Topological Spaces homeomorphicClosure of Topological SpacesRing with spectrum homeomorphic to a given...

Why is the SNP putting so much emphasis on currency plans?

How to back up a running Linode server?

Stark VS Thanos

How did Captain America use this power?

Floor tile layout process?

Applying a function to a nested list

Junior developer struggles: how to communicate with management?

Pressure to defend the relevance of one's area of mathematics

Accidentally deleted the "/usr/share" folder

A non-technological, repeating, phenomenon in the sky, holding its position in the sky for hours

Power LED from 3.3V Power Pin without Resistor

Entropy as a function of temperature: is temperature well defined?

Write to EXCEL from SQL DB using VBA script

Can PCs use nonmagical armor and weapons looted from monsters?

Meaning of "individuandum"

Why do money exchangers give different rates to different bills

CRT Oscilloscope - part of the plot is missing

How long can a 35mm film be used/stored before it starts to lose its quality after expiry?

Selecting a secure PIN for building access

Why was Germany not as successful as other Europeans in establishing overseas colonies?

Is lying to get "gardening leave" fraud?

Why is Thanos so tough at the beginning of "Avengers: Endgame"?

How can I close a gap between my fence and my neighbor's that's on his side of the property line?

Unexpected email from Yorkshire Bank



Topological Spaces homeomorphic


Closure of Topological SpacesRing with spectrum homeomorphic to a given topological spaceIs there a general way to tell whether two topological spaces are homeomorphic?Do topological spaces deserve their name?What topological properties are invariant under diffeomorphism?Quotient spaces not homeomorphicProof that $S^1$ isn't homeomorphic to $[0, 1]$Homeomorphism of union of infinitely many topological spacesSets homeomorphic to convex setsHomeomorphic Topological Spaces with the Subspace Topology













1












$begingroup$


I am new on this topology stuff, so came to ask for help, my intuition tells me that I need to use a topological invariant to see when they are homeomorphic but I am not quite sure how to proceed.



Prove that $Bbb R$$cong Bbb {R^n}$ if only if n=1 and $Bbb {S^1} cong Bbb {S^n}$ if and only if n=1






general-topology







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am new on this topology stuff, so came to ask for help, my intuition tells me that I need to use a topological invariant to see when they are homeomorphic but I am not quite sure how to proceed.



    Prove that $Bbb R$$cong Bbb {R^n}$ if only if n=1 and $Bbb {S^1} cong Bbb {S^n}$ if and only if n=1






    general-topology







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am new on this topology stuff, so came to ask for help, my intuition tells me that I need to use a topological invariant to see when they are homeomorphic but I am not quite sure how to proceed.



      Prove that $Bbb R$$cong Bbb {R^n}$ if only if n=1 and $Bbb {S^1} cong Bbb {S^n}$ if and only if n=1






      general-topology







      share|cite|improve this question









      $endgroup$




      I am new on this topology stuff, so came to ask for help, my intuition tells me that I need to use a topological invariant to see when they are homeomorphic but I am not quite sure how to proceed.



      Prove that $Bbb R$$cong Bbb {R^n}$ if only if n=1 and $Bbb {S^1} cong Bbb {S^n}$ if and only if n=1






      general-topology




      general-topology






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 hours ago









      Frank SambeFrank Sambe

      174




      174






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          A more elementary way than what Chris Custer suggests is this: if we remove a point from $mathbb{R}$, we get a disconnected set. But if $n>1$, $mathbb{R}^n$ minus one point remains connected (it is easy to prove this). For the other part, note that $mathbb{S}^n$ minus one point is homeomorphic to $mathbb{R}^n$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            You're right. Connectedness is a little more elementary.
            $endgroup$
            – Chris Custer
            1 hour ago



















          2












          $begingroup$

          Suppose there is such homeomorphism $f:Bbb RtoBbb R^n$, then setting $A:=(-infty,0)cup(0,infty)$ it is enough to check that $Bbb R^nsetminus {f(0)}$ is connected for $n>1$ but $A$ is disconnected, hence $f^{-1}$ cannot be continuous, so it is not an homeomorphism.



