function only contains jump discontinuity but is not piecewise continuousWhat is the type of discontinuity of...

Is every open circuit a capacitor?

Is there a math equivalent to the conditional ternary operator?

« Rendre » et « render » (the meaning)

Where is this quote about overcoming the impossible said in "Interstellar"?

What's the best tool for cutting holes into duct work?

The (Easy) Road to Code

Why is it "take a leak?"

How do we objectively assess if a dialogue sounds unnatural or cringy?

Should I use HTTPS on a domain that will only be used for redirection?

Effect of "wrong" driver with slightly long RS-485 stubs

What can I do if someone tampers with my SSH public key?

Should we avoid writing fiction about historical events without extensive research?

School performs periodic password audits. Is my password compromised?

Split a number into equal parts given the number of parts

Caulking a corner instead of taping with joint compound?

Deal the cards to the players

Difference between 'stomach' and 'uterus'

The need of reserving one's ability in job interviews

Is being socially reclusive okay for a graduate student?

Is there a way to find out the age of climbing ropes?

Why won't the strings command stop?

Why can't we use freedom of speech and expression to incite people to rebel against government?

What is the purpose of a disclaimer like "this is not legal advice"?

How does insurance birth control work in the United States?



function only contains jump discontinuity but is not piecewise continuous


What is the type of discontinuity of $frac{exp x-1}{exp|x|-1}$ at $0$? Is it piecewise continuous?Nonpiecewise Function Defined at a Point but Not Continuous ThereJump discontinuity is not a Lebesgue pointOnly removable/jump discontinuitiesProving discontinuity of a piecewise functionPiecewise continuous contours with discontinuity only at end pointsIntegration with piecewise defined function and jump discontinuitytotal variation of piecewise constant average functionDoes making a jump discontinuity undefined actually make the function continuous?Is there a function only contains jump discontinuity but is not piecewise continuous













2












$begingroup$


Does there exists a function defined on $[0,1] rightarrow mathbb{R}$ such that it contains only jump discontinuity but it is not piecewise continuous?



Jump discontinuity at a point $a$ means $lim_{x rightarrow a^{-}}f(x)$ and
$lim_{x rightarrow a^{+}}f(x)$ both exists but not equal



piecewise continuous means every finite subinterval only contains a finite number of discontinuous points and they are all jump discontinuity



My first thought is Dirichlet function and but it appears that it is not the function that I am looking for.....










share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Does there exists a function defined on $[0,1] rightarrow mathbb{R}$ such that it contains only jump discontinuity but it is not piecewise continuous?



    Jump discontinuity at a point $a$ means $lim_{x rightarrow a^{-}}f(x)$ and
    $lim_{x rightarrow a^{+}}f(x)$ both exists but not equal



    piecewise continuous means every finite subinterval only contains a finite number of discontinuous points and they are all jump discontinuity



    My first thought is Dirichlet function and but it appears that it is not the function that I am looking for.....










    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Does there exists a function defined on $[0,1] rightarrow mathbb{R}$ such that it contains only jump discontinuity but it is not piecewise continuous?



      Jump discontinuity at a point $a$ means $lim_{x rightarrow a^{-}}f(x)$ and
      $lim_{x rightarrow a^{+}}f(x)$ both exists but not equal



      piecewise continuous means every finite subinterval only contains a finite number of discontinuous points and they are all jump discontinuity



      My first thought is Dirichlet function and but it appears that it is not the function that I am looking for.....










      share|cite|improve this question









      $endgroup$




      Does there exists a function defined on $[0,1] rightarrow mathbb{R}$ such that it contains only jump discontinuity but it is not piecewise continuous?



      Jump discontinuity at a point $a$ means $lim_{x rightarrow a^{-}}f(x)$ and
      $lim_{x rightarrow a^{+}}f(x)$ both exists but not equal



      piecewise continuous means every finite subinterval only contains a finite number of discontinuous points and they are all jump discontinuity



      My first thought is Dirichlet function and but it appears that it is not the function that I am looking for.....







      real-analysis calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 13 hours ago









      JoeJoe

      414




      414






















          2 Answers
          2






          active

          oldest

          votes


















          7












          $begingroup$

          OK, next thought - the function $f(frac pq)=frac1q$ and zero elsewhere. That's closer; it has limits of zero everywhere. But then each rational is a removable discontinuity, not a jump discontinuity. Closer, but still not it.



