceed? I am really having
trouble to visualize that the Cauchy sequence $(x_n)$ converges to $(y)$.






metric-spaces complete-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    if $(x_n)$ is Cauchy, then so are the coordinate sequences $(x^{(j)}_n)$. So if your scalar field is complete, you get a candidate for the limit sequence.
    $endgroup$
    – Matematleta
    Jan 6 at 19:01












  • $begingroup$
    Yea I understood up to the limit point convergence. I have written this in my question. $(y)$ is that sequence. But I have to prove that $(x_n)$ converges to $(y)$.@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 19:04










  • $begingroup$
    Then, I am not sure what you are asking but I have posted a sketch of a proof. If it is not what you are looking for, let me know and I will be happy to delete it.
    $endgroup$
    – Matematleta
    Jan 6 at 19:53












  • $begingroup$
    Please show me how the Cauchy sequence $(x_n)$ converges to $y$.@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:01














0












0








0





$begingroup$


I want to prove $l^1$ is complete. I want to prove that any Cauchy sequence $(x_n)$ is Convergent to $(y)$ whose $y_i$ is the limit of the sequence $x_{(n,i)}$ . I want to prove this part first then I would show that $||y||$ is bounded not the other way around.



Can anyone please help me to proceed? I am really having
trouble to visualize that the Cauchy sequence $(x_n)$ converges to $(y)$.






metric-spaces complete-spaces







share|cite|improve this question









$endgroup$




I want to prove $l^1$ is complete. I want to prove that any Cauchy sequence $(x_n)$ is Convergent to $(y)$ whose $y_i$ is the limit of the sequence $x_{(n,i)}$ . I want to prove this part first then I would show that $||y||$ is bounded not the other way around.



Can anyone please help me to proceed? I am really having
trouble to visualize that the Cauchy sequence $(x_n)$ converges to $(y)$.






metric-spaces complete-spaces




metric-spaces complete-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 18:51









cmicmi

1,136312




1,136312












  • $begingroup$
    if $(x_n)$ is Cauchy, then so are the coordinate sequences $(x^{(j)}_n)$. So if your scalar field is complete, you get a candidate for the limit sequence.
    $endgroup$
    – Matematleta
    Jan 6 at 19:01












  • $begingroup$
    Yea I understood up to the limit point convergence. I have written this in my question. $(y)$ is that sequence. But I have to prove that $(x_n)$ converges to $(y)$.@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 19:04










  • $begingroup$
    Then, I am not sure what you are asking but I have posted a sketch of a proof. If it is not what you are looking for, let me know and I will be happy to delete it.
    $endgroup$
    – Matematleta
    Jan 6 at 19:53












  • $begingroup$
    Please show me how the Cauchy sequence $(x_n)$ converges to $y$.@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:01


















  • $begingroup$
    if $(x_n)$ is Cauchy, then so are the coordinate sequences $(x^{(j)}_n)$. So if your scalar field is complete, you get a candidate for the limit sequence.
    $endgroup$
    – Matematleta
    Jan 6 at 19:01












  • $begingroup$
    Yea I understood up to the limit point convergence. I have written this in my question. $(y)$ is that sequence. But I have to prove that $(x_n)$ converges to $(y)$.@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 19:04










  • $begingroup$
    Then, I am not sure what you are asking but I have posted a sketch of a proof. If it is not what you are looking for, let me know and I will be happy to delete it.
    $endgroup$
    – Matematleta
    Jan 6 at 19:53












  • $begingroup$
    Please show me how the Cauchy sequence $(x_n)$ converges to $y$.@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:01
















$begingroup$
if $(x_n)$ is Cauchy, then so are the coordinate sequences $(x^{(j)}_n)$. So if your scalar field is complete, you get a candidate for the limit sequence.
$endgroup$
– Matematleta
Jan 6 at 19:01






$begingroup$
if $(x_n)$ is Cauchy, then so are the coordinate sequences $(x^{(j)}_n)$. So if your scalar field is complete, you get a candidate for the limit sequence.
$endgroup$
– Matematleta
Jan 6 at 19:01














$begingroup$
Yea I understood up to the limit point convergence. I have written this in my question. $(y)$ is that sequence. But I have to prove that $(x_n)$ converges to $(y)$.@Matematleta
$endgroup$
– cmi
Jan 6 at 19:04




$begingroup$
Yea I understood up to the limit point convergence. I have written this in my question. $(y)$ is that sequence. But I have to prove that $(x_n)$ converges to $(y)$.@Matematleta
$endgroup$
– cmi
Jan 6 at 19:04












