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.

Today is the Center

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

I probably found a bug with the sudo apt install function

How is it possible to have an ability score that is less than 3?

"You are your self first supporter", a more proper way to say it

Why Is Death Allowed In the Matrix?

How did the USSR manage to innovate in an environment characterized by government censorship and high bureaucracy?

Japan - Plan around max visa duration

Do any Labour MPs support no-deal?

Can I make popcorn with any corn?

N.B. ligature in Latex

Example of a relative pronoun

Book about a traveler who helps planets in need

Theorems that impeded progress

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

How to report a triplet of septets in NMR tabulation?

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

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

When blogging recipes, how can I support both readers who want the narrative/journey and ones who want the printer-friendly recipe?

Infinite past with a beginning?

The magic money tree problem

Why can't I see bouncing of a switch on an oscilloscope?

Copenhagen passport control - US citizen

Prevent a directory in /tmp from being deleted



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 $f(x_0)inoverlinef(A)$.






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



    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









    $endgroup$














      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 $f(x_0)inoverlinef(A)$.






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



      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




      general-topology continuity






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 5 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 5 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 5 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

                              Bett Inhaltsverzeichnis Geschichte | Bettformen | Bettgrößen | Andere Bezeichnungen | Bettenmangel | Betten in der bildenden Kunst | Schlafmedizinische Gesichtspunkte | Siehe auch | Literatur | Weblinks | Einzelnachweise | NavigationsmenüBett, Bettstatt, BettstelleCommons: BettBabybetten: Anwendung, Ausstattungsmerkmale und VergleichskriterienWasserbetten. Vorurteile im TestHapfnNursch10.1007/s11818-012-0584-74006250-8AKS4329276-8

                              Image
                              Bett MöbelstückSchlafenMatratzeBettwäscheBettwareSofaalten ÄgyptenGriechenRömernMückennetzAssyrerMederPerserTeppichenPerlmutterElfenbeinOdysseushomerischenLeinentüchernRömerSchilfHeuFedernGänsenSchwänenKissenStickereienpurpurnerSchaubettLaubStrohsäckenMatratzenMittelalterBaldachinAlkovenPelze als DeckenLeintuchseidenenStrohUnterbettenMatratzenKemenateHängelampenNachtlichtBoudoirsAlkovenBettnischenRenaissancefrz.leverfrz.coucherLudwig XIV.FederbettenPlumeausMassenproduktionSäuglingeLattenrostsWiegeStubenwagenLattenrostMatratzeDoppelbettHochbettStockbettNachtschränkchenKaserneJugendherbergenSeemannsheimeWohncontainerBaustellenSaisonarbeiternGefängnissenStraflagernSchlafkojenFaltstuhlLazarettenEschenTeakholzMoskitonetzenTablettKarpfenanglerKufenLiegestuhlTSchaumstoffstückFutonbettJapanMatratzeBettdeckenwestlichen WeltFutonBoxspringbettAmerikaMatratzeSchlittenbettKahnbettBaldachinintimerMittelalterAlkovenMü...
                              Read more

                              Luksemburg Sisukord Nimi | Asend | Loodus | Riigikord | Haldusjaotus | Rahvastik | Riigikaitse | Majandus | Taristu | Ajalugu | Eesti ja Luksemburgi suhted | Haridus | Kultuur | Vaata ka | Viited | Välislingid | Navigeerimismenüü50° N, 6° EÜlevaade Luksemburgi kaitsealadest.Luksemburgi rahvaarv. Statistikaamet.World Bank'i andmebaasÜlevaade Luksemburgi loodusest.Ülevaade Luksemburgi metsadest.Guy Colling. "Red List of the Vascular Plants of Luxembourg." Travaux scientifiques du Musée national d’histoire naturelle Luxembourg. 2005.Luxembourg’s biodiversity at risk.Maailma kahepaiksete andmebaas.Denis Lepage. "Luxembourg." Avibase.Ülevaade temperatuuridest. Luksemburgi meteoroloogiateenistus.Ülevaade Luksemburgist. Euroopa Liidu esinduse koduleht.Système politique. TerritoireÜlevaade Luksemburgi rahvastikust. Luksemburgi statistikaamet.Luksemburgi rahvastik. Luksemburgi statistikaamet.The World FactbookMonique Borsenberger, Paul Dickes. "Religions au Luxembourg. Quelle évolution entre 1999-2008". Luksemburgi statistikaamet. 2011.Luksemburgi peapiiskopkond. Catholic-Hierarchy.Luksemburgi armee koduleht.Luksemburgi armee relvastus.Eesti Välisministeerium.Luksemburgi rahvastik. Luksemburgi statistikaamet.Luksemburgi Eesti Seltsi koduleht.Helen Eelrand. "Raadio, mis muutis maailma." Eesti Päevaleht. 13. märts 2004.Ülevaade Luksemburgi haridussüsteemist.Ülevaade Luksemburgi keskkoolidest.Luksemburgr