          The next idea after that: let's build an increasing function with jumps at every rational. Let $g$ be an enumeration of the rationals; for each rational $r$, $g(r)$ is a different positive integer $n$. Then, define
          $$f(x) = sum_{rinmathbb{Q},rle x}frac1{g(r)^2+g(r)}$$
          Since $sum_n frac1{n^2+n}$ converges (to $1$), that sum is finite for every $x$.

          Choose some arbitrary $x$ and $epsilon>0$. Let $n$ be such that $epsilongefrac1n$. There are only finitely many values $r_1,r_2,dots,r_n$ with $g(r_i)le n$. If we choose $delta$ such that $(x,x+delta)$ contains none of these $r_i$, then for $yin (x,x+delta)$,
          $$f(y)-f(x)=sum_{rinmathbb{Q},x<rle y}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
          From that, $lim_{yto x^+}f(y)=f(x)$ for all $x$. We have limits from the right.



          For limits from the left, consider the variant function
          $$f^*(x)=sum_{rinmathbb{Q},r< x}frac1{g(r)^2+g(r)}$$
          This $f^*$ is equal to $f$ except at the rationals, where $f(r)-f^*(r)=frac1{g(r)^2+g(r)}$. Again, choose arbitrary $x$ and $epsilon>0$, and let $n$ be such that $epsilonge frac1n$. Find $delta$ such that $(x-delta,x)$ contains none of the $n$ points $r_i$ with $g(r_i)le n$. Then, for $yin (x-delta,x)$,
          $$f^*(x)-f(y) = sum_{rinmathbb{Q},yle r< x}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
          From that, $lim_{yto x^-}f(y) = f^*(x)$ for all $x$, and we have limits from the left.



          Of course, these limits $lim_{yto x^+}f(y)=f(x)$ and $lim_{yto x^-}f(y) = f^*(x)$ differ for every rational $x$, so there's a jump discontinuity at every rational.



          With $f$ discontinuous at a dense set of points, it fails to be continuous on any interval, and can't be a piecewise continuous function. Done. We have our example.



          I defined this as a function from $mathbb{R}$ to $mathbb{R}$, but it's easy to get a function on a smaller interval. Restricting $f$ works, as does using an enumeration of the rationals in that smaller interval.






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$

            $$f(x)=sum_{k=1}^infty 2^{-k} ( 2^kx-lfloor 2^kxrfloor)$$ has a jump discontinuity at every $frac{n}{2^k}$ and it is continuous everywhere else



            $$g(x)=sum_{k=1}^infty 2^{-k} ( lfloor 2^kxrfloor-2lfloor 2^{k+1}xrfloor)$$ is easier to see : if $x in [frac{N}{2^k},frac{N+1}{2^k}]$ then $g(x) = frac{N}{2^k}+ O(2^{-k})$






            share|cite|improve this answer











            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3138401%2ffunction-only-contains-jump-discontinuity-but-is-not-piecewise-continuous%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              7












              $begingroup$

              OK, next thought - the function $f(frac pq)=frac1q$ and zero elsewhere. That's closer; it has limits of zero everywhere. But then each rational is a removable discontinuity, not a jump discontinuity. Closer, but still not it.



              The next idea after that: let's build an increasing function with jumps at every rational. Let $g$ be an enumeration of the rationals; for each rational $r$, $g(r)$ is a different positive integer $n$. Then, define
              $$f(x) = sum_{rinmathbb{Q},rle x}frac1{g(r)^2+g(r)}$$
              Since $sum_n frac1{n^2+n}$ converges (to $1$), that sum is finite for every $x$.

              Choose some arbitrary $x$ and $epsilon>0$. Let $n$ be such that $epsilongefrac1n$. There are only finitely many values $r_1,r_2,dots,r_n$ with $g(r_i)le n$. If we choose $delta$ such that $(x,x+delta)$ contains none of these $r_i$, then for $yin (x,x+delta)$,
              $$f(y)-f(x)=sum_{rinmathbb{Q},x<rle y}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
              From that, $lim_{yto x^+}f(y)=f(x)$ for all $x$. We have limits from the right.



