Any open subset of $\Bbb R$ is a countable union of disjoint ...

Let U be an open set in R. Then U is a countable union of disjoint intervals. This question has probably been asked. MathematicsStackExchangeisaquestionandanswersiteforpeoplestudyingmathatanylevelandprofessionalsinrelatedfields.Itonlytakesaminutetosignup. Signuptojointhiscommunity Anybodycanaskaquestion Anybodycananswer Thebestanswersarevotedupandrisetothetop Home Public Questions Tags Users Unanswered Teams StackOverflowforTeams –Startcollaboratingandsharingorganizationalknowledge. CreateafreeTeam WhyTeams? Teams CreatefreeTeam Teams Q&Aforwork Connectandshareknowledgewithinasinglelocationthatisstructuredandeasytosearch. Learnmore Anyopensubsetof$\BbbR$isacountableunionofdisjointopenintervals AskQuestion Asked 9years,2monthsago Modified 19daysago Viewed 102ktimes 256 221 $\begingroup$ Let$U$beanopensetin$\mathbbR$.Then$U$isacountableunionofdisjointintervals. Thisquestionhasprobablybeenasked.However,Iamnotinterestedinjustgettingtheanswertoit.Rather,Iaminterestedincollectingasmanydifferentproofsofitwhichareasdiverseaspossible.Aprofessortoldmethattherearemany.So,Iinviteeveryonewhohasseenproofsofthisfacttosharethemwiththecommunity.Ithinkitisaresultworthknowinghowtoproveinmanydifferentwaysandhavingapostthatcombinesasmanyofthemaspossiblewill,nodoubt,bequiteuseful.Aftertwodays,Iwillplaceabountyonthisquestiontoattractasmanypeopleaspossible.Ofcourse,anycomments,corrections,suggestions,linkstopapers/notesetc.aremorethanwelcome. real-analysisgeneral-topologybig-listfaq Share Cite Follow editedJun8,2020at3:30 communitywiki 7revs,5users60%ArcticChar $\endgroup$ 14 108 $\begingroup$ Firstproofthatcomesinmind:if$O$isanopensetand$x\inO$thenthereexistsaninterval$I$suchthat$x\inI\subsetO$.Ifthereexistsonesuchinterval,thenthereexistsone'largest'intervalwhichcontains$x$(theunionofallsuchintervals).Denoteby$\{I_\alpha\}$thefamilyofallsuchmaximalintervals.Firstallintervals$I_\alpha$arepairwisedisjoint(otherwisetheywouldn'tbemaximal)andeveryintervalcontainsarationalnumber,andthereforetherecanonlybeacountablenumberofintervalsinthefamily. $\endgroup$ – BeniBogosel Mar2,2013at0:12 1 $\begingroup$ Sincethisisabig-listquestion,IamconvertingittoCW. $\endgroup$ – robjohn ♦ Mar2,2013at9:01 $\begingroup$ Oh,OK!Thanks! $\endgroup$ – OrestXherija Mar3,2013at1:04 $\begingroup$ Icannotunderstandhoweach$I_a$isdisjoint.Ifoneintervalcontainsxandanotherintervalalsocontainsx,aren'ttheyintersecting? $\endgroup$ – mesllo May29,2014at1:41 $\begingroup$ Yesbuttheirunionwouldthenbeanotherintervalthatcontainsan$I_a$andthereforemustbeequaltoitbymaximality. $\endgroup$ – GregoryGrant Mar15,2015at16:36  |  Show9morecomments 16Answers 16 Sortedby: Resettodefault Highestscore(default) Datemodified(newestfirst) Datecreated(oldestfirst) 176 +100 $\begingroup$ Here’sonetogetthingsstarted. Let$U$beanon-emptyopensubsetof$\BbbR$.For$x,y\inU$define$x\simy$iff$\big[\min\{x,y\},\max\{x,y\}\big]\subseteqU$.It’seasilycheckedthat$\sim$isanequivalencerelationon$U$whoseequivalenceclassesarepairwisedisjointopenintervalsin$\BbbR$.