Proving that $g(x)$ defined as $x^2$ on the rationals and $x^4$ on irrationals, is discontinuous at $2$












1












$begingroup$


Define $g: mathbb{R} to mathbb{R}$ by $$g(x) =begin{cases} x^2 & text{if } x text{ is rational}, \\ x^4 & text{if } x text{ is irrational}. end{cases}$$ Prove that $g$ is discontinuous at $x = 2$.



Solution Attempt:



Assume instead that $g$ is continuous at 2. Then since $g(2)= 4$, $;;displaystyle lim_{xto2}g(x)=4$;



so taking $epsilon=1$, there is a $delta>0$ such that if $0<|x-2|<delta$, then $big|g(x)-4big|<1$.



Therefore if $x$ is irrational and $2-delta<x<2$, then
$big|x^4-4big|<1$



$-1 <x^4-4<1implies 3<x^4<5implies x^4>3 implies x = 3^frac{1}{4} implies x < 2$ Contradiction






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  • 1




    $begingroup$
    Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
    $endgroup$
    – Julian Mejia
    2 days ago








  • 1




    $begingroup$
    $x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
    $endgroup$
    – Julian Mejia
    2 days ago
















1












$begingroup$


Define $g: mathbb{R} to mathbb{R}$ by $$g(x) =begin{cases} x^2 & text{if } x text{ is rational}, \\ x^4 & text{if } x text{ is irrational}. end{cases}$$ Prove that $g$ is discontinuous at $x = 2$.



Solution Attempt:



Assume instead that $g$ is continuous at 2. Then since $g(2)= 4$, $;;displaystyle lim_{xto2}g(x)=4$;



so taking $epsilon=1$, there is a $delta>0$ such that if $0<|x-2|<delta$, then $big|g(x)-4big|<1$.



Therefore if $x$ is irrational and $2-delta<x<2$, then
$big|x^4-4big|<1$



$-1 <x^4-4<1implies 3<x^4<5implies x^4>3 implies x = 3^frac{1}{4} implies x < 2$ Contradiction






real-analysis







share|cite|improve this question









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SHajs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    $begingroup$
    Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
    $endgroup$
    – Julian Mejia
    2 days ago








  • 1




    $begingroup$
    $x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
    $endgroup$
    – Julian Mejia
    2 days ago














1












1








1





$begingroup$


Define $g: mathbb{R} to mathbb{R}$ by $$g(x) =begin{cases} x^2 & text{if } x text{ is rational}, \\ x^4 & text{if } x text{ is irrational}. end{cases}$$ Prove that $g$ is discontinuous at $x = 2$.



Solution Attempt:



Assume instead that $g$ is continuous at 2. Then since $g(2)= 4$, $;;displaystyle lim_{xto2}g(x)=4$;



so taking $epsilon=1$, there is a $delta>0$ such that if $0<|x-2|<delta$, then $big|g(x)-4big|<1$.



Therefore if $x$ is irrational and $2-delta<x<2$, then
$big|x^4-4big|<1$



$-1 <x^4-4<1implies 3<x^4<5implies x^4>3 implies x = 3^frac{1}{4} implies x < 2$ Contradiction






real-analysis







share|cite|improve this question









New contributor




SHajs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$




Define $g: mathbb{R} to mathbb{R}$ by $$g(x) =begin{cases} x^2 & text{if } x text{ is rational}, \\ x^4 & text{if } x text{ is irrational}. end{cases}$$ Prove that $g$ is discontinuous at $x = 2$.



Solution Attempt:



Assume instead that $g$ is continuous at 2. Then since $g(2)= 4$, $;;displaystyle lim_{xto2}g(x)=4$;



so taking $epsilon=1$, there is a $delta>0$ such that if $0<|x-2|<delta$, then $big|g(x)-4big|<1$.



