Show that $min M$ is unique












0












$begingroup$


Let $S$ be a subspace of a normed linear space, $X$ and $x_0in Xbackslash S$. Consider the subspace spanned by $M,$ i.e.
begin{align} M:=[Scup {x_0}]={m=x+alpha,x_0:,xin S,;text{for some};alphainBbb{R}} end{align}
I want to show that $m$ is unique.



MY TRIAL



Let $min M,$ then there exists $alphainBbb{R} $ such that $m=x+alpha,x_0.$ Suppose we have another representation, then there exists $betain Bbb{R},;betaneq alpha,$ such that $m=x+beta,x_0.$ Thus,
begin{align} x_0=0 ;text{which implies};x_0in S,;text{contradiction, since }; x_0notin S.end{align}
Please, I'm I right? If not, could you please, provide an alternative proof?






linear-algebra functional-analysis analysis proof-writing proof-explanation







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm not 100% sure I understand the question. As I understand it, you're trying to show that, if you have $x + alpha x_0 = y + beta x_0$, where $x, y in S$, then $x = y$ and $alpha = beta$?
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:01










  • $begingroup$
    @Theo Bendit: That's true! That almost shows that I'm not correct!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:02










  • $begingroup$
    You're almost correct; instead consider $x - y = (beta - alpha)x_0$.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:03










  • $begingroup$
    @Theo Bendit: Okay, let me try that!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:04










  • $begingroup$
    If you get, write an answer. :-)
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:05
















0












$begingroup$


Let $S$ be a subspace of a normed linear space, $X$ and $x_0in Xbackslash S$. Consider the subspace spanned by $M,$ i.e.
begin{align} M:=[Scup {x_0}]={m=x+alpha,x_0:,xin S,;text{for some};alphainBbb{R}} end{align}
I want to show that $m$ is unique.



MY TRIAL



Let $min M,$ then there exists $alphainBbb{R} $ such that $m=x+alpha,x_0.$ Suppose we have another representation, then there exists $betain Bbb{R},;betaneq alpha,$ such that $m=x+beta,x_0.$ Thus,
begin{align} x_0=0 ;text{which implies};x_0in S,;text{contradiction, since }; x_0notin S.end{align}
Please, I'm I right? If not, could you please, provide an alternative proof?






linear-algebra functional-analysis analysis proof-writing proof-explanation







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm not 100% sure I understand the question. As I understand it, you're trying to show that, if you have $x + alpha x_0 = y + beta x_0$, where $x, y in S$, then $x = y$ and $alpha = beta$?
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:01










  • $begingroup$
    @Theo Bendit: That's true! That almost shows that I'm not correct!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:02










  • $begingroup$
    You're almost correct; instead consider $x - y = (beta - alpha)x_0$.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:03










  • $begingroup$
    @Theo Bendit: Okay, let me try that!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:04










  • $begingroup$
    If you get, write an answer. :-)
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:05














0












0








0


1



$begingroup$


Let $S$ be a subspace of a normed linear space, $X$ and $x_0in Xbackslash S$. Consider the subspace spanned by $M,$ i.e.
begin{align} M:=[Scup {x_0}]={m=x+alpha,x_0:,xin S,;text{for some};alphainBbb{R}} end{align}
I want to show that $m$ is unique.



MY TRIAL



Let $min M,$ then there exists $alphainBbb{R} $ such that $m=x+alpha,x_0.$ Suppose we have another representation, then there exists $betain Bbb{R},;betaneq alpha,$ such that $m=x+beta,x_0.$ Thus,
begin{align} x_0=0 ;text{which implies};x_0in S,;text{contradiction, since }; x_0notin S.end{align}
Please, I'm I right? If not, could you please, provide an alternative proof?






linear-algebra functional-analysis analysis proof-writing proof-explanation







share|cite|improve this question











$endgroup$




Let $S$ be a subspace of a normed linear space, $X$ and $x_0in Xbackslash S$. Consider the subspace spanned by $M,$ i.e.
begin{align} M:=[Scup {x_0}]={m=x+alpha,x_0:,xin S,;text{for some};alphainBbb{R}} end{align}
I want to show that $m$ is unique.