          For the other case you can proceed analogously noticing that $Bbb S^nsetminus{a}congBbb R^n$ via stereographic projection for any chosen $ainBbb S^n$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            $[a,b]$ is compact.
            $endgroup$
            – Chris Custer
            1 hour ago










          • $begingroup$
            @ChrisCuster right, good catch. Fixed
            $endgroup$
            – Masacroso
            1 hour ago



















          1












          $begingroup$

          One nice way is to use the invariant of homology. $H_n(S^n)congBbb Z$ and $H_m(S^n)cong0,,mneq n,0$.



          $Bbb R^n$ can then be done using the one-point compactification.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Considering "I am new to this topology stuff", this is probably too much although the quickest.
            $endgroup$
            – IAmNoOne
            1 hour ago










          • $begingroup$
            hahhaha yes, it is too much to me!
            $endgroup$
            – Frank Sambe
            1 hour ago












          Your Answer








          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%2f3207587%2ftopological-spaces-homeomorphic%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          A more elementary way than what Chris Custer suggests is this: if we remove a point from $mathbb{R}$, we get a disconnected set. But if $n>1$, $mathbb{R}^n$ minus one point remains connected (it is easy to prove this). For the other part, note that $mathbb{S}^n$ minus one point is homeomorphic to $mathbb{R}^n$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            You're right. Connectedness is a little more elementary.
            $endgroup$
            – Chris Custer
            1 hour ago
















          3












          $begingroup$

          A more elementary way than what Chris Custer suggests is this: if we remove a point from $mathbb{R}$, we get a disconnected set. But if $n>1$, $mathbb{R}^n$ minus one point remains connected (it is easy to prove this). For the other part, note that $mathbb{S}^n$ minus one point is homeomorphic to $mathbb{R}^n$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            You're right. Connectedness is a little more elementary.
            $endgroup$
            – Chris Custer
            1 hour ago














          3












          3








          3





          $begingroup$

          A more elementary way than what Chris Custer suggests is this: if we remove a point from $mathbb{R}$, we get a disconnected set. But if $n>1$, $mathbb{R}^n$ minus one point remains connected (it is easy to prove this). For the other part, note that $mathbb{S}^n$ minus one point is homeomorphic to $mathbb{R}^n$.






          share|cite|improve this answer









          $endgroup$



          A more elementary way than what Chris Custer suggests is this: if we remove a point from $mathbb{R}$, we get a disconnected set. But if $n>1$, $mathbb{R}^n$ minus one point remains connected (it is easy to prove this). For the other part, note that $mathbb{S}^n$ minus one point is homeomorphic to $mathbb{R}^n$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          JustDroppedInJustDroppedIn

          2,467420




          2,467420








          • 1




            $begingroup$
            You're right. Connectedness is a little more elementary.
            $endgroup$
            – Chris Custer
            1 hour ago














          • 1




            $begingroup$
            You're right. Connectedness is a little more elementary.
            $endgroup$
            – Chris Custer
            1 hour ago








          1




          1




          $begingroup$
          You're right. Connectedness is a little more elementary.
          $endgroup$
          – Chris Custer
          1 hour ago




          $begingroup$
          You're right. Connectedness is a little more elementary.
          $endgroup$
          – Chris Custer
          1 hour ago











          2












          $begingroup$

          Suppose there is such homeomorphism $f:Bbb RtoBbb R^n$, then setting $A:=(-infty,0)cup(0,infty)$ it is enough to check that $Bbb R^nsetminus {f(0)}$ is connected for $n>1$ but $A$ is disconnected, hence $f^{-1}$ cannot be continuous, so it is not an homeomorphism.



          For the other case you can proceed analogously noticing that $Bbb S^nsetminus{a}congBbb R^n$ via stereographic projection for any chosen $ainBbb S^n$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            $[a,b]$ is compact.
            $endgroup$
            – Chris Custer
            1 hour ago










          • $begingroup$
            @ChrisCuster right, good catch. Fixed
            $endgroup$
            – Masacroso
            1 hour ago
















          2












          $begingroup$

          Suppose there is such homeomorphism $f:Bbb RtoBbb R^n$, then setting $A:=(-infty,0)cup(0,infty)$ it is enough to check that $Bbb R^nsetminus {f(0)}$ is connected for $n>1$ but $A$ is disconnected, hence $f^{-1}$ cannot be continuous, so it is not an homeomorphism.



