Continuity at a point in terms of closure The 2019 Stack Overflow Developer Survey Results Are...

Is bread bad for ducks?

Geography at the pixel level

Springs with some finite mass

How to make payment on the internet without leaving a money trail?

What is this 4-propeller plane?

I see my dog run

Why isn't airport relocation done gradually?

How long do I have to send payment?

Deadlock Graph and Interpretation, solution to avoid

Access elements in std::string where positon of string is greater than its size

Landlord wants to switch my lease to a "Land contract" to "get back at the city"

Is there a name of the flying bionic bird?

What do the Banks children have against barley water?

How to manage monthly salary

Are there any other methods to apply to solving simultaneous equations?

If the Wish spell is used to duplicate the effect of Simulacrum, are existing duplicates destroyed?

Are USB sockets on wall outlets live all the time, even when the switch is off?

How come people say “Would of”?

How to reverse every other sublist of a list?

The difference between dialogue marks

Why do UK politicians seemingly ignore opinion polls on Brexit?

Could JWST stay at L2 "forever"?

What is the use of option -o in the useradd command?

How was Skylab's orbit inclination chosen?



Continuity at a point in terms of closure



The 2019 Stack Overflow Developer Survey Results Are InSet Closure Union and IntersectionContinuity of a function through adherence of subsetsConvex set with empty interior is nowhere dense?Is preimage of closure equal to closure of preimage under continuous topological maps?How to show the logical equivalence of the following two definitions of continuity in a topological space?Given $A subseteq X$ in the discrete and the trivial topology, find closure of $A$Show two notions of dense are equivalentEquivalent definitions of continuity at a pointAbout continuity and clousureEquivalent definition of irreducible topological subspace.












4












$begingroup$


If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverline{A} implies f(x_0)inoverline{f(A)}$.



I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!



Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^{-1}(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverline{A}$, we have $Acap f^{-1}(V)neqvarnothing$. Let $xin Acap f^{-1}(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverline{f(A)}$.






general-topology continuity







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverline{A} implies f(x_0)inoverline{f(A)}$.



    I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!



    Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^{-1}(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverline{A}$, we have $Acap f^{-1}(V)neqvarnothing$. Let $xin Acap f^{-1}(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverline{f(A)}$.






    general-topology continuity







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverline{A} implies f(x_0)inoverline{f(A)}$.



      I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!



      Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^{-1}(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverline{A}$, we have $Acap f^{-1}(V)neqvarnothing$. Let $xin Acap f^{-1}(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverline{f(A)}$.






      general-topology continuity







      share|cite|improve this question









      $endgroup$




      If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverline{A} implies f(x_0)inoverline{f(A)}$.



      I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!



      Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^{-1}(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverline{A}$, we have $Acap f^{-1}(V)neqvarnothing$. Let $xin Acap f^{-1}(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverline{f(A)}$.






      general-topology continuity




      general-topology continuity






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 days ago









      BlondCaféBlondCafé

      364




      364






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overline{f^{-1}(Ysetminus V)}$, then $f(x_o)in overline{f(f^{-1}(Ysetminus V))}subset overline{Ysetminus V}= Ysetminus V$, a contradiction, so $x_0 notin overline{f^{-1}(Ysetminus V)}$. Then, if $U = Xsetminus overline{f^{-1}(Ysetminus V)}$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
            We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.



            Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^{-1}[Ysetminus V] neq emptyset$.



            It follows that then $x_0 in overline{f^{-1}[Ysetminus V]}$ and so the assumption on $f$ would imply that $y=f(x_0) in overline{f[f^{-1}[Ysetminus V]]}$. But $f[f^{-1}[B]] subseteq B$ for any $B$ so we'd deduce that $y in overline{Ysetminus V} = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.






            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%2f3178435%2fcontinuity-at-a-point-in-terms-of-closure%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









              3












              $begingroup$

              Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overline{f^{-1}(Ysetminus V)}$, then $f(x_o)in overline{f(f^{-1}(Ysetminus V))}subset overline{Ysetminus V}= Ysetminus V$, a contradiction, so $x_0 notin overline{f^{-1}(Ysetminus V)}$. Then, if $U = Xsetminus overline{f^{-1}(Ysetminus V)}$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overline{f^{-1}(Ysetminus V)}$, then $f(x_o)in overline{f(f^{-1}(Ysetminus V))}subset overline{Ysetminus V}= Ysetminus V$, a contradiction, so $x_0 notin overline{f^{-1}(Ysetminus V)}$. Then, if $U = Xsetminus overline{f^{-1}(Ysetminus V)}$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overline{f^{-1}(Ysetminus V)}$, then $f(x_o)in overline{f(f^{-1}(Ysetminus V))}subset overline{Ysetminus V}= Ysetminus V$, a contradiction, so $x_0 notin overline{f^{-1}(Ysetminus V)}$. Then, if $U = Xsetminus overline{f^{-1}(Ysetminus V)}$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.