Therefore if $x$ is irrational and $2-delta<x<2$, then
$big|x^4-4big|<1$



$-1 <x^4-4<1implies 3<x^4<5implies x^4>3 implies x = 3^frac{1}{4} implies x < 2$ Contradiction






real-analysis




real-analysis






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edited 2 days ago









Asaf Karagila♦

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  • 1




    $begingroup$
    Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
    $endgroup$
    – Julian Mejia
    2 days ago








  • 1




    $begingroup$
    $x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
    $endgroup$
    – Julian Mejia
    2 days ago














  • 1




    $begingroup$
    Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
    $endgroup$
    – Julian Mejia
    2 days ago








  • 1




    $begingroup$
    $x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
    $endgroup$
    – Julian Mejia
    2 days ago








1




1




$begingroup$
Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
$endgroup$
– Julian Mejia
2 days ago






$begingroup$
Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
$endgroup$
– Julian Mejia
2 days ago






1




1




$begingroup$
$x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
$endgroup$
– Julian Mejia
2 days ago




$begingroup$
$x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
$endgroup$
– Julian Mejia
2 days ago










4 Answers
4






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$begingroup$

Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
$$-sqrt[4]{5} < x < sqrt[4]{5}.$$
If you had chosen instead an irrational $x$ such that
$$2 - min{delta, 2 - sqrt[4]{5}} < x < 2,$$
then you'd have $x > 2 - (2 - sqrt[4]{5}) = sqrt[4]{5}$, a contradiction.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    I think a simpe way is to construct two sequences $a_n$ and $b_n$, where $a_n$ are rational and $b_n$ are irrational, such that $lim_{ntoinfty}g(a_n)=2^2$ and $lim_{ntoinfty}g(b_n)=2^4$.



    Then we can say $lim_{xto2^+}g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.




    Put $a_n = 2 + frac{1}{n}$ and $b_n=2+frac{pi}{n}$.







    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      An option?



      Sequential definition of continuity .



      $x_n=2(4n)^{(1/(4n))}.$



      $lim_{n rightarrow infty}x_n=2.$



      $lim_{n rightarrow infty}g(x_n)=$



      $lim_{n rightarrow infty} 2^4(4n)^{(1/n)}=$



      $lim_{n rightarrow infty}2^4 cdot 4^{(1/n)} cdot n^{(1/n)}=$



      $=2^4 not =g(2)$.






      share|cite|improve this answer











      $endgroup$





















        0












        $begingroup$

        You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take



        $$left{a_n:=2pifrac n{pi n+1}right}_{ninBbb N}$$



        Observe the above sequence is irrational:



        $$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,text{so};;lim_{ntoinfty}g(a_n)=lim_{ntoinfty}16pi^4frac{n^4}{(npi+1)^4}=16pi^4frac1{pi^4}=16$$



        yet $;g(2)=2^2=4;$ , so the function is not continuous at $;2;$ ...






        share|cite|improve this answer









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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
          $$-sqrt[4]{5} < x < sqrt[4]{5}.$$
          If you had chosen instead an irrational $x$ such that
          $$2 - min{delta, 2 - sqrt[4]{5}} < x < 2,$$
          then you'd have $x > 2 - (2 - sqrt[4]{5}) = sqrt[4]{5}$, a contradiction.






          share|cite|improve this answer









          $endgroup$


















            1












            $begingroup$

            Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
            $$-sqrt[4]{5} < x < sqrt[4]{5}.$$
            If you had chosen instead an irrational $x$ such that
            $$2 - min{delta, 2 - sqrt[4]{5}} < x < 2,$$
            then you'd have $x > 2 - (2 - sqrt[4]{5}) = sqrt[4]{5}$, a contradiction.






            share|cite|improve this answer









            $endgroup$
















              1












              1








              1





              $begingroup$

              Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
              $$-sqrt[4]{5} < x < sqrt[4]{5}.$$
              If you had chosen instead an irrational $x$ such that
              $$2 - min{delta, 2 - sqrt[4]{5}} < x < 2,$$
              then you'd have $x > 2 - (2 - sqrt[4]{5}) = sqrt[4]{5}$, a contradiction.






              share|cite|improve this answer









              $endgroup$



              Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
              $$-sqrt[4]{5} < x < sqrt[4]{5}.$$
              If you had chosen instead an irrational $x$ such that
              $$2 - min{delta, 2 - sqrt[4]{5}} < x < 2,$$
              then you'd have $x > 2 - (2 - sqrt[4]{5}) = sqrt[4]{5}$, a contradiction.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 2 days ago









              Theo BenditTheo Bendit

              20.8k12355




              20.8k12355























                  1












                  $begingroup$

                  I think a simpe way is to construct two sequences $a_n$ and $b_n$, where $a_n$ are rational and $b_n$ are irrational, such that $lim_{ntoinfty}g(a_n)=2^2$ and $lim_{ntoinfty}g(b_n)=2^4$.