(Thetermintervalhereincludesunboundedintervals,i.e.,rays.)Let$\mathscr{I}$bethesetof$\sim$-classes.Clearly$U=\bigcup_{I\in\mathscr{I}}I$.Foreach$I\in\mathscr{I}$choosearational$q_I\inI$;themap$\mathscr{I}\to\BbbQ:I\mapstoq_I$isinjective,so$\mathscr{I}$iscountable. Avariantofthesamebasicideaistolet$\mathscr{I}$bethesetofopenintervalsthataresubsetsof$U$.For$I,J\in\mathscr{I}$define$I\simJ$iffthereare$I_0=I,I_1,\dots,I_n=J\in\mathscr{I}$suchthat$I_k\capI_{k+1}\ne\varnothing$for$k=0,\dots,n-1$.Then$\sim$isanequivalencerelationon$\mathscr{I}$.For$I\in\mathscr{I}$let$[I]$bethe$\sim$-classof$I$.Then$\left\{\bigcup[I]:I\in\mathscr{I}\right\}$isadecompositionof$U$intopairwisedisjointopenintervals. BothoftheseargumentsgeneralizetoanyLOTS(=LinearlyOrderedTopologicalSpace),i.e.,anylinearlyorderedset$\langleX,\le\rangle$withthetopologygeneratedbythesubbaseofopenrays$(\leftarrow,x)$and$(x,\to)$:if$U$isanon-emptyopensubsetof$X$,then$U$istheunionofafamilyofpairwisedisjointopenintervals.Ingeneralthefamilyneednotbecountable,ofcourse. Share Cite Follow editedJun26,2015at14:00 communitywiki 4revs,3users80%BrianM.Scott $\endgroup$ 10 22 $\begingroup$ IlikethisanswerLOTS.:-)And+1. $\endgroup$ – coffeemath Mar2,2013at1:30 2 $\begingroup$ Thisanswerisveryclear.Itdependsontheaxiomofchoicethough,arethereanyconstructivevariantsofthisargument? $\endgroup$ – Bunder Mar2,2013at8:28 8 $\begingroup$ @Bunder:No,itdoesn’tdependontheaxiomofchoice.TheonlyplacewhereyoumighteventhinkthatitdidiswhenIshowedthatthereareonlycountablymanyintervals,buttherationalscanbeexplicitlywell-ordered,sonochoiceisnecessaryeventhere. $\endgroup$ – BrianM.Scott Mar2,2013at12:24 1 $\begingroup$ Whyaretheequivalenceclassesopen? $\endgroup$ – IntegrateThis Sep20,2018at22:14 3 $\begingroup$ @IntegrateThis.Let$S$bea$\sim$-equivalenceclass.Forany$x\inS$thereexists$r>0$suchthat$(-r+x,r+x)\subsetU$so$\forally\in(-r+x,r+x)\;(y\simx).$So$(-r+x,r+x)\subset\{y\inU:y\simx\}=S.$ $\endgroup$ – DanielWainfleet Sep24,2018at20:38  |  Show5morecomments 64 $\begingroup$ Theseanswersallseemtobevariationsononeanother,butI'vefoundeachonesofartobeatleastalittlecryptic.Here'smyversion/adaptation. Let$U\subseteq\mathbb{R}$beopenandlet$x\inU$.Either$x$isrationalorirrational.If$x$isrational,define \begin{align}I_x=\bigcup\limits_{\substack{I\text{anopeninterval}\\x~\in~I~\subseteq~U}}I,\end{align} which,asaunionofnon-disjointopenintervals(each$I$contains$x$),isanopenintervalsubsetto$U$.If$x$isirrational,byopennessof$U$thereis$\varepsilon>0$suchthat$(x-\varepsilon,x+\varepsilon)\subseteqU$,andthereexistsrational$y\in(x-\varepsilon,x+\varepsilon)\subseteqI_y$(bythedefinitionof$I_y$).Hence$x\inI_y$.Soany$x\inU$isin$I_q$forsome$q\inU\cap\mathbb{Q}$,andso \begin{align}U\subseteq\bigcup\limits_{q~\in~U\cap~\mathbb{Q}}I_q.