          For the other case you can proceed analogously noticing that $Bbb S^nsetminus{a}congBbb R^n$ via stereographic projection for any chosen $ainBbb S^n$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            $[a,b]$ is compact.
            $endgroup$
            – Chris Custer
            1 hour ago










          • $begingroup$
            @ChrisCuster right, good catch. Fixed
            $endgroup$
            – Masacroso
            1 hour ago














          2












          2








          2





          $begingroup$

          Suppose there is such homeomorphism $f:Bbb RtoBbb R^n$, then setting $A:=(-infty,0)cup(0,infty)$ it is enough to check that $Bbb R^nsetminus {f(0)}$ is connected for $n>1$ but $A$ is disconnected, hence $f^{-1}$ cannot be continuous, so it is not an homeomorphism.



          For the other case you can proceed analogously noticing that $Bbb S^nsetminus{a}congBbb R^n$ via stereographic projection for any chosen $ainBbb S^n$.






          share|cite|improve this answer











          $endgroup$



          Suppose there is such homeomorphism $f:Bbb RtoBbb R^n$, then setting $A:=(-infty,0)cup(0,infty)$ it is enough to check that $Bbb R^nsetminus {f(0)}$ is connected for $n>1$ but $A$ is disconnected, hence $f^{-1}$ cannot be continuous, so it is not an homeomorphism.



          For the other case you can proceed analogously noticing that $Bbb S^nsetminus{a}congBbb R^n$ via stereographic projection for any chosen $ainBbb S^n$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 2 hours ago









          MasacrosoMasacroso

          13.3k41749




          13.3k41749








          • 1




            $begingroup$
            $[a,b]$ is compact.
            $endgroup$
            – Chris Custer
            1 hour ago










          • $begingroup$
            @ChrisCuster right, good catch. Fixed
            $endgroup$
            – Masacroso
            1 hour ago














          • 1




            $begingroup$
            $[a,b]$ is compact.
            $endgroup$
            – Chris Custer
            1 hour ago










          • $begingroup$
            @ChrisCuster right, good catch. Fixed
            $endgroup$
            – Masacroso
            1 hour ago








          1




          1




          $begingroup$
          $[a,b]$ is compact.
          $endgroup$
          – Chris Custer
          1 hour ago




          $begingroup$
          $[a,b]$ is compact.
          $endgroup$
          – Chris Custer
          1 hour ago












          $begingroup$
          @ChrisCuster right, good catch. Fixed
          $endgroup$
          – Masacroso
          1 hour ago




          $begingroup$
          @ChrisCuster right, good catch. Fixed
          $endgroup$
          – Masacroso
          1 hour ago











          1












          $begingroup$

          One nice way is to use the invariant of homology. $H_n(S^n)congBbb Z$ and $H_m(S^n)cong0,,mneq n,0$.



          $Bbb R^n$ can then be done using the one-point compactification.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Considering "I am new to this topology stuff", this is probably too much although the quickest.
            $endgroup$
            – IAmNoOne
            1 hour ago










          • $begingroup$
            hahhaha yes, it is too much to me!
            $endgroup$
            – Frank Sambe
            1 hour ago
















          1












          $begingroup$

          One nice way is to use the invariant of homology. $H_n(S^n)congBbb Z$ and $H_m(S^n)cong0,,mneq n,0$.



          $Bbb R^n$ can then be done using the one-point compactification.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Considering "I am new to this topology stuff", this is probably too much although the quickest.
            $endgroup$
            – IAmNoOne
            1 hour ago










          • $begingroup$
            hahhaha yes, it is too much to me!
            $endgroup$
            – Frank Sambe
            1 hour ago














          1












          1








          1





          $begingroup$

          One nice way is to use the invariant of homology. $H_n(S^n)congBbb Z$ and $H_m(S^n)cong0,,mneq n,0$.



          $Bbb R^n$ can then be done using the one-point compactification.






          share|cite|improve this answer









          $endgroup$



          One nice way is to use the invariant of homology. $H_n(S^n)congBbb Z$ and $H_m(S^n)cong0,,mneq n,0$.