                              Image
                              RiigidEuroopa maadLuksemburg merepiiritariikEuroopasMadalmaadeHollandiBelgiagaBeneluxiPrantsusmaaSaksamaaLuxembourgikindlusestSaksa keeles963SiegfriedileläänestpõhjastidastRheinland-PfalziSaarimaalõunastLorraineMoselleMeurthe-et-Moselle'i departemangudArdennideMoseli jõejõedmetsagaharilik pöökharilik tammhall nelkharilik kortslehtmageveekalukahepaikseidpäriskonnalisiroomajaidimetajaidsaarmasNatura 20001964EuroopasÜlevaade Luksemburgi kaitsealadest.parasvöötme kliimasuvejaanuarveebruarjuuliaugustkonstitutsiooniline monarhiaTäidesaatvat võimuSeadusandlik võim2013põhiseadus18421868NATOEuroopa Liidu1. jaanuari20171985letseburgi keelprantsuse keelkodakondsusePortugaliPrantsusmaaItaaliakatoliiklust187019882013ilmikpreestritajateenistusBosnia ja Hertsegoviinaskindral2000politseijõudLiechtensteiniKataritöötusSaksamaaBelgiaPrantsusmaaHollandmineraalidHiinaeurotsooniga1999euro2002sadamKeldi2. sajandil e...
                              Read more

                              Valle di Casies Indice Geografia fisica | Origini del nome | Storia | Società | Amministrazione | Sport | Note | Bibliografia | Voci correlate | Altri progetti | Collegamenti esterni | Menu di navigazione46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)Sito istituzionaleAstat Censimento della popolazione 2011 - Determinazione della consistenza dei tre gruppi linguistici della Provincia Autonoma di Bolzano-Alto Adige - giugno 2012Numeri e fattiValle di CasiesDato IstatTabella dei gradi/giorno dei Comuni italiani raggruppati per Regione e Provincia26 agosto 1993, n. 412Heraldry of the World: GsiesStatistiche I.StatValCasies.comWikimedia CommonsWikimedia CommonsValle di CasiesSito ufficialeValle di CasiesMM14870458910042978-6

                              Image
                              Valle di Casies tedescocomune italianoprovincia autonoma di BolzanoTrentino-Alto AdigeValVal Pusteriario CasiesMonguelfoAustriatoponimo117811891299tedescoXII secolo1929municipio1985ted.cappuccinoJoachim HaspingerAndreas HoferbavaresifrancesinapoleonicibronzoOthmar Winklerbunkersbarramento Planca di SopraVallo Alpino in Alto Adigesci di fondoGran fondo della Val Casiesskiliftpiste da slittino naturale Valle di Casies Da Wikipedia, l'enciclopedia libera. Jump to navigation Jump to search Valle di Casies comune ( IT ) Valle di Casies ( DE ) Gsies Una veduta della vallata Localizzazione Stato   Italia Regione Trentino-Alto Adige Provincia Bolzano Amministrazione Sindaco Kurti Taschler (SVP) dal 10-5-2015 (1º mandato) Lingue ufficiali Tedesco, Italiano Territorio Coordinate 46°46′N 12°11′E  /  46.766667°N 12.183333°E 46.766667; 12.183333  ( Valle di Casies ) Coordinate: 46°46...
                              Read more