$begingroup$
Then, I am not sure what you are asking but I have posted a sketch of a proof. If it is not what you are looking for, let me know and I will be happy to delete it.
$endgroup$
– Matematleta
Jan 6 at 19:53






$begingroup$
Then, I am not sure what you are asking but I have posted a sketch of a proof. If it is not what you are looking for, let me know and I will be happy to delete it.
$endgroup$
– Matematleta
Jan 6 at 19:53














$begingroup$
Please show me how the Cauchy sequence $(x_n)$ converges to $y$.@Matematleta
$endgroup$
– cmi
Jan 6 at 20:01




$begingroup$
Please show me how the Cauchy sequence $(x_n)$ converges to $y$.@Matematleta
$endgroup$
– cmi
Jan 6 at 20:01










1 Answer
1






active

oldest

votes


















1












$begingroup$

Sketch: picking up from our comments, we have $x^{(j)}_nto y_j$ for each $jin mathbb N$.



Step $1$: show that $y=(y_j)in ell^1:$



$|y|:=lim_{Jto infty}sum^J_{j=1}|y_j|=lim_{Jto infty}(sum^J_{j=1}lim_{nto infty}|x^{(j)}_n|)=lim_{Jto infty}lim_{nto infty}(sum^J_{j=1}|x^{(j)}_n|).$



Now, $sum^J_{j=1}|x^{(j)}_n|le sum^{infty}_{j=1}|x^{(j)}_n|=|x_n|le M<infty $ because the original sequence $(x_n)$ is Cauchy, hence bounded. So,



$|y|le lim_{nto infty}M=M.$



Step $2:$ Prove that $|x_n-y|to 0:$



Let $epsilon>0$ and choose an integer $N$ such that $l,n>NRightarrow |x_l-x_n|<epsilon.$



Then, $sum^J_{j=1}|x_n^{(j)}-x_l^{(j)}|le sum^{infty}_{j=1}|x_n^{(j)}-x_l^{(j)}|=|x_n-x_l|<epsilon.$



I'll leave the rest to you: keep $n>N$, fix $J$ and let $lto infty.$ Finally, let $Jto infty.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have problem in the second part. Please complete it. @Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:00












  • $begingroup$
    And I have seen this proof earlier. It is kind of escaping proof. It can not be visualized.@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:05










  • $begingroup$
    The Cauchy criterion? What? All I am doing is proving that the logical candidate for the limit of the $(x_n)$ is indeed the limit. To do this, you need to show two things 1). it's in $ell^1$. and 2) the sequence $(x_n)$ actually converges to it. I think my sketch is a good roadmap. Suggestion: write the sequence $(x_n)$ in matrix form to see what is going on. The fact that $(x_n)$ is Cauchy in each coordinate is really all you need to know.
    $endgroup$
    – Matematleta
    Jan 6 at 20:06












  • $begingroup$
    without using limit can you show it ?@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:43






  • 1




    $begingroup$
    Your answer helped me to understand what I wanted to understand.@Matematleta
    $endgroup$
    – cmi
    Jan 9 at 4:14











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

Sketch: picking up from our comments, we have $x^{(j)}_nto y_j$ for each $jin mathbb N$.



Step $1$: show that $y=(y_j)in ell^1:$



$|y|:=lim_{Jto infty}sum^J_{j=1}|y_j|=lim_{Jto infty}(sum^J_{j=1}lim_{nto infty}|x^{(j)}_n|)=lim_{Jto infty}lim_{nto infty}(sum^J_{j=1}|x^{(j)}_n|).$



Now, $sum^J_{j=1}|x^{(j)}_n|le sum^{infty}_{j=1}|x^{(j)}_n|=|x_n|le M<infty $ because the original sequence $(x_n)$ is Cauchy, hence bounded. So,



$|y|le lim_{nto infty}M=M.$



Step $2:$ Prove that $|x_n-y|to 0:$



Let $epsilon>0$ and choose an integer $N$ such that $l,n>NRightarrow |x_l-x_n|<epsilon.$



Then, $sum^J_{j=1}|x_n^{(j)}-x_l^{(j)}|le sum^{infty}_{j=1}|x_n^{(j)}-x_l^{(j)}|=|x_n-x_l|<epsilon.$



I'll leave the rest to you: keep $n>N$, fix $J$ and let $lto infty.$ Finally, let $Jto infty.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have problem in the second part. Please complete it. @Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:00












