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$Bbb R cong Bbb {R^n}$ iff $n=1$


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 and 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 and 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 and 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 and 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








      edited 5 mins ago









      user21820

      40.5k544163




      40.5k544163










      asked 3 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
            2 hours 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
            2 hours ago










          • $begingroup$
            @ChrisCuster right, good catch. Fixed
            $endgroup$
            – Masacroso
            2 hours ago



















          0












          $begingroup$

          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
            2 hours ago










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












          Your Answer








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          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
            2 hours 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
            2 hours 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 2 hours ago









          JustDroppedInJustDroppedIn

          2,467420




          2,467420








          • 1




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














          • 1




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








          1




          1




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




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










          • $begingroup$
            @ChrisCuster right, good catch. Fixed
            $endgroup$
            – Masacroso
            2 hours 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
            2 hours ago










          • $begingroup$
            @ChrisCuster right, good catch. Fixed
            $endgroup$
            – Masacroso
            2 hours 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 2 hours ago

























          answered 2 hours ago









          MasacrosoMasacroso

          13.3k41749




          13.3k41749








          • 1




            $begingroup$
            $[a,b]$ is compact.
            $endgroup$
            – Chris Custer
            2 hours ago










          • $begingroup$
            @ChrisCuster right, good catch. Fixed
            $endgroup$
            – Masacroso
            2 hours ago














          • 1




            $begingroup$
            $[a,b]$ is compact.
            $endgroup$
            – Chris Custer
            2 hours ago










          • $begingroup$
            @ChrisCuster right, good catch. Fixed
            $endgroup$
            – Masacroso
            2 hours ago








          1




          1




          $begingroup$
          $[a,b]$ is compact.
          $endgroup$
          – Chris Custer
          2 hours ago




          $begingroup$
          $[a,b]$ is compact.
          $endgroup$
          – Chris Custer
          2 hours ago












          $begingroup$
          @ChrisCuster right, good catch. Fixed
          $endgroup$
          – Masacroso
          2 hours ago




          $begingroup$
          @ChrisCuster right, good catch. Fixed
          $endgroup$
          – Masacroso
          2 hours ago











          0












          $begingroup$

          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
            2 hours ago










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
















          0












          $begingroup$

          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
            2 hours ago










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














          0












          0








          0





          $begingroup$

          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 3 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
            2 hours ago










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


















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










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
















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




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












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




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


















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