MY TRIAL



Let $min M,$ then there exists $alphainBbb{R} $ such that $m=x+alpha,x_0.$ Suppose we have another representation, then there exists $betain Bbb{R},;betaneq alpha,$ such that $m=x+beta,x_0.$ Thus,
begin{align} x_0=0 ;text{which implies};x_0in S,;text{contradiction, since }; x_0notin S.end{align}
Please, I'm I right? If not, could you please, provide an alternative proof?






linear-algebra functional-analysis analysis proof-writing proof-explanation




linear-algebra functional-analysis analysis proof-writing proof-explanation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 7:06







Omojola Micheal

















asked Jan 10 at 4:54









Omojola MichealOmojola Micheal

1,920324




1,920324












  • $begingroup$
    I'm not 100% sure I understand the question. As I understand it, you're trying to show that, if you have $x + alpha x_0 = y + beta x_0$, where $x, y in S$, then $x = y$ and $alpha = beta$?
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:01










  • $begingroup$
    @Theo Bendit: That's true! That almost shows that I'm not correct!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:02










  • $begingroup$
    You're almost correct; instead consider $x - y = (beta - alpha)x_0$.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:03










  • $begingroup$
    @Theo Bendit: Okay, let me try that!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:04










  • $begingroup$
    If you get, write an answer. :-)
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:05


















  • $begingroup$
    I'm not 100% sure I understand the question. As I understand it, you're trying to show that, if you have $x + alpha x_0 = y + beta x_0$, where $x, y in S$, then $x = y$ and $alpha = beta$?
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:01










  • $begingroup$
    @Theo Bendit: That's true! That almost shows that I'm not correct!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:02










  • $begingroup$
    You're almost correct; instead consider $x - y = (beta - alpha)x_0$.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:03










  • $begingroup$
    @Theo Bendit: Okay, let me try that!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:04










  • $begingroup$
    If you get, write an answer. :-)
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:05
















$begingroup$
I'm not 100% sure I understand the question. As I understand it, you're trying to show that, if you have $x + alpha x_0 = y + beta x_0$, where $x, y in S$, then $x = y$ and $alpha = beta$?
$endgroup$
– Theo Bendit
Jan 10 at 5:01




$begingroup$
I'm not 100% sure I understand the question. As I understand it, you're trying to show that, if you have $x + alpha x_0 = y + beta x_0$, where $x, y in S$, then $x = y$ and $alpha = beta$?
$endgroup$
– Theo Bendit
Jan 10 at 5:01












$begingroup$
@Theo Bendit: That's true! That almost shows that I'm not correct!
$endgroup$
– Omojola Micheal
Jan 10 at 5:02




$begingroup$
@Theo Bendit: That's true! That almost shows that I'm not correct!
$endgroup$
– Omojola Micheal
Jan 10 at 5:02












$begingroup$
You're almost correct; instead consider $x - y = (beta - alpha)x_0$.
$endgroup$
– Theo Bendit
Jan 10 at 5:03




$begingroup$
You're almost correct; instead consider $x - y = (beta - alpha)x_0$.
$endgroup$
– Theo Bendit
Jan 10 at 5:03












$begingroup$
@Theo Bendit: Okay, let me try that!
$endgroup$
– Omojola Micheal
Jan 10 at 5:04




$begingroup$
@Theo Bendit: Okay, let me try that!
$endgroup$
– Omojola Micheal
Jan 10 at 5:04












$begingroup$
If you get, write an answer. :-)
$endgroup$
– Theo Bendit
Jan 10 at 5:05




$begingroup$
If you get, write an answer. :-)
$endgroup$
– Theo Bendit
Jan 10 at 5:05










1 Answer
1






active

oldest

votes


















1












$begingroup$

Credits to @Theo Bendit for the hints.



Corrected:



Let $x_0in Xbackslash S$ be arbitrary. Suppose that $min M, $ then there exists $xin S;text{and};alphainBbb{R} $ such that $m=x+alpha,x_0.$ Assume there is another representation, then there exists $yin S;text{and};betain Bbb{R}$ such that $m=y+beta,x_0,$ where $xneq y$ or $alphaneqbeta$. If $xneq y$, then $beta neq alpha.$ Otherwise, if $beta neq alpha$ then $x=y$ or $xneq y.$ In both cases, $beta neq alpha$. Thus,
begin{align} left(x+alpha,x_0=y+beta,x_0right)&implies (x-y)=(beta-alpha),x_0\&implies x_0=dfrac{1}{beta-alpha}(x-y)in S,;text{where};beta-alphaneq 0;text{and};x-yin S,\&implies x_0in S,;text{since };S;text{is a subspace of a normed linear space }\&impliestext{a contradiction }end{align}
Hence, the representation $m=x+alpha x_0,;$ for $;min M$ is unique.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You should make the final implication more clear. What's wrong with $alpha - beta neq 0$? You should be able to use this assumption to conclude $x_0 in S$ (though you won't be able to conclude $x_0 = 0$).
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:17










  • $begingroup$
    @Theo Bendit: Okay, thanks for the guidiance.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:18










  • $begingroup$
    Hint: subspaces are closed under subtraction and scalar multipliaction.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:38










  • $begingroup$
    @Theo Bendit: Okay, thanks, once again, for the guidiance. You have my respect!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:51








  • 1




    $begingroup$
    @Theo Bendit: Thanks a lot for your time and efforts.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 6:11











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Credits to @Theo Bendit for the hints.



