Continuity at a point in terms of closureSet 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.

Is it possible to make sharp wind that can cut stuff from afar?

How to report a triplet of septets in NMR tabulation?

How old can references or sources in a thesis be?

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

What Brexit solution does the DUP want?

The magic money tree problem

declaring a variable twice in IIFE

Why are 150k or 200k jobs considered good when there are 300k+ births a month?

How can the DM most effectively choose 1 out of an odd number of players to be targeted by an attack or effect?

How is it possible for user's password to be changed after storage was encrypted? (on OS X, Android)

What makes Graph invariants so useful/important?

What do you call something that goes against the spirit of the law, but is legal when interpreting the law to the letter?

Chess with symmetric move-square

Do airline pilots ever risk not hearing communication directed to them specifically, from traffic controllers?

New order #4: World

What would the Romans have called "sorcery"?

Why did the Germans forbid the possession of pet pigeons in Rostov-on-Don in 1941?

What is the command to reset a PC without deleting any files

Is there really no realistic way for a skeleton monster to move around without magic?

Can I make popcorn with any corn?

Are tax years 2016 & 2017 back taxes deductible for tax year 2018?

Should I join office cleaning event for free?

Possibly bubble sort algorithm

Why don't electron-positron collisions release infinite energy?



Continuity at a point in terms of closure


Set 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 $Asubseteq X land x_0inoverlineA implies f(x_0)inoverlinef(A)$.



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 $x_0inoverlineA$, 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 $f(x_0)inoverlinef(A)$.










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 $Asubseteq X land x_0inoverlineA implies f(x_0)inoverlinef(A)$.



    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 $x_0inoverlineA$, 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 $f(x_0)inoverlinef(A)$.










    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 $Asubseteq X land x_0inoverlineA implies f(x_0)inoverlinef(A)$.



      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 $x_0inoverlineA$, 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 $f(x_0)inoverlinef(A)$.










      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 $Asubseteq X land x_0inoverlineA implies f(x_0)inoverlinef(A)$.



      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 $x_0inoverlineA$, 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 $f(x_0)inoverlinef(A)$.







      general-topology continuity






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 9 hours 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 $x_0 in overlinef^-1(Ysetminus V)$, then $f(x_o)in overlinef(f^-1(Ysetminus V))subset overlineYsetminus V= Ysetminus V$, a contradiction, so $x_0 notin overlinef^-1(Ysetminus V)$. Then, if $U = Xsetminus overlinef^-1(Ysetminus V)$ 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 $x_0 in overlinef^-1[Ysetminus V]$ and so the assumption on $f$ would imply that $y=f(x_0) in overlinef[f^-1[Ysetminus V]]$. But $f[f^-1[B]] subseteq B$ for any $B$ so we'd deduce that $y in overlineYsetminus V = Ysetminus V$ 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 $x_0 in overlinef^-1(Ysetminus V)$, then $f(x_o)in overlinef(f^-1(Ysetminus V))subset overlineYsetminus V= Ysetminus V$, a contradiction, so $x_0 notin overlinef^-1(Ysetminus V)$. Then, if $U = Xsetminus overlinef^-1(Ysetminus V)$ 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 $x_0 in overlinef^-1(Ysetminus V)$, then $f(x_o)in overlinef(f^-1(Ysetminus V))subset overlineYsetminus V= Ysetminus V$, a contradiction, so $x_0 notin overlinef^-1(Ysetminus V)$. Then, if $U = Xsetminus overlinef^-1(Ysetminus V)$ 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 $x_0 in overlinef^-1(Ysetminus V)$, then $f(x_o)in overlinef(f^-1(Ysetminus V))subset overlineYsetminus V= Ysetminus V$, a contradiction, so $x_0 notin overlinef^-1(Ysetminus V)$. Then, if $U = Xsetminus overlinef^-1(Ysetminus V)$ 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 $x_0 in overlinef^-1(Ysetminus V)$, then $f(x_o)in overlinef(f^-1(Ysetminus V))subset overlineYsetminus V= Ysetminus V$, a contradiction, so $x_0 notin overlinef^-1(Ysetminus V)$. Then, if $U = Xsetminus overlinef^-1(Ysetminus V)$ 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 9 hours ago









                  guchiheguchihe

                  20718




                  20718





















                      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 $x_0 in overlinef^-1[Ysetminus V]$ and so the assumption on $f$ would imply that $y=f(x_0) in overlinef[f^-1[Ysetminus V]]$. But $f[f^-1[B]] subseteq B$ for any $B$ so we'd deduce that $y in overlineYsetminus V = Ysetminus V$ 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 $x_0 in overlinef^-1[Ysetminus V]$ and so the assumption on $f$ would imply that $y=f(x_0) in overlinef[f^-1[Ysetminus V]]$. But $f[f^-1[B]] subseteq B$ for any $B$ so we'd deduce that $y in overlineYsetminus V = Ysetminus V$ 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 $x_0 in overlinef^-1[Ysetminus V]$ and so the assumption on $f$ would imply that $y=f(x_0) in overlinef[f^-1[Ysetminus V]]$. But $f[f^-1[B]] subseteq B$ for any $B$ so we'd deduce that $y in overlineYsetminus V = Ysetminus V$ 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 $x_0 in overlinef^-1[Ysetminus V]$ and so the assumption on $f$ would imply that $y=f(x_0) in overlinef[f^-1[Ysetminus V]]$. But $f[f^-1[B]] subseteq B$ for any $B$ so we'd deduce that $y in overlineYsetminus V = Ysetminus V$ 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 9 hours ago









                          Henno BrandsmaHenno Brandsma

                          115k349125




                          115k349125



























                              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

                              Can not update quote_id field of “quote_item” table magento 2Magento 2.1 - We can't remove the item. (Shopping Cart doesnt allow us to remove items before becomes empty)Add value for custom quote item attribute using REST apiREST API endpoint v1/carts/cartId/items always returns error messageCorrect way to save entries to databaseHow to remove all associated quote objects of a customer completelyMagento 2 - Save value from custom input field to quote_itemGet quote_item data using quote id and product id filter in Magento 2How to set additional data to quote_item table from controller in Magento 2?What is the purpose of additional_data column in quote_item table in magento2Set Custom Price to Quote item magento2 from controller

                              How to solve knockout JS error in Magento 2 Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?(Magento2) knockout.js:3012 Uncaught ReferenceError: Unable to process bindingUnable to process binding Knockout.js magento 2Cannot read property `scopeLabel` of undefined on Product Detail PageCan't get Customer Data on frontend in Magento 2Magento2 Order Summary - unable to process bindingKO templates are not loading in Magento 2.1 applicationgetting knockout js error magento 2Product grid not load -— Unable to process binding Knockout.js magento 2Product form not loaded in magento2Uncaught ReferenceError: Unable to process binding “if: function()return (isShowLegend()) ” magento 2

                              Nissan Patrol Зміст Перше покоління — 4W60 (1951-1960) | Друге покоління — 60 series (1960-1980) | Третє покоління (1980–2002) | Четверте покоління — Y60 (1987–1998) | П'яте покоління — Y61 (1997–2013) | Шосте покоління — Y62 (2010- ) | Посилання | Зноски | Навігаційне менюОфіційний український сайтТест-драйв Nissan Patrol 2010 7-го поколінняNissan PatrolКак мы тестировали Nissan Patrol 2016рвиправивши або дописавши її