  • $begingroup$
    And I have seen this proof earlier. It is kind of escaping proof. It can not be visualized.@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:05










  • $begingroup$
    The Cauchy criterion? What? All I am doing is proving that the logical candidate for the limit of the $(x_n)$ is indeed the limit. To do this, you need to show two things 1). it's in $ell^1$. and 2) the sequence $(x_n)$ actually converges to it. I think my sketch is a good roadmap. Suggestion: write the sequence $(x_n)$ in matrix form to see what is going on. The fact that $(x_n)$ is Cauchy in each coordinate is really all you need to know.
    $endgroup$
    – Matematleta
    Jan 6 at 20:06












  • $begingroup$
    without using limit can you show it ?@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:43






  • 1




    $begingroup$
    Your answer helped me to understand what I wanted to understand.@Matematleta
    $endgroup$
    – cmi
    Jan 9 at 4:14
















1












$begingroup$

Sketch: picking up from our comments, we have $x^{(j)}_nto y_j$ for each $jin mathbb N$.



Step $1$: show that $y=(y_j)in ell^1:$



$|y|:=lim_{Jto infty}sum^J_{j=1}|y_j|=lim_{Jto infty}(sum^J_{j=1}lim_{nto infty}|x^{(j)}_n|)=lim_{Jto infty}lim_{nto infty}(sum^J_{j=1}|x^{(j)}_n|).$



Now, $sum^J_{j=1}|x^{(j)}_n|le sum^{infty}_{j=1}|x^{(j)}_n|=|x_n|le M<infty $ because the original sequence $(x_n)$ is Cauchy, hence bounded. So,



$|y|le lim_{nto infty}M=M.$



Step $2:$ Prove that $|x_n-y|to 0:$



Let $epsilon>0$ and choose an integer $N$ such that $l,n>NRightarrow |x_l-x_n|<epsilon.$



Then, $sum^J_{j=1}|x_n^{(j)}-x_l^{(j)}|le sum^{infty}_{j=1}|x_n^{(j)}-x_l^{(j)}|=|x_n-x_l|<epsilon.$



I'll leave the rest to you: keep $n>N$, fix $J$ and let $lto infty.$ Finally, let $Jto infty.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have problem in the second part. Please complete it. @Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:00












  • $begingroup$
    And I have seen this proof earlier. It is kind of escaping proof. It can not be visualized.@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:05










  • $begingroup$
    The Cauchy criterion? What? All I am doing is proving that the logical candidate for the limit of the $(x_n)$ is indeed the limit. To do this, you need to show two things 1). it's in $ell^1$. and 2) the sequence $(x_n)$ actually converges to it. I think my sketch is a good roadmap. Suggestion: write the sequence $(x_n)$ in matrix form to see what is going on. The fact that $(x_n)$ is Cauchy in each coordinate is really all you need to know.
    $endgroup$
    – Matematleta
    Jan 6 at 20:06












  • $begingroup$
    without using limit can you show it ?@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:43






  • 1




    $begingroup$
    Your answer helped me to understand what I wanted to understand.@Matematleta
    $endgroup$
    – cmi
    Jan 9 at 4:14














1












1








1





$begingroup$

Sketch: picking up from our comments, we have $x^{(j)}_nto y_j$ for each $jin mathbb N$.



Step $1$: show that $y=(y_j)in ell^1:$



$|y|:=lim_{Jto infty}sum^J_{j=1}|y_j|=lim_{Jto infty}(sum^J_{j=1}lim_{nto infty}|x^{(j)}_n|)=lim_{Jto infty}lim_{nto infty}(sum^J_{j=1}|x^{(j)}_n|).$



Now, $sum^J_{j=1}|x^{(j)}_n|le sum^{infty}_{j=1}|x^{(j)}_n|=|x_n|le M<infty $ because the original sequence $(x_n)$ is Cauchy, hence bounded. So,



$|y|le lim_{nto infty}M=M.$



Step $2:$ Prove that $|x_n-y|to 0:$



Let $epsilon>0$ and choose an integer $N$ such that $l,n>NRightarrow |x_l-x_n|<epsilon.$



Then, $sum^J_{j=1}|x_n^{(j)}-x_l^{(j)}|le sum^{infty}_{j=1}|x_n^{(j)}-x_l^{(j)}|=|x_n-x_l|<epsilon.$



I'll leave the rest to you: keep $n>N$, fix $J$ and let $lto infty.$ Finally, let $Jto infty.$






share|cite|improve this answer









$endgroup$



Sketch: picking up from our comments, we have $x^{(j)}_nto y_j$ for each $jin mathbb N$.