              For limits from the left, consider the variant function
              $$f^*(x)=sum_{rinmathbb{Q},r< x}frac1{g(r)^2+g(r)}$$
              This $f^*$ is equal to $f$ except at the rationals, where $f(r)-f^*(r)=frac1{g(r)^2+g(r)}$. Again, choose arbitrary $x$ and $epsilon>0$, and let $n$ be such that $epsilonge frac1n$. Find $delta$ such that $(x-delta,x)$ contains none of the $n$ points $r_i$ with $g(r_i)le n$. Then, for $yin (x-delta,x)$,
              $$f^*(x)-f(y) = sum_{rinmathbb{Q},yle r< x}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
              From that, $lim_{yto x^-}f(y) = f^*(x)$ for all $x$, and we have limits from the left.



              Of course, these limits $lim_{yto x^+}f(y)=f(x)$ and $lim_{yto x^-}f(y) = f^*(x)$ differ for every rational $x$, so there's a jump discontinuity at every rational.



              With $f$ discontinuous at a dense set of points, it fails to be continuous on any interval, and can't be a piecewise continuous function. Done. We have our example.



              I defined this as a function from $mathbb{R}$ to $mathbb{R}$, but it's easy to get a function on a smaller interval. Restricting $f$ works, as does using an enumeration of the rationals in that smaller interval.






              share|cite|improve this answer









              $endgroup$


















                7












                $begingroup$

                OK, next thought - the function $f(frac pq)=frac1q$ and zero elsewhere. That's closer; it has limits of zero everywhere. But then each rational is a removable discontinuity, not a jump discontinuity. Closer, but still not it.



                The next idea after that: let's build an increasing function with jumps at every rational. Let $g$ be an enumeration of the rationals; for each rational $r$, $g(r)$ is a different positive integer $n$. Then, define
                $$f(x) = sum_{rinmathbb{Q},rle x}frac1{g(r)^2+g(r)}$$
                Since $sum_n frac1{n^2+n}$ converges (to $1$), that sum is finite for every $x$.

                Choose some arbitrary $x$ and $epsilon>0$. Let $n$ be such that $epsilongefrac1n$. There are only finitely many values $r_1,r_2,dots,r_n$ with $g(r_i)le n$. If we choose $delta$ such that $(x,x+delta)$ contains none of these $r_i$, then for $yin (x,x+delta)$,
                $$f(y)-f(x)=sum_{rinmathbb{Q},x<rle y}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
                From that, $lim_{yto x^+}f(y)=f(x)$ for all $x$. We have limits from the right.



                For limits from the left, consider the variant function
                $$f^*(x)=sum_{rinmathbb{Q},r< x}frac1{g(r)^2+g(r)}$$
                This $f^*$ is equal to $f$ except at the rationals, where $f(r)-f^*(r)=frac1{g(r)^2+g(r)}$. Again, choose arbitrary $x$ and $epsilon>0$, and let $n$ be such that $epsilonge frac1n$. Find $delta$ such that $(x-delta,x)$ contains none of the $n$ points $r_i$ with $g(r_i)le n$. Then, for $yin (x-delta,x)$,
                $$f^*(x)-f(y) = sum_{rinmathbb{Q},yle r< x}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
                From that, $lim_{yto x^-}f(y) = f^*(x)$ for all $x$, and we have limits from the left.



                Of course, these limits $lim_{yto x^+}f(y)=f(x)$ and $lim_{yto x^-}f(y) = f^*(x)$ differ for every rational $x$, so there's a jump discontinuity at every rational.



                With $f$ discontinuous at a dense set of points, it fails to be continuous on any interval, and can't be a piecewise continuous function. Done. We have our example.



