Equivalent statements about a metric space (homeomorphism, continuity, cluster points, clopen sets and...












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Problem. Let $M$ a metric space with metric $d$. Prove that the following conditions are equivalent.



(a) $M$ is homeomorphic to $M$ with discrete metric.



(b) Every function $f:M to M$ is continuous



(c) Every bijection $g: M to M$ is a homeomorphism



(d) $M$ has no cluster points



(e) Every subset of $M$ is clopen



(f) Every compact subset of $M$ is finite





  • [(a) $Longrightarrow$ (b)] Given $epsilon > 0$ take $delta < 1$ so, $d_{M}(x,y) < 1$ implies $x = y$ and so, $d(f(x),f(y)) = 0 < epsilon$.


  • [(b) $Longrightarrow$ (c)] $g: M to M$ is continuous and, since $g^{-1}:M to M$ is a bijection too, $g^{-1}$ is continuous.


  • [(d) $Longrightarrow$ (e)] Let $S$ be a subset of $M$. Since $M$ has no cluster points, $S$ is closed. Moreover, if $p in S$, we can take ball $B_{r}(p)$ with $r$ small such that $B_{r}(p) cap S = {p}$. Then $S$ is open too.


  • [(f) $Longrightarrow$ (a)] If the metric is not discrete, take $p$ a cluster point of $M$. Thus, for every $n in mathbb{N}$, considering $B_{1/n}(p)$, we obtain a convergente sequence, that is, $p_{n} to p$. But, every convergent sequence is bounded and every subsequence converges to $p$, therefore, ${p_{n}}_{n}cup {p}$ is compact, a contradiction.



Is there a mistake?



For (c) $Longrightarrow$ (d) and (e) $Longrightarrow$ (f) I have no idea.



Can someone help me?






real-analysis general-topology metric-spaces







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    $begingroup$



    Problem. Let $M$ a metric space with metric $d$. Prove that the following conditions are equivalent.



    (a) $M$ is homeomorphic to $M$ with discrete metric.



    (b) Every function $f:M to M$ is continuous



    (c) Every bijection $g: M to M$ is a homeomorphism



    (d) $M$ has no cluster points



    (e) Every subset of $M$ is clopen



    (f) Every compact subset of $M$ is finite





    • [(a) $Longrightarrow$ (b)] Given $epsilon > 0$ take $delta < 1$ so, $d_{M}(x,y) < 1$ implies $x = y$ and so, $d(f(x),f(y)) = 0 < epsilon$.


    • [(b) $Longrightarrow$ (c)] $g: M to M$ is continuous and, since $g^{-1}:M to M$ is a bijection too, $g^{-1}$ is continuous.


    • [(d) $Longrightarrow$ (e)] Let $S$ be a subset of $M$. Since $M$ has no cluster points, $S$ is closed. Moreover, if $p in S$, we can take ball $B_{r}(p)$ with $r$ small such that $B_{r}(p) cap S = {p}$. Then $S$ is open too.


    • [(f) $Longrightarrow$ (a)] If the metric is not discrete, take $p$ a cluster point of $M$. Thus, for every $n in mathbb{N}$, considering $B_{1/n}(p)$, we obtain a convergente sequence, that is, $p_{n} to p$. But, every convergent sequence is bounded and every subsequence converges to $p$, therefore, ${p_{n}}_{n}cup {p}$ is compact, a contradiction.



    Is there a mistake?



    For (c) $Longrightarrow$ (d) and (e) $Longrightarrow$ (f) I have no idea.



    Can someone help me?






    real-analysis general-topology metric-spaces







    share|cite|improve this question









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      1








      1





      $begingroup$



      Problem. Let $M$ a metric space with metric $d$. Prove that the following conditions are equivalent.



      (a) $M$ is homeomorphic to $M$ with discrete metric.



      (b) Every function $f:M to M$ is continuous



      (c) Every bijection $g: M to M$ is a homeomorphism



      (d) $M$ has no cluster points



      (e) Every subset of $M$ is clopen



      (f) Every compact subset of $M$ is finite





      • [(a) $Longrightarrow$ (b)] Given $epsilon > 0$ take $delta < 1$ so, $d_{M}(x,y) < 1$ implies $x = y$ and so, $d(f(x),f(y)) = 0 < epsilon$.


      • [(b) $Longrightarrow$ (c)] $g: M to M$ is continuous and, since $g^{-1}:M to M$ is a bijection too, $g^{-1}$ is continuous.