                  Then we can say $lim_{xto2^+}g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.




                  Put $a_n = 2 + frac{1}{n}$ and $b_n=2+frac{pi}{n}$.







                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    I think a simpe way is to construct two sequences $a_n$ and $b_n$, where $a_n$ are rational and $b_n$ are irrational, such that $lim_{ntoinfty}g(a_n)=2^2$ and $lim_{ntoinfty}g(b_n)=2^4$.



                    Then we can say $lim_{xto2^+}g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.




                    Put $a_n = 2 + frac{1}{n}$ and $b_n=2+frac{pi}{n}$.







                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      I think a simpe way is to construct two sequences $a_n$ and $b_n$, where $a_n$ are rational and $b_n$ are irrational, such that $lim_{ntoinfty}g(a_n)=2^2$ and $lim_{ntoinfty}g(b_n)=2^4$.



                      Then we can say $lim_{xto2^+}g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.




                      Put $a_n = 2 + frac{1}{n}$ and $b_n=2+frac{pi}{n}$.







                      share|cite|improve this answer









                      $endgroup$



                      I think a simpe way is to construct two sequences $a_n$ and $b_n$, where $a_n$ are rational and $b_n$ are irrational, such that $lim_{ntoinfty}g(a_n)=2^2$ and $lim_{ntoinfty}g(b_n)=2^4$.



                      Then we can say $lim_{xto2^+}g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.




                      Put $a_n = 2 + frac{1}{n}$ and $b_n=2+frac{pi}{n}$.








                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 2 days ago









                      zongxiang yizongxiang yi

                      352110




                      352110























                          1












                          $begingroup$

                          An option?



                          Sequential definition of continuity .



                          $x_n=2(4n)^{(1/(4n))}.$



                          $lim_{n rightarrow infty}x_n=2.$



                          $lim_{n rightarrow infty}g(x_n)=$



                          $lim_{n rightarrow infty} 2^4(4n)^{(1/n)}=$



                          $lim_{n rightarrow infty}2^4 cdot 4^{(1/n)} cdot n^{(1/n)}=$



                          $=2^4 not =g(2)$.






                          share|cite|improve this answer











                          $endgroup$


















                            1












                            $begingroup$

                            An option?



                            Sequential definition of continuity .



                            $x_n=2(4n)^{(1/(4n))}.$



                            $lim_{n rightarrow infty}x_n=2.$



                            $lim_{n rightarrow infty}g(x_n)=$



                            $lim_{n rightarrow infty} 2^4(4n)^{(1/n)}=$



                            $lim_{n rightarrow infty}2^4 cdot 4^{(1/n)} cdot n^{(1/n)}=$



                            $=2^4 not =g(2)$.






                            share|cite|improve this answer











                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              An option?



                              Sequential definition of continuity .



                              $x_n=2(4n)^{(1/(4n))}.$



                              $lim_{n rightarrow infty}x_n=2.$



                              $lim_{n rightarrow infty}g(x_n)=$



                              $lim_{n rightarrow infty} 2^4(4n)^{(1/n)}=$



                              $lim_{n rightarrow infty}2^4 cdot 4^{(1/n)} cdot n^{(1/n)}=$



                              $=2^4 not =g(2)$.






                              share|cite|improve this answer











                              $endgroup$



                              An option?



                              Sequential definition of continuity .



                              $x_n=2(4n)^{(1/(4n))}.$



                              $lim_{n rightarrow infty}x_n=2.$



                              $lim_{n rightarrow infty}g(x_n)=$



                              $lim_{n rightarrow infty} 2^4(4n)^{(1/n)}=$



                              $lim_{n rightarrow infty}2^4 cdot 4^{(1/n)} cdot n^{(1/n)}=$



                              $=2^4 not =g(2)$.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited 2 days ago

























                              answered 2 days ago









                              Peter SzilasPeter Szilas

                              11.9k2822




                              11.9k2822























                                  0












                                  $begingroup$

                                  You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take



                                  $$left{a_n:=2pifrac n{pi n+1}right}_{ninBbb N}$$



                                  Observe the above sequence is irrational:



                                  $$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,text{so};;lim_{ntoinfty}g(a_n)=lim_{ntoinfty}16pi^4frac{n^4}{(npi+1)^4}=16pi^4frac1{pi^4}=16$$



                                  yet $;g(2)=2^2=4;$ , so the function is not continuous at $;2;$ ...