                I defined this as a function from $mathbb{R}$ to $mathbb{R}$, but it's easy to get a function on a smaller interval. Restricting $f$ works, as does using an enumeration of the rationals in that smaller interval.






                share|cite|improve this answer









                $endgroup$
















                  7












                  7








                  7





                  $begingroup$

                  OK, next thought - the function $f(frac pq)=frac1q$ and zero elsewhere. That's closer; it has limits of zero everywhere. But then each rational is a removable discontinuity, not a jump discontinuity. Closer, but still not it.



                  The next idea after that: let's build an increasing function with jumps at every rational. Let $g$ be an enumeration of the rationals; for each rational $r$, $g(r)$ is a different positive integer $n$. Then, define
                  $$f(x) = sum_{rinmathbb{Q},rle x}frac1{g(r)^2+g(r)}$$
                  Since $sum_n frac1{n^2+n}$ converges (to $1$), that sum is finite for every $x$.

                  Choose some arbitrary $x$ and $epsilon>0$. Let $n$ be such that $epsilongefrac1n$. There are only finitely many values $r_1,r_2,dots,r_n$ with $g(r_i)le n$. If we choose $delta$ such that $(x,x+delta)$ contains none of these $r_i$, then for $yin (x,x+delta)$,
                  $$f(y)-f(x)=sum_{rinmathbb{Q},x<rle y}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
                  From that, $lim_{yto x^+}f(y)=f(x)$ for all $x$. We have limits from the right.



                  For limits from the left, consider the variant function
                  $$f^*(x)=sum_{rinmathbb{Q},r< x}frac1{g(r)^2+g(r)}$$
                  This $f^*$ is equal to $f$ except at the rationals, where $f(r)-f^*(r)=frac1{g(r)^2+g(r)}$. Again, choose arbitrary $x$ and $epsilon>0$, and let $n$ be such that $epsilonge frac1n$. Find $delta$ such that $(x-delta,x)$ contains none of the $n$ points $r_i$ with $g(r_i)le n$. Then, for $yin (x-delta,x)$,
                  $$f^*(x)-f(y) = sum_{rinmathbb{Q},yle r< x}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
                  From that, $lim_{yto x^-}f(y) = f^*(x)$ for all $x$, and we have limits from the left.



                  Of course, these limits $lim_{yto x^+}f(y)=f(x)$ and $lim_{yto x^-}f(y) = f^*(x)$ differ for every rational $x$, so there's a jump discontinuity at every rational.



                  With $f$ discontinuous at a dense set of points, it fails to be continuous on any interval, and can't be a piecewise continuous function. Done. We have our example.



                  I defined this as a function from $mathbb{R}$ to $mathbb{R}$, but it's easy to get a function on a smaller interval. Restricting $f$ works, as does using an enumeration of the rationals in that smaller interval.






                  share|cite|improve this answer









                  $endgroup$



                  OK, next thought - the function $f(frac pq)=frac1q$ and zero elsewhere. That's closer; it has limits of zero everywhere. But then each rational is a removable discontinuity, not a jump discontinuity. Closer, but still not it.



                  The next idea after that: let's build an increasing function with jumps at every rational. Let $g$ be an enumeration of the rationals; for each rational $r$, $g(r)$ is a different positive integer $n$. Then, define
                  $$f(x) = sum_{rinmathbb{Q},rle x}frac1{g(r)^2+g(r)}$$
                  Since $sum_n frac1{n^2+n}$ converges (to $1$), that sum is finite for every $x$.

                  Choose some arbitrary $x$ and $epsilon>0$. Let $n$ be such that $epsilongefrac1n$. There are only finitely many values $r_1,r_2,dots,r_n$ with $g(r_i)le n$. If we choose $delta$ such that $(x,x+delta)$ contains none of these $r_i$, then for $yin (x,x+delta)$,
                  $$f(y)-f(x)=sum_{rinmathbb{Q},x<rle y}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
                  From that, $lim_{yto x^+}f(y)=f(x)$ for all $x$. We have limits from the right.