Corrected:



Let $x_0in Xbackslash S$ be arbitrary. Suppose that $min M, $ then there exists $xin S;text{and};alphainBbb{R} $ such that $m=x+alpha,x_0.$ Assume there is another representation, then there exists $yin S;text{and};betain Bbb{R}$ such that $m=y+beta,x_0,$ where $xneq y$ or $alphaneqbeta$. If $xneq y$, then $beta neq alpha.$ Otherwise, if $beta neq alpha$ then $x=y$ or $xneq y.$ In both cases, $beta neq alpha$. Thus,
begin{align} left(x+alpha,x_0=y+beta,x_0right)&implies (x-y)=(beta-alpha),x_0\&implies x_0=dfrac{1}{beta-alpha}(x-y)in S,;text{where};beta-alphaneq 0;text{and};x-yin S,\&implies x_0in S,;text{since };S;text{is a subspace of a normed linear space }\&impliestext{a contradiction }end{align}
Hence, the representation $m=x+alpha x_0,;$ for $;min M$ is unique.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You should make the final implication more clear. What's wrong with $alpha - beta neq 0$? You should be able to use this assumption to conclude $x_0 in S$ (though you won't be able to conclude $x_0 = 0$).
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:17










  • $begingroup$
    @Theo Bendit: Okay, thanks for the guidiance.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:18










  • $begingroup$
    Hint: subspaces are closed under subtraction and scalar multipliaction.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:38










  • $begingroup$
    @Theo Bendit: Okay, thanks, once again, for the guidiance. You have my respect!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:51








  • 1




    $begingroup$
    @Theo Bendit: Thanks a lot for your time and efforts.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 6:11
















1












$begingroup$

Credits to @Theo Bendit for the hints.



Corrected:



Let $x_0in Xbackslash S$ be arbitrary. Suppose that $min M, $ then there exists $xin S;text{and};alphainBbb{R} $ such that $m=x+alpha,x_0.$ Assume there is another representation, then there exists $yin S;text{and};betain Bbb{R}$ such that $m=y+beta,x_0,$ where $xneq y$ or $alphaneqbeta$. If $xneq y$, then $beta neq alpha.$ Otherwise, if $beta neq alpha$ then $x=y$ or $xneq y.$ In both cases, $beta neq alpha$. Thus,
begin{align} left(x+alpha,x_0=y+beta,x_0right)&implies (x-y)=(beta-alpha),x_0\&implies x_0=dfrac{1}{beta-alpha}(x-y)in S,;text{where};beta-alphaneq 0;text{and};x-yin S,\&implies x_0in S,;text{since };S;text{is a subspace of a normed linear space }\&impliestext{a contradiction }end{align}
Hence, the representation $m=x+alpha x_0,;$ for $;min M$ is unique.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You should make the final implication more clear. What's wrong with $alpha - beta neq 0$? You should be able to use this assumption to conclude $x_0 in S$ (though you won't be able to conclude $x_0 = 0$).
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:17










  • $begingroup$
    @Theo Bendit: Okay, thanks for the guidiance.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:18










  • $begingroup$
    Hint: subspaces are closed under subtraction and scalar multipliaction.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:38










  • $begingroup$
    @Theo Bendit: Okay, thanks, once again, for the guidiance. You have my respect!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:51








  • 1




    $begingroup$
    @Theo Bendit: Thanks a lot for your time and efforts.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 6:11














1












1








1





$begingroup$

Credits to @Theo Bendit for the hints.



Corrected:



Let $x_0in Xbackslash S$ be arbitrary. Suppose that $min M, $ then there exists $xin S;text{and};alphainBbb{R} $ such that $m=x+alpha,x_0.$ Assume there is another representation, then there exists $yin S;text{and};betain Bbb{R}$ such that $m=y+beta,x_0,$ where $xneq y$ or $alphaneqbeta$. If $xneq y$, then $beta neq alpha.$ Otherwise, if $beta neq alpha$ then $x=y$ or $xneq y.$ In both cases, $beta neq alpha$. Thus,
begin{align} left(x+alpha,x_0=y+beta,x_0right)&implies (x-y)=(beta-alpha),x_0\&implies x_0=dfrac{1}{beta-alpha}(x-y)in S,;text{where};beta-alphaneq 0;text{and};x-yin S,\&implies x_0in S,;text{since };S;text{is a subspace of a normed linear space }\&impliestext{a contradiction }end{align}
Hence, the representation $m=x+alpha x_0,;$ for $;min M$ is unique.






share|cite|improve this answer











$endgroup$



Credits to @Theo Bendit for the hints.