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take



                                    $$left{a_n:=2pifrac n{pi n+1}right}_{ninBbb N}$$



                                    Observe the above sequence is irrational:



                                    $$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,text{so};;lim_{ntoinfty}g(a_n)=lim_{ntoinfty}16pi^4frac{n^4}{(npi+1)^4}=16pi^4frac1{pi^4}=16$$



                                    yet $;g(2)=2^2=4;$ , so the function is not continuous at $;2;$ ...






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take



                                      $$left{a_n:=2pifrac n{pi n+1}right}_{ninBbb N}$$



                                      Observe the above sequence is irrational:



                                      $$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,text{so};;lim_{ntoinfty}g(a_n)=lim_{ntoinfty}16pi^4frac{n^4}{(npi+1)^4}=16pi^4frac1{pi^4}=16$$



                                      yet $;g(2)=2^2=4;$ , so the function is not continuous at $;2;$ ...






                                      share|cite|improve this answer









                                      $endgroup$



                                      You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take



                                      $$left{a_n:=2pifrac n{pi n+1}right}_{ninBbb N}$$



                                      Observe the above sequence is irrational:



                                      $$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,text{so};;lim_{ntoinfty}g(a_n)=lim_{ntoinfty}16pi^4frac{n^4}{(npi+1)^4}=16pi^4frac1{pi^4}=16$$



                                      yet $;g(2)=2^2=4;$ , so the function is not continuous at $;2;$ ...







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 2 days ago









                                      DonAntonioDonAntonio

                                      180k1494233




                                      180k1494233






















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                                          How did Captain America manage to do this?

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                                          2 How does Captain America lift Thor's hammer? In Avengers: Age of Ultron , he is not worthy but how he is worthy in this film. plot-explanation marvel-cinematic-universe avengers-age-of-ultron avengers-endgame share | improve this question edited 19 mins ago A J ♦ 42.7k 16 229 246 asked 39 mins ago Andi AR Andi AR 325 1 4 14
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                                          迪纳利

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                                          迪纳利峰 位于迪纳利国家公园和保留区的迪纳利峰（旧名麦金利峰） 最高点 海拔 20,237英尺（6,168米）   NGVD 29 . [1] [2] 相對高度 20,073英尺（6,118米） 列表 各洲最高峰列表 特独立山峰 各国最高点列表 美国最高点  排名第一 [3] 坐标 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley ) 坐标： 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley )   地理 位置   美國阿拉斯加州迪纳利自治市镇 迪纳利国家公园和保留区 山脈 阿拉斯加山脉 攀山 首次登頂 （ 英语 ： First ascent ） 1913年6月7日 哈德逊·史塔克 哈里·卡斯坦斯 沃尔特·哈珀 罗伯特·塔特姆 最簡路線 西坡路线（攀登冰川/雪） 迪纳利 （北阿萨巴斯卡语支、 英语： Denali ， / d ᵻ ˈ n ɑː l i / ，直译“高山”），1917年至2015年官方名称为 麦金利山 （另译作 麦金利峰 ， 英语： Mount McKinley ），位于阿拉斯加州东南部、阿拉斯加山脉中段，海拔6194米，是北美洲最高峰，也是美国的最高峰。它的俄语名称是 Большая Гора ， 罗马化：Bolshaya Gora ，意思是“大山”。 迪纳利山地区拥有变幻莫测的高山风、典型的北极植被以及野生动植物。这里大部分地区终年积雪，山间经常浓雾不断，雾气在皑皑白雪中缭绕弥漫时，几百米之外的景物便不可见。夏季，迪纳利山的青青山坡上鲜花盛开，紫色的杜鹃和精巧的铃状石南花随处可见。迪纳利山还是世界登山爱好者的汇集之地。每年5月到7月有数百人登山，只有一半不到的人能登顶成功。一次登山约花3个星期。 2015年8月31日，美國政府宣佈將山名由“麦金利山”改回原名“迪纳利山”。美國內政部發表聲
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                                          南乌拉尔铁路局

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                                          南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
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