Step $1$: show that $y=(y_j)in ell^1:$



$|y|:=lim_{Jto infty}sum^J_{j=1}|y_j|=lim_{Jto infty}(sum^J_{j=1}lim_{nto infty}|x^{(j)}_n|)=lim_{Jto infty}lim_{nto infty}(sum^J_{j=1}|x^{(j)}_n|).$



Now, $sum^J_{j=1}|x^{(j)}_n|le sum^{infty}_{j=1}|x^{(j)}_n|=|x_n|le M<infty $ because the original sequence $(x_n)$ is Cauchy, hence bounded. So,



$|y|le lim_{nto infty}M=M.$



Step $2:$ Prove that $|x_n-y|to 0:$



Let $epsilon>0$ and choose an integer $N$ such that $l,n>NRightarrow |x_l-x_n|<epsilon.$



Then, $sum^J_{j=1}|x_n^{(j)}-x_l^{(j)}|le sum^{infty}_{j=1}|x_n^{(j)}-x_l^{(j)}|=|x_n-x_l|<epsilon.$



I'll leave the rest to you: keep $n>N$, fix $J$ and let $lto infty.$ Finally, let $Jto infty.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 6 at 19:52









MatematletaMatematleta

11.6k2920




11.6k2920












  • $begingroup$
    I have problem in the second part. Please complete it. @Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:00












  • $begingroup$
    And I have seen this proof earlier. It is kind of escaping proof. It can not be visualized.@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:05










  • $begingroup$
    The Cauchy criterion? What? All I am doing is proving that the logical candidate for the limit of the $(x_n)$ is indeed the limit. To do this, you need to show two things 1). it's in $ell^1$. and 2) the sequence $(x_n)$ actually converges to it. I think my sketch is a good roadmap. Suggestion: write the sequence $(x_n)$ in matrix form to see what is going on. The fact that $(x_n)$ is Cauchy in each coordinate is really all you need to know.
    $endgroup$
    – Matematleta
    Jan 6 at 20:06












  • $begingroup$
    without using limit can you show it ?@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:43






  • 1




    $begingroup$
    Your answer helped me to understand what I wanted to understand.@Matematleta
    $endgroup$
    – cmi
    Jan 9 at 4:14


















  • $begingroup$
    I have problem in the second part. Please complete it. @Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:00












  • $begingroup$
    And I have seen this proof earlier. It is kind of escaping proof. It can not be visualized.@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:05










  • $begingroup$
    The Cauchy criterion? What? All I am doing is proving that the logical candidate for the limit of the $(x_n)$ is indeed the limit. To do this, you need to show two things 1). it's in $ell^1$. and 2) the sequence $(x_n)$ actually converges to it. I think my sketch is a good roadmap. Suggestion: write the sequence $(x_n)$ in matrix form to see what is going on. The fact that $(x_n)$ is Cauchy in each coordinate is really all you need to know.
    $endgroup$
    – Matematleta
    Jan 6 at 20:06












  • $begingroup$
    without using limit can you show it ?@Matematleta
    $endgroup$
    – cmi
    Jan 6 at 20:43






  • 1




    $begingroup$
    Your answer helped me to understand what I wanted to understand.@Matematleta
    $endgroup$
    – cmi
    Jan 9 at 4:14
















$begingroup$
I have problem in the second part. Please complete it. @Matematleta
$endgroup$
– cmi
Jan 6 at 20:00






$begingroup$
I have problem in the second part. Please complete it. @Matematleta
$endgroup$
– cmi
Jan 6 at 20:00














$begingroup$
And I have seen this proof earlier. It is kind of escaping proof. It can not be visualized.@Matematleta
$endgroup$
– cmi
Jan 6 at 20:05




$begingroup$
And I have seen this proof earlier. It is kind of escaping proof. It can not be visualized.@Matematleta
$endgroup$
– cmi
Jan 6 at 20:05












$begingroup$
The Cauchy criterion? What? All I am doing is proving that the logical candidate for the limit of the $(x_n)$ is indeed the limit. To do this, you need to show two things 1). it's in $ell^1$. and 2) the sequence $(x_n)$ actually converges to it. I think my sketch is a good roadmap. Suggestion: write the sequence $(x_n)$ in matrix form to see what is going on. The fact that $(x_n)$ is Cauchy in each coordinate is really all you need to know.
$endgroup$
– Matematleta
Jan 6 at 20:06