Corrected:



Let $x_0in Xbackslash S$ be arbitrary. Suppose that $min M, $ then there exists $xin S;text{and};alphainBbb{R} $ such that $m=x+alpha,x_0.$ Assume there is another representation, then there exists $yin S;text{and};betain Bbb{R}$ such that $m=y+beta,x_0,$ where $xneq y$ or $alphaneqbeta$. If $xneq y$, then $beta neq alpha.$ Otherwise, if $beta neq alpha$ then $x=y$ or $xneq y.$ In both cases, $beta neq alpha$. Thus,
begin{align} left(x+alpha,x_0=y+beta,x_0right)&implies (x-y)=(beta-alpha),x_0\&implies x_0=dfrac{1}{beta-alpha}(x-y)in S,;text{where};beta-alphaneq 0;text{and};x-yin S,\&implies x_0in S,;text{since };S;text{is a subspace of a normed linear space }\&impliestext{a contradiction }end{align}
Hence, the representation $m=x+alpha x_0,;$ for $;min M$ is unique.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 10 at 6:44

























answered Jan 10 at 5:14









Omojola MichealOmojola Micheal

1,920324




1,920324








  • 1




    $begingroup$
    You should make the final implication more clear. What's wrong with $alpha - beta neq 0$? You should be able to use this assumption to conclude $x_0 in S$ (though you won't be able to conclude $x_0 = 0$).
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:17










  • $begingroup$
    @Theo Bendit: Okay, thanks for the guidiance.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:18










  • $begingroup$
    Hint: subspaces are closed under subtraction and scalar multipliaction.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:38










  • $begingroup$
    @Theo Bendit: Okay, thanks, once again, for the guidiance. You have my respect!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:51








  • 1




    $begingroup$
    @Theo Bendit: Thanks a lot for your time and efforts.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 6:11














  • 1




    $begingroup$
    You should make the final implication more clear. What's wrong with $alpha - beta neq 0$? You should be able to use this assumption to conclude $x_0 in S$ (though you won't be able to conclude $x_0 = 0$).
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:17










  • $begingroup$
    @Theo Bendit: Okay, thanks for the guidiance.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:18










  • $begingroup$
    Hint: subspaces are closed under subtraction and scalar multipliaction.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:38










  • $begingroup$
    @Theo Bendit: Okay, thanks, once again, for the guidiance. You have my respect!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:51








  • 1




    $begingroup$
    @Theo Bendit: Thanks a lot for your time and efforts.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 6:11








1




1




$begingroup$
You should make the final implication more clear. What's wrong with $alpha - beta neq 0$? You should be able to use this assumption to conclude $x_0 in S$ (though you won't be able to conclude $x_0 = 0$).
$endgroup$
– Theo Bendit
Jan 10 at 5:17




$begingroup$
You should make the final implication more clear. What's wrong with $alpha - beta neq 0$? You should be able to use this assumption to conclude $x_0 in S$ (though you won't be able to conclude $x_0 = 0$).
$endgroup$
– Theo Bendit
Jan 10 at 5:17












$begingroup$
@Theo Bendit: Okay, thanks for the guidiance.
$endgroup$
– Omojola Micheal
Jan 10 at 5:18




$begingroup$
@Theo Bendit: Okay, thanks for the guidiance.
$endgroup$
– Omojola Micheal
Jan 10 at 5:18












$begingroup$
Hint: subspaces are closed under subtraction and scalar multipliaction.
$endgroup$
– Theo Bendit
Jan 10 at 5:38




$begingroup$
Hint: subspaces are closed under subtraction and scalar multipliaction.
$endgroup$
– Theo Bendit
Jan 10 at 5:38












$begingroup$
@Theo Bendit: Okay, thanks, once again, for the guidiance. You have my respect!
$endgroup$
– Omojola Micheal
Jan 10 at 5:51






$begingroup$
@Theo Bendit: Okay, thanks, once again, for the guidiance. You have my respect!
$endgroup$
– Omojola Micheal
Jan 10 at 5:51






1




1




$begingroup$
@Theo Bendit: Thanks a lot for your time and efforts.
$endgroup$
– Omojola Micheal
Jan 10 at 6:11




$begingroup$
@Theo Bendit: Thanks a lot for your time and efforts.
$endgroup$
– Omojola Micheal
Jan 10 at 6:11


















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Human spaceflight

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Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
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Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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張江高科駅

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張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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