                  For limits from the left, consider the variant function
                  $$f^*(x)=sum_{rinmathbb{Q},r< x}frac1{g(r)^2+g(r)}$$
                  This $f^*$ is equal to $f$ except at the rationals, where $f(r)-f^*(r)=frac1{g(r)^2+g(r)}$. Again, choose arbitrary $x$ and $epsilon>0$, and let $n$ be such that $epsilonge frac1n$. Find $delta$ such that $(x-delta,x)$ contains none of the $n$ points $r_i$ with $g(r_i)le n$. Then, for $yin (x-delta,x)$,
                  $$f^*(x)-f(y) = sum_{rinmathbb{Q},yle r< x}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
                  From that, $lim_{yto x^-}f(y) = f^*(x)$ for all $x$, and we have limits from the left.



                  Of course, these limits $lim_{yto x^+}f(y)=f(x)$ and $lim_{yto x^-}f(y) = f^*(x)$ differ for every rational $x$, so there's a jump discontinuity at every rational.



                  With $f$ discontinuous at a dense set of points, it fails to be continuous on any interval, and can't be a piecewise continuous function. Done. We have our example.



                  I defined this as a function from $mathbb{R}$ to $mathbb{R}$, but it's easy to get a function on a smaller interval. Restricting $f$ works, as does using an enumeration of the rationals in that smaller interval.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 12 hours ago









                  jmerryjmerry

                  12.5k1628




                  12.5k1628























                      4












                      $begingroup$

                      $$f(x)=sum_{k=1}^infty 2^{-k} ( 2^kx-lfloor 2^kxrfloor)$$ has a jump discontinuity at every $frac{n}{2^k}$ and it is continuous everywhere else



                      $$g(x)=sum_{k=1}^infty 2^{-k} ( lfloor 2^kxrfloor-2lfloor 2^{k+1}xrfloor)$$ is easier to see : if $x in [frac{N}{2^k},frac{N+1}{2^k}]$ then $g(x) = frac{N}{2^k}+ O(2^{-k})$






                      share|cite|improve this answer











                      $endgroup$


















                        4












                        $begingroup$

                        $$f(x)=sum_{k=1}^infty 2^{-k} ( 2^kx-lfloor 2^kxrfloor)$$ has a jump discontinuity at every $frac{n}{2^k}$ and it is continuous everywhere else



                        $$g(x)=sum_{k=1}^infty 2^{-k} ( lfloor 2^kxrfloor-2lfloor 2^{k+1}xrfloor)$$ is easier to see : if $x in [frac{N}{2^k},frac{N+1}{2^k}]$ then $g(x) = frac{N}{2^k}+ O(2^{-k})$






                        share|cite|improve this answer











                        $endgroup$
















                          4












                          4








                          4





                          $begingroup$

                          $$f(x)=sum_{k=1}^infty 2^{-k} ( 2^kx-lfloor 2^kxrfloor)$$ has a jump discontinuity at every $frac{n}{2^k}$ and it is continuous everywhere else



                          $$g(x)=sum_{k=1}^infty 2^{-k} ( lfloor 2^kxrfloor-2lfloor 2^{k+1}xrfloor)$$ is easier to see : if $x in [frac{N}{2^k},frac{N+1}{2^k}]$ then $g(x) = frac{N}{2^k}+ O(2^{-k})$






                          share|cite|improve this answer











                          $endgroup$



                          $$f(x)=sum_{k=1}^infty 2^{-k} ( 2^kx-lfloor 2^kxrfloor)$$ has a jump discontinuity at every $frac{n}{2^k}$ and it is continuous everywhere else



                          $$g(x)=sum_{k=1}^infty 2^{-k} ( lfloor 2^kxrfloor-2lfloor 2^{k+1}xrfloor)$$ is easier to see : if $x in [frac{N}{2^k},frac{N+1}{2^k}]$ then $g(x) = frac{N}{2^k}+ O(2^{-k})$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 11 hours ago

























                          answered 11 hours ago









                          reunsreuns

                          21.1k21250




                          21.1k21250






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3138401%2ffunction-only-contains-jump-discontinuity-but-is-not-piecewise-continuous%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              El tren de la libertad Índice Antecedentes "Porque yo decido" Desarrollo de la...

                              Puerta de Hutt Referencias Enlaces externos Menú de navegación15°58′00″S 5°42′00″O /...

                              Castillo d'Acher Características Menú de navegación