\end{align} But$I_q\subseteqU$foreach$q\inU\cap\mathbb{Q}$;thus \begin{align}U=\bigcup\limits_{q~\in~U\cap~\mathbb{Q}}I_q,\end{align} whichisacountableunionofopenintervals. Share Cite Follow editedMay3at23:14 communitywiki 2revs,2users94%Stromael $\endgroup$ 2 1 $\begingroup$ Hewantedacountableunionofdisjointopenintervals.If$U$issimplyanopeninterval,then$U=I_q=I_{\tilde{q}}$forall$q,\tilde{q}\in\mathcal{Q}\capU$. $\endgroup$ – JoséSiqueira Oct15,2013at16:15 8 $\begingroup$ Thatitisadisjointunionfollowsfromthedefinitionof$I_q$:if$x\inI_q\capI_p$then$I_q\cupI_p\subseteqI_q$and$I_p$;henceif$I_q\neqI_p$then$I_q\capI_p=\emptyset$.Strictlyspeakingoneshouldthrowawayallrepeated$I_q$s,andonecandefinitelydothiswithoutdestroyingthecountabilityoftheunion. $\endgroup$ – Stromael Oct21,2013at11:14 Addacomment  |  22 $\begingroup$ Inalocallyconnectedspace$X$,allconnectedcomponentsofopensetsareopen.Thisisinfactequivalenttobeinglocallyconnected. Proof:(onedirection)let$O$beanopensubsetofalocallyconnectedspace$X$.Let$C$beacomponentof$O$(asa(sub)spaceinitsownright).Let$x\inC$.Thenlet$U_x$beaconnectedneighbourhoodof$x$in$X$suchthat$U_x\subsetO$,whichcanbedoneas$O$isopenandtheconnectedneighbourhoodsformalocalbase.Then$U_x,C\subsetO$arebothconnectedandintersect(in$x$)sotheirunion$U_x\cupC\subsetO$isaconnectedsubsetof$O$containing$x$,sobymaximalityofcomponents$U_x\cupC\subsetC$.Butthen$U_x$witnessesthat$x$isaninteriorpointof$C$,andthisshowsallpointsof$C$areinteriorpoints,hence$C$isopen(ineither$X$or$O$,that'sequivalent). Now$\mathbb{R}$islocallyconnected(openintervalsformalocalbaseofconnectedsets)andsoeveryopensetifadisjointunionofitscomponents,whichareopenconnectedsubsetsof$\mathbb{R}$,henceareopenintervals(potentiallyofinfinite"length",i.e.segments).Thattherearecountablymanyofthematmost,followsfromthealreadygiven"rationalineveryinterval"argument. Share Cite Follow answeredMar2,2013at7:06 communitywiki HennoBrandsma $\endgroup$ 2 $\begingroup$ "connected=>interval"alsorequiresaproof(usuallyusingtheintermediatestep"path-connected"). $\endgroup$ – MartinBrandenburg Mar4,2013at19:29 1 $\begingroup$ Pathconnectedisnotneeded.Itwouldbecircular,asoneneedsconnectednessofintervalstoseethatpath-connectedimpliesconnected...Ifasetisnotorderconvex(sox0\}$and$\beta_x=\sup\{\beta\in\mathbbR:(\alpha_x,\beta)\subseteqU\}$. Then$\displaystyleU=\bigcup_{x\inU}(\alpha_x,\beta_x)$where$\{(\alpha_x,\beta_x):x\inU\}$isadisjointfamilyofopenintervals. Theintervalsappearingintheunionaredisjointinthesensethateverytime$x,y\inU$with$x0\},$$$$\alpha_y=\inf\{\overline\alpha\leqx:(\overline\alpha,y+\overline\epsilon)\subseteqU,\text{forsome}\overline\epsilon>0\},$$andthesearethesame;sothenalso$\beta_x$and$\beta_y$arethesame. Share Cite Follow editedDec27,2021at1:55 communitywiki 4revs,2users70%P.. $\endgroup$ 3 1 $\begingroup$ Areyousuretheunioniscountablehere? $\endgroup$ – user10444 Sep11,2014at16:55 3 $\begingroup$ @user10444:Itiscountableyes!Ifyouagreethat$\{(\alpha_x,\beta_x):x\inU\}$isadisjointfamilyofopenintervalsthenyoucanseeitbychoosing$r_x\in\mathbbQ\cap(\alpha_x,\beta_x)$forall$x\inU$.Then$\{r_x:x\inU\}$iscountableright?Notethattheintervals$\{(\alpha_x,\beta_x):x\inU\}$arenotdistinct! $\endgroup$ – P.. Sep11,2014at19:03 3 $\begingroup$ Recall,wecan"choose"suchan$r_{x}\in(\alpha_{x},\beta_{x})$suchthat$r_{x}\in\mathbb{Q}$bytheAxiomofChoiceandbythedensityof$\mathbb{Q}$in$\mathbb{R}$. $\endgroup$ – Procore Aug30,2017at20:29 Addacomment  |  15 $\begingroup$ Thisproofisanextendedversionoftheniceproofproposedby Stromaelanditservesbestforbeginnerswhowanttounderstandeverydetail(thatonethatforanyestablishedmathematicianlogicallyseemstrivial)oftheproof. $\textbf{Proof:}$ Let$U\subseteq\mathbb{R}$beopenandlet$x\inU$.ThenEither$x$isrationalor$x$isirrational. Suppose$x$isrational,thendefine \begin{align}I_x=\bigcup\limits_{\substack{I\text{anopeninterval}\\x~\in~I~\subseteq~U}}I,\end{align} Claim:$I_x$isinterval,$I_x$isopenand$I_x\subseteqU$ Definition:Anintervalisasubset$I\subseteq\mathbb{R}$suchthat,forall$ax$:If$c>x$thenwehavethateither$a0$suchthat$(x-\varepsilon,x+\varepsilon)\subseteqU$,andbythepropertyofrealnumbersthatforanyirrationalnumberthereexistsasequenceofrationalunmbersthatconvergestothatirrationalnumber,thereexistsrational$y\in(x-\varepsilon,x+\varepsilon)$.Thenbyconstruction$(x-\varepsilon,x+\varepsilon)\subseteqI_y$.Hence$x\inI_y$.Soany$x\inU$isin$I_q$forsome$q\inU\cap\mathbb{Q}$,andso \begin{align}U\subseteq\bigcup\limits_{q~\in~U\cap~\mathbb{Q}}I_q.\end{align} But$I_q\subseteqU$foreach$q\inU\cap\mathbb{Q}$;thus \begin{align}U=\bigcup\limits_{q~\in~U\cap~\mathbb{Q}}I_q,\end{align} whichisacountableunionofopenintervals. Nowlet'sshowthatintervals$\{I_q\}~\q\inU\cap\mathbb{Q}$aredisjoint.Supposethereis$i,j,\inU\cap\mathbb{Q}$suchthat$I_i\capI_j\neq\emptyset$then$I_i\subseteqI_q$and$I_j\subseteqI_q$forsome$q\inU\cap\mathbb{Q}$ Henceweconstructeddisjointintervals$\{I_q\}~\q\inU\cap\mathbb{Q}$thatareenumeratedbyrationalnumbersin$U$andwhoseunionis$U$.Sinceanysubsetofrationalnumbersiscountable,$\{I_q\}~\q\inU\cap\mathbb{Q}$iscountableaswell.Thisfinishestheproof. Share Cite Follow editedMay3at23:15 communitywiki 4revs,4users96%G.T. $\endgroup$ 3 1 $\begingroup$ Whatistheguaranteeofthestatement'Denote$I_a$tobeanintervalsuchthat$x\inI_a$'? $\endgroup$ – user464147 Dec30,2018at14:47 $\begingroup$ Alsotomethefirstpartofthisanswerseemedobscure,Itriedtoclarifyitheremath.stackexchange.com/questions/3040319/… $\endgroup$ – JackJ. Jan17,2019at9:28 $\begingroup$ MostprobablyIamverylateforthiscommentbutstillIwillclarify.Byconstruction$I_x$issuchthat$x\in$every$I$thatforms$I_x$.ThenIsaythat$a\inI_x$thisimpliesthat$a$mustbeamemberofoneofthe$I$sthatform$I_x$.Why?Becauseifthisisnotthecasethen$a$isnotin$I_x$.ThenIjusttakeandcall$I_a$oneofthe$I$sthatcontainthat$a$.Since$x\in$everyI$x$willbeamemberofthatparticular$I$thatwedefinedas$I_a$Letmeknowifthisiswrongandwhy. $\endgroup$ – G.T. Sep5,2020at7:15 Addacomment  |  9 $\begingroup$ Avariantoftheusualproofwiththeequivalencerelation,whichtradesintheeaseofconstructingtheintervalswiththeeaseofprovingcountability(notthateitherishard...): Definethesameequivalencerelation,butonlyon$\mathbbQ\capU$: $q_1\simq_2$iff$(q_1,q_2)\subsetU$(or$(q_2,q_1)\subsetU$,whichevermakessense). Fromeachequivalencyclass$C$,producetheopeninterval$(\infC,\supC)\subsetU$(where$\infC$isdefinedtobe$-\infty$incase$C$isnotboundedfrombelow,and$\supC=\infty$incase$C$isnotboundedfromabove). Theamountofequivalenceclassesisclearlycountable,since$\mathbbQ\capU$iscountable. Share Cite Follow answeredMar2,2013at1:28 communitywiki YoniRozenshein $\endgroup$ Addacomment  |  8 $\begingroup$ Let$U$beanopensubsetof$\mathbb{R}$.Let$P$betheposetconsistingofcollections$\mathcal{A}$ofdisjointopenintervalswherewesay$\mathcal{A}\le\mathcal{A}'$ifeachofthesetsin$\mathcal{A}$isasubsetofsomeopenintervalin$\mathcal{A}'$.Everychain$C$inthisposethasanupperbound,namely$$\mathcal{B}=\left\{\bigcup\left\{J\in\bigcup\bigcupC:I\subseteqJ\right\}:I\in\bigcup\bigcupC\right\}.$$ ThereforebyZorn'slemmatheposet$P$hasamaximalelement$\mathcal{M}$.Weclaimthattheunionoftheintervalsin$\mathcal{M}$isallof$U$.Supposetowardacontradictionthatthereisareal$x\inU$thatisnotcontainedinanyoftheintervalsin$\mathcal{M}$.Because$U$isopenwecantakeanopeninterval$I$with$x\inI\subseteqU$. Thentheset $$\mathcal{M}'=\{J\in\mathcal{M}:J\capI=\emptyset\}\cup\left\{I\cup\bigcup\{J\in\mathcal{M}:J\capI\ne\emptyset\}\right\}$$ isacollectionofdisjointopenintervalsand isabove$\mathcal{M}$intheposet$P$,contradictingthemaximalityof$\mathcal{M}$.Itremainstoobservethat$\mathcal{M}$iscountable,whichfollowsfromthefactthatitselementscontaindistinctrationalnumbers. Notethattheonlywayinwhichanythingaboutorder(orconnectedness)isusedistoseethat$I\cup\bigcup\{J\in\mathcal{M}:J\capI\ne\emptyset\}$isaninterval. Share Cite Follow editedMar8,2013at2:17 communitywiki 2revsTrevorWilson $\endgroup$ 3 $\begingroup$ Andyes,beforeyouask,Iknowthisproofissilly. $\endgroup$ – TrevorWilson Mar8,2013at2:21 2 $\begingroup$ Ithinkthisproofconceptuallyconnectsmanythings,includingwell-ordering,ordertheoryandsoon.So,Ireallybenefited. $\endgroup$ – user64066 Oct22,2013at20:19 $\begingroup$ @user64066Thanks,I'mgladtohearit. $\endgroup$ – TrevorWilson Oct22,2013at20:30 Addacomment  |  7 $\begingroup$ $\mathbb{R}$withstandardtopologyissecond-countablespace. Forasecond-countablespacewitha(notnecessarilycountable)base,anyopensetcanbewrittenasacountableunionofbasicopenset. Givenanybaseforasecondcountablespace,iseveryopensetthecountableunionofbasicopensets? Clearly,collectionofopenintervalsisabaseforthestandardtopology. Henceanyopensetin$\mathbb{R}$canbewrittenascountableunionofopenintervals. Ifanytwoofexploitedopenintervalsoverlap,mergethem.Thenwehavedisjointunionofopenintervals,whichisstillcountable. Share Cite Follow editedApr13,2017at12:19 communitywiki 2revsGuldam $\endgroup$ Addacomment  |  6 $\begingroup$ Let$G$beanonemptyopensetin$\mathbb{R}$.Write$a\simb$iftheclosedinterval$[a,b]$or$[b,a]$if$b\delta_1\geq\gamma_1,\beta>\delta_2\geq\gamma_2$s.t.