$begingroup$
The Cauchy criterion? What? All I am doing is proving that the logical candidate for the limit of the $(x_n)$ is indeed the limit. To do this, you need to show two things 1). it's in $ell^1$. and 2) the sequence $(x_n)$ actually converges to it. I think my sketch is a good roadmap. Suggestion: write the sequence $(x_n)$ in matrix form to see what is going on. The fact that $(x_n)$ is Cauchy in each coordinate is really all you need to know.
$endgroup$
– Matematleta
Jan 6 at 20:06














$begingroup$
without using limit can you show it ?@Matematleta
$endgroup$
– cmi
Jan 6 at 20:43




$begingroup$
without using limit can you show it ?@Matematleta
$endgroup$
– cmi
Jan 6 at 20:43




1




1




$begingroup$
Your answer helped me to understand what I wanted to understand.@Matematleta
$endgroup$
– cmi
Jan 9 at 4:14




$begingroup$
Your answer helped me to understand what I wanted to understand.@Matematleta
$endgroup$
– cmi
Jan 9 at 4:14


















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Måne

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För jordens måne, se Månen. Några månar i solsystemet i skala till vår måne En måne , naturlig satellit eller drabant är en himlakropp som kretsar kring en planet eller en asteroid i ett solsystem. Ordet i bestämd form – månen – avser planeten jordens måne. Det finns inte i något fall belagt att någon måne har egna månar. De flesta stora månar i solsystemet är bundna i synkron rotation med en och samma sida vänd mot planeten. Ett exempel på en måne som inte beter sig på detta vis, är Hyperion som kaotiskt roterar kring Saturnus. En asteroidmåne är en måne som kretsar kring en asteroid. En måne kan ha bildats samtidigt som moderplaneten eller senare infångats av planetens gravitationskraft. I vårt eget solsystem känner vi i nuläget till 173 månar runt sex av de åtta planeterna [ 1 ] : Jorden – månen Mars – Phobos och Deimos Jupiter – Ganymedes (som är solsystemets största måne), Io, Europa, Callisto och 59 andra kända månar Saturnus – Titan (den enda måne...
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Storängen

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För stadsdelen i Mariehamn, se Storängen, Åland. För ett naturreservat i Norrtälje kommun, se Storängens naturreservat. Storängen från luften i september 1967, vy mot väst. Till vänster syns Järlasjön och Saltsjöbanans spår. Till höger löper Värmdövägen och längst ut till höger syns anläggningsarbetena för Värmdöleden. Storängen är ett villaområde i centrala Nacka i Stockholms län. Området är beläget strax sydost om Nacka Forum vid Saltsjöbanan mellan Lillängen och Saltsjö-Duvnäs. Storängens villasamhälle grundades 1904 av "Tjänstemännens Egnahemsförening vid Storängen" och de första villorna började byggas samma år. Redan 1909 var 107 tomter bebyggda av totalt runt 160. [ 1 ] År 2013 fanns drygt 150 villor med lika många familjer i området. Egnahemsföreningen existerar fortfarande och fungerar idag som intresseförening för de boende i Storängen. Storängen anses vara ett av Sveriges bäst bevarade villasamhällen från 1900-talets början, tack vare sina tidstypis...
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VLT Carioca

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VLT Carioca Informações Proprietário Prefeitura do Rio de Janeiro Local [[Rio de Janeiro, RJ]] País {{country data Brasil| country flag2| name = Brasil| variant = | size = }} Tipo de transporte Bonde (Veículo leve sobre trilhos) Número de linhas 2 (+1 em construção) Número de estações 26 (+3 em construção) Tráfego 80 mil passageiros/dia [ 1 ] (dezembro/2018) Website www.vltrio.com.br Funcionamento Início de funcionamento 5 de junho de 2016 (2 anos) Operadora(s) Concessionária do VLT Carioca S.A. Número de veículos 32 Alstom Citadis Comprimento dos veículos 44 m (144 ft) Dados técnicos Extensão do sistema 28 km (17,4 mi) [ 2 ] Frequência 5–10 minutos Bitola Bitola padrão/internacional 1 435 mm (4,71 ft) Eletrificação APS, 750V DC Velocidade média 15 km/h (9,32 mph) [ 2 ] Velocidade máxima 50 km/h (31,1 mph) ...
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