                  share|cite|improve this answer









                  $endgroup$



                  Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overline{f^{-1}(Ysetminus V)}$, then $f(x_o)in overline{f(f^{-1}(Ysetminus V))}subset overline{Ysetminus V}= Ysetminus V$, a contradiction, so $x_0 notin overline{f^{-1}(Ysetminus V)}$. Then, if $U = Xsetminus overline{f^{-1}(Ysetminus V)}$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 days ago









                  guchiheguchihe

                  21919




                  21919























                      2












                      $begingroup$

                      It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
                      We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.



                      Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^{-1}[Ysetminus V] neq emptyset$.



                      It follows that then $x_0 in overline{f^{-1}[Ysetminus V]}$ and so the assumption on $f$ would imply that $y=f(x_0) in overline{f[f^{-1}[Ysetminus V]]}$. But $f[f^{-1}[B]] subseteq B$ for any $B$ so we'd deduce that $y in overline{Ysetminus V} = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
                        We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.



                        Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^{-1}[Ysetminus V] neq emptyset$.



                        It follows that then $x_0 in overline{f^{-1}[Ysetminus V]}$ and so the assumption on $f$ would imply that $y=f(x_0) in overline{f[f^{-1}[Ysetminus V]]}$. But $f[f^{-1}[B]] subseteq B$ for any $B$ so we'd deduce that $y in overline{Ysetminus V} = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
                          We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.



                          Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^{-1}[Ysetminus V] neq emptyset$.



                          It follows that then $x_0 in overline{f^{-1}[Ysetminus V]}$ and so the assumption on $f$ would imply that $y=f(x_0) in overline{f[f^{-1}[Ysetminus V]]}$. But $f[f^{-1}[B]] subseteq B$ for any $B$ so we'd deduce that $y in overline{Ysetminus V} = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.






                          share|cite|improve this answer









                          $endgroup$



                          It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
                          We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.



                          Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^{-1}[Ysetminus V] neq emptyset$.



                          It follows that then $x_0 in overline{f^{-1}[Ysetminus V]}$ and so the assumption on $f$ would imply that $y=f(x_0) in overline{f[f^{-1}[Ysetminus V]]}$. But $f[f^{-1}[B]] subseteq B$ for any $B$ so we'd deduce that $y in overline{Ysetminus V} = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 2 days ago









                          Henno BrandsmaHenno Brandsma

                          116k349127




                          116k349127






























                              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%2f3178435%2fcontinuity-at-a-point-in-terms-of-closure%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

                              Why not use the yoke to control yaw, as well as pitch and roll? Announcing the arrival of...

                              Image
                              Does the Mueller report show a conspiracy between Russia and the Trump Campaign? Relating to the President and obstruction, were Mueller's conclusions preordained? How much damage would a cupful of neutron star matter do to the Earth? Nose gear failure in single prop aircraft: belly landing or nose landing? Should a wizard buy fine inks every time he want to copy spells into his spellbook? Can you force honesty by using the Speak with Dead and Zone of Truth spells together? A trigonometry question from STEP examination Special flights What is the "studentd" process? systemd and copy (/bin/cp): no such file or directory...
                              Read more

                              Couldn't open a raw socket. Error: Permission denied (13) (nmap)Is it possible to run networking commands...

                              Image
                              Why didn't Boeing produce its own regional jet? What is the most common color to indicate the input-field is disabled? Theorists sure want true answers to this! What is the opposite of "eschatology"? What do you call someone who asks many questions? Is there a hemisphere-neutral way of specifying a season? Is this draw by repetition? Bug? String.split chopping off empty trailing value Is it inappropriate for a student to attend their mentor's dissertation defense? How dangerous is XSS Why do I get negative height? Is it possible to map ...
                              Read more

                              VNC viewer RFB protocol error: bad desktop size 0x0I Cannot Type the Key 'd' (lowercase) in VNC Viewer...

                              Image
                              risk of flooding in petra in november How to get directions in deep space? How could a planet have erratic days? 15% tax on $7.5k earnings. Is that right? Did the UK lift the requirement for registering SIM cards? Is it allowed to activate the ability of multiple planeswalkers in a single turn? Stack Interview Code methods made from class Node and Smart Pointers Non-trope happy ending? What is the highest possible scrabble score for placing a single tile Open a doc from terminal, but not by its name The IT department bottlenecks progress, how should I handle this? ...
                              Read more