          $Bbb R^n$ can then be done using the one-point compactification.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Chris CusterChris Custer

          14.9k3827




          14.9k3827












          • $begingroup$
            Considering "I am new to this topology stuff", this is probably too much although the quickest.
            $endgroup$
            – IAmNoOne
            1 hour ago










          • $begingroup$
            hahhaha yes, it is too much to me!
            $endgroup$
            – Frank Sambe
            1 hour ago


















          • $begingroup$
            Considering "I am new to this topology stuff", this is probably too much although the quickest.
            $endgroup$
            – IAmNoOne
            1 hour ago










          • $begingroup$
            hahhaha yes, it is too much to me!
            $endgroup$
            – Frank Sambe
            1 hour ago
















          $begingroup$
          Considering "I am new to this topology stuff", this is probably too much although the quickest.
          $endgroup$
          – IAmNoOne
          1 hour ago




          $begingroup$
          Considering "I am new to this topology stuff", this is probably too much although the quickest.
          $endgroup$
          – IAmNoOne
          1 hour ago












          $begingroup$
          hahhaha yes, it is too much to me!
          $endgroup$
          – Frank Sambe
          1 hour ago




          $begingroup$
          hahhaha yes, it is too much to me!
          $endgroup$
          – Frank Sambe
          1 hour ago


















          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%2f3207587%2ftopological-spaces-homeomorphic%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

          Gersau Kjelder | Navigasjonsmeny46°59′0″N 8°31′0″E46°59′0″N...

          Image
          Tidlegare republikkarSkipingar i 1433Kommunar i SchwyzDistrikt i SchwyzNedleggingar i 1798Stader ved Vierwaldstättersee kommunekantonenSchwyzSveitskm²VierwaldstätterseeLauerz Gersau Kommunevåpenet til Gersau Plassering Styresmakter Land   Sveits Kanton kantonen Schwyz Distrikt Gersau Språk tysk Geografi Flatevidd  - kommune 23,70 km² Innbyggjarar  - kommune (desember 2010)    - folketettleik 2 094   88,4 /km² Koordinatar 46°59′0″N 8°31′0″E  Koordinatar: 46°59′0″N 8°31′0″E  Høgd over havet 435 moh Diverse annan informasjon Postnummer 6442 Nettstad: www.gersau.ch Freistaat und Republik Gersau Fristaten og republikken Gersau By-republikk og alliert protektorat i Det gamle sveitsiske eidsforundet ←   ← 1433–1798 1814–1818 →   → Hovudstad Gersau Styreform Republikk Historisk periode Tidleg moderne tid  - Kjøpte fridom frå
          Read more

          Hestehale Innhaldsliste Hestehale på kvinner | Hestehale på menn | Galleri | Sjå òg |...

          Image
          Frisyrar frisyrehårhårstrikkhårklemmerøyrethesthovudverkeuropeisk motejobbtreningfletterparykkarungdomsopprørmotkulturmusikararmotorsykkelentusiastarpolitietpolitiskforretningverksemdfiskeflette Hestehale sett bakfrå. Hestehale på ein mannleg indianarleiar. Hestehale er ein frisyre der ein samlar det meste eller alt av langt hår [1] i bakhovudet med ein hårstrikk eller liknande. Avhengig av mote kan det òg vere vanleg å sette opp hestehalen med hårklemmer eller å ha hestehalen hengande på den eine sida av hovudet ved å samle håret ved øyret. Hestehale er ein vanleg frisyre for jenter og for gutar med langt hår. Han har fått namnet sitt etter likskapen med halen på ein hest. For stramme hestehaler kan føre til hårtap, [2] i nokre høve også hovudverk. [3] Innhaldsliste 1 Hestehale på kvinner 2 Hestehale på menn 3 Galleri 4 Sjå òg 5 Bakgrunnsstoff 6 Kjelder Hestehale på kvinner | Ungjente med hestehale som held håret unna ansiktet
          Read more

          What is the “three and three hundred thousand syndrome”?Who wrote the book Arena?What five creatures were...

          Image
          What happens when the centripetal force is equal and opposite to the centrifugal force? Why is "la Gestapo" feminine? Nested Dynamic SOQL Query How are passwords stolen from companies if they only store hashes? Exposing a company lying about themselves in a tightly knit industry: Is my career at risk on the long run? Can "few" be used as a subject? If so, what is the rule? Why are there no stars visible in cislunar space? categorizing a variable turns it from insignificant to significant Do I need an EFI partition for each 18.04 ubuntu I have on my HD? What are the consequences of changing the number of hours in
          Read more