$J^{\gamma_1}_{\delta_1}\capJ^{\gamma_2}_{\delta_2}\neq\emptyset$,meaningthateither$J^{\gamma_1}_{\delta_2}=J^{\gamma_2}_{\delta_2}$or$J^{\gamma_1}_{\delta_1}=J^{\gamma_2}_{\delta_1}$thus,$\forall\beta>\epsilon\geq\delta_1,\delta_2$,$J^{\gamma_1}_{\epsilon}=J^{\gamma_2}_{\epsilon}$andsincewearetalkinghereaboutnon-decreasingsequences,thiswillcontradict$\widetilde{J}_{\gamma_1}\neq\widetilde{J}_{\gamma_2}$).Andif$I_\beta$isn'tdisjointofall$\widetilde{J}_\gamma$,Thenwecantake$J_\beta^\gamma=\widetilde{J}_\gamma$forall$\gamma0$suchthat$(y-\epsilon,y+\epsilon)\subseteqU$. Setsofrealnumbersareconnectedifftheyareintervals,singletonsorempty. $(y-\epsilon,y+\epsilon)$anintervalhenceitisconnected. Thereforesince$U_y$isthelargestconnectedsubsetof$U$containing$y$wemusthave$(y-\epsilon,y+\epsilon)\subseteqU_y=U_x$. Thisshowsthat$U_x$isopenforall$x$. $U_x$openandconnectedimpliesthat$U_x$mustbeanopeninterval. Also$\mathbb{Q}$densein$\mathbb{R}$, so$\forallx\inU$,$U_x\cap\mathbb{Q}\neq\varnothing$and$U_x=U_q$forsome$q\in\mathbb{Q}$. Sowecanwrite$\{U_x\}_{x\inU}=\{U_q\}_{q\inS}$forsome$S\subseteq\mathbb{Q}$. $\mathbb{Q}$iscountableso$S$isatmostcountable. Inconclusion,wehavejustshownthattheunionoftheconnectedcomponentsof$U$isadisjointunionofopenintervalsthatequals$U$andisatmostcountable. Share Cite Follow answeredMay31,2017at22:25 communitywiki NathanA.S. $\endgroup$ Addacomment  |  4 $\begingroup$ Theproofthateveryopensetisadisjointunionofcountablymanyopenintervalsreliesonthreefacts: $\BbbR$islocally-connected $\BbbR$isccc Theopenconnectedsetsin$\BbbR$areopenintervals Let$U\subseteq\BbbR$beopen.Thenthereisacollectionofdisjoint,open,connectedsets$\{G_\alpha\}_{\alpha\inA}$suchthat$U=\bigcup_{\alpha\inA}G_\alpha$.Since$\BbbR$isccc,thecollection$\{G_\alpha\}$isatmostcountable.Sincetheopenconnectedsets$\BbbR$areopenintervals,$\{G_\alpha\}$isacountablecollectionofdisjoint,openintervals. Thefirsttwofactsallowustoseesomegeneralizations.Namelyanyopensetinalocally-connected,cccspaceisacountabledisjointunionofconnectedopensets.ThisappliestoanyEuclideanspace.AlthoughopenconnectedsubsetsofEuclideanspacearemorecomplicatedthanopenintervals,theyarestillrelativelywell-behaved. Share Cite Follow answeredJul31,2014at18:16 communitywiki user123641 $\endgroup$ 4 $\begingroup$ Whatdoes"Risccc"mean? $\endgroup$ – ChristianBueno Feb24,2015at8:39 $\begingroup$ Itmeans"satisfiesthecountable-chaincondition"whichreallyconcernsitselfwithanti-chains.Itmeansthateverycollectionofdisjointopensetsisatmostcountable.Everyseparablespaceisccc. $\endgroup$ – user123641 Feb25,2015at0:56 $\begingroup$ Okthanksforclearingthatup. $\endgroup$ – ChristianBueno Feb25,2015at5:18 $\begingroup$ howtodecompose(0,1}intoacountableunionofdisjointopenintervals? $\endgroup$ – Bearandbunny Mar27,2015at15:31 Addacomment  |  4 $\begingroup$ Thefollowingiscertainlynotthequickestapproachtoaproof,butwhenthisquestionwasfirstposedtomeinclass,myfirstintuitionwastousesomeelementarygraphtheory: Let$U$beanopensetof$\mathbb{R}$.Asweknow,$\mathbb{R}$hasacountablebasis$\mathcal{B}$comprisedofconnectedopensetsandsowemaywrite$U=\bigcup_{n\inI}U_n$,whereforeach$n$wehave$U_n\in\mathcal{B}$and$I$issomecountableindexset. Let$G$betheintersectiongraphof$\{U_n\}$.Thatistosay,thevertexsetof$G$issimply$\{U_n\}$andthereisanedgebetween$U_i$and$U_j$ifftheyhavenonemptyintersection.It'seasytoconvinceyourselfthat: Thisgraphmusthavecountablymanygraphically-connectedcomponents(otherwisewe'dhaveuncountablymanyverticeswhichisimpossible). Theintersectiongraphof$A\subseteq\{U_n\}$isgraphically-connectediffforanytwo$V,W\inA$thereisasequence$V=U_{n_1},U_{n_2},\ldots,U_{n_k}=W$suchthat$U_{n_i}\capU_{n_{i+1}}\neq\varnothing$. Theunion$\bigcupA$isaconnectedsetof$\mathbb{R}$whenevertheintersectiongraphof$A$isgraphically-connected. Thus,whenwetaketheunionofalltheverticeswithinagraphically-connectedcomponent,foreverycomponent,weobtaincountably-manyconnectedopensets.Theunionofthesesetsisofcourse$U$itself.Sincetheconnectedopensetsof$\mathbb{R}$areintervals(includingrays),we'redone. SideNote:Thiswouldalsoworkin$\mathbb{R}^n$oringeneral,anytopologicalspace$X$thathasacountablebasiscomprisedofconnectedsets.Well,solongaswereplacecountableunionofdisjointopenintervalswithcountableunionofdisjointopenconnectedsets. Share Cite Follow answeredFeb24,2015at8:38 communitywiki ChristianBueno $\endgroup$ Addacomment  |  4 $\begingroup$ Theballswithradii$\frac{1}{n}$andcenteratarationalnumberformabasisfortheeuclideantopology.Thisfamilyiscountablesince$\mathbbN\times\mathbbQ\equiv\mathbbN$andwehaveacountablebasis$(B_\lambda)_{\,\lambda\in\mathbbN\times\mathbbQ}$ofopenintervalsfor$\mathbbR$. Let$U$beanonemptyopensetin$\mathbbR$;wecanexpressitascountableunionofopenballsfrom$(B_\lambda)$. Also, $\tag1\text{}$ $\quad$Iftwoopenintervalshaveanonemptyintersection,thentheirunionisalsoanopeninterval. Soif$U$isafiniteunionofthe$B_\lambda$,itisaneasymattertocombinethe$B_\lambda$,ifnecessary,andwriting$U$asadisjointunionofafinitenumberofopenintervals.Soassume,WLOG,that $\tag2U=\bigcup_{\,n\in\mathbbN\,}B_n$. Wedefinearelationonour(new)indexset$\mathbbN$with$m\simn$if$B_m\capB_n\ne\emptyset$orthereisisafinite'nonemptyintersection$B\text{-}$chain'connecting$B_m$with$B_n$.Itiseasytoseethatthispartitions$\mathbbN$andthattakingthecorrespondingunionsofthe$B_n$overanindex$\lambda\text{-}$blockgivesapartitionof$U$.Also,using(1),wecanshowthatwehavealsoexpressed$U$asacountableunionofdisjointopenintervals. Share Cite Follow editedNov6,2017at19:32 communitywiki 3revsMikeMathMan $\endgroup$ Addacomment  |  3 $\begingroup$ Moreofaquestionthananswer.Iamachemistturnedpharmacist,whowishestohavestudiedmathematics.IamtryingtoworkthroughRudin'sPrinciplesofMathematicalAnalysis.Ienvyyouallwhoareinvolvedinmathforacareer. Cansomeonegivemefeedbackonmyattemptataproofin$\mathbbR?Completelynoviceandnotatallpretty,butisitsound? segment:=openintervalinR. Lemma:disjointsegmentsinRareseparated(proofnotshown). ItfollowsfromthelemmathatanopenconnectedsubsetofRcannotbetheunionofdisjointsegments. LetEbeanopensubsetofR.SinceRisseparablebyRudinprob2.22,thereexistsasubset,D,ofRthatiscountableanddenseinR.AssumeEisconnected,whichincludesE=R,thenEistheunionofanatmostcountablecollectionofopensegments,containingonlyE. SupposeEisseparated.ThenEistheunionofacollectionofdisjointsegments,includingthepossibilityofsegmentsunboundedaboveorbelow. Ifthecollectionisfinite,thenitisatmostcountable. Assumethecollectionofsegmentsisinfinite.BecauseDisdenseinRandEiscontainedinR,everyopensubsetofEcontainsapointofD.TheneachoftheinfinitelymanydisjointsegmentscontainsauniquepointofD.SinceDiscountable,aone-to-onecorrespondencebetweenauniquepointofDandthesegmentisestablished.Thisimpliesthattherearecountablymanydisjointsegmentsinthecollection. ThereforeEistheunionoffinitelymanyorcountablymany ,henceatmostcountablymany,disjointsegments. Share Cite Follow editedSep24,2018at17:43 communitywiki 3revs,2users97%RJM $\endgroup$ 6 1 $\begingroup$ Moreofaquestionthananswer.Isupposethiswouldbebetterpostedasaquestionthanasananswer,then.SupposeEisseparated.ThenEistheunionofacollectionofdisjointsegments.It'snotobvious(tome,atleast)wherethisfollowsfrominthegivencontext. $\endgroup$ – dxiv Oct9,2016at5:43 1 $\begingroup$ Iknow.Justlearninghowthingsworkonthissite.WhichIthinkisquiteamazing. $\endgroup$ – RJM Oct9,2016at5:48 $\begingroup$ @dxivSinceEseparated(ornotconnected),wouldithavetoatleastbetheunionof2separatedsets.Sinceitisopen,theyhavetobeopensubsets,whichimpliesopenintervalinR1.Thankyou! $\endgroup$ – RJM Oct9,2016at5:53 $\begingroup$ @dxiviwillnotkeepbuggingyou.Anytipsonhowtobecomemathematician-likeonone'sowninessentiallymatgematicaisolation? $\endgroup$ – RJM Oct9,2016at6:00 $\begingroup$ IMHOthatlatterimpliesopenintervalinR1isapartthatneedstobeprovedinthiscontext.SeeforexampleIntervalsareconnectedandtheonlyconnectedsetsinR.Anyway,asIsaid,yoursappearstobebettersuitedasaquestionthanasananswer.P.S.Keeppluggingawayuntilitallstartsmakingsense;-) $\endgroup$ – dxiv Oct9,2016at6:00  |  Show1morecomment YourAnswer ThanksforcontributingananswertoMathematicsStackExchange!Pleasebesuretoanswerthequestion.Providedetailsandshareyourresearch!Butavoid…Askingforhelp,clarification,orrespondingtootheranswers.Makingstatementsbasedonopinion;backthemupwithreferencesorpersonalexperience.UseMathJaxtoformatequations.MathJaxreference.Tolearnmore,seeourtipsonwritinggreatanswers. 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