      • [(d) $Longrightarrow$ (e)] Let $S$ be a subset of $M$. Since $M$ has no cluster points, $S$ is closed. Moreover, if $p in S$, we can take ball $B_{r}(p)$ with $r$ small such that $B_{r}(p) cap S = {p}$. Then $S$ is open too.


      • [(f) $Longrightarrow$ (a)] If the metric is not discrete, take $p$ a cluster point of $M$. Thus, for every $n in mathbb{N}$, considering $B_{1/n}(p)$, we obtain a convergente sequence, that is, $p_{n} to p$. But, every convergent sequence is bounded and every subsequence converges to $p$, therefore, ${p_{n}}_{n}cup {p}$ is compact, a contradiction.



      Is there a mistake?



      For (c) $Longrightarrow$ (d) and (e) $Longrightarrow$ (f) I have no idea.



      Can someone help me?






      real-analysis general-topology metric-spaces







      share|cite|improve this question









      $endgroup$





      Problem. Let $M$ a metric space with metric $d$. Prove that the following conditions are equivalent.



      (a) $M$ is homeomorphic to $M$ with discrete metric.



      (b) Every function $f:M to M$ is continuous



      (c) Every bijection $g: M to M$ is a homeomorphism



      (d) $M$ has no cluster points



      (e) Every subset of $M$ is clopen



      (f) Every compact subset of $M$ is finite





      • [(a) $Longrightarrow$ (b)] Given $epsilon > 0$ take $delta < 1$ so, $d_{M}(x,y) < 1$ implies $x = y$ and so, $d(f(x),f(y)) = 0 < epsilon$.


      • [(b) $Longrightarrow$ (c)] $g: M to M$ is continuous and, since $g^{-1}:M to M$ is a bijection too, $g^{-1}$ is continuous.


      • [(d) $Longrightarrow$ (e)] Let $S$ be a subset of $M$. Since $M$ has no cluster points, $S$ is closed. Moreover, if $p in S$, we can take ball $B_{r}(p)$ with $r$ small such that $B_{r}(p) cap S = {p}$. Then $S$ is open too.


      • [(f) $Longrightarrow$ (a)] If the metric is not discrete, take $p$ a cluster point of $M$. Thus, for every $n in mathbb{N}$, considering $B_{1/n}(p)$, we obtain a convergente sequence, that is, $p_{n} to p$. But, every convergent sequence is bounded and every subsequence converges to $p$, therefore, ${p_{n}}_{n}cup {p}$ is compact, a contradiction.



      Is there a mistake?



      For (c) $Longrightarrow$ (d) and (e) $Longrightarrow$ (f) I have no idea.



      Can someone help me?






      real-analysis general-topology metric-spaces




      real-analysis general-topology metric-spaces






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      asked Dec 21 '18 at 17:00









      Lucas CorrêaLucas Corrêa

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          Not a mistake but easier: (d), $M$ has no cluster points, can be translated as



          $$forall x in M: exists r_x >0: B(x,r_x) = {x}tag{1}$$



          which says that all singleton sets are open, hence all sets are open. Hence $(e)$.



          $(e) implies (f)$ is easy: we know all singletons are open, so a compact set can be disjointly covered by its singletons. There is but one subcover (the cover itself, as we cannot omit one) and if it's finite so is the compact set.



          $(f) implies (a)$: your argument shows that all convergent sequences in $M$ are eventually constant (or the sequence with its limit is an infinite compact subset). This implies condition $(1)$ above by standard arguments, and thus also $(a)$.



          For $(c) implies (d)$, suppose $M$ has a cluster point $p$. Then, by standard arguments, we can find a sequence $x_n to p$ where all $x_n neq p$. Let $g$ be any bijection that is the identity except that we interchange $p$ and, say, $x_1$. Then the fact that $g$ is a homeomorphism would imply that $x_n to x_1$ also, which is impossible by unicity of limits.






          share|cite|improve this answer











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          • $begingroup$
            I got it! For (e) implies (f), I've tried using some sequence argument, but using cover gets easier. Thank you!
            $endgroup$
            – Lucas Corrêa
            Dec 21 '18 at 18:59











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          $begingroup$

          Not a mistake but easier: (d), $M$ has no cluster points, can be translated as



          $$forall x in M: exists r_x >0: B(x,r_x) = {x}tag{1}$$



          which says that all singleton sets are open, hence all sets are open. Hence $(e)$.



          $(e) implies (f)$ is easy: we know all singletons are open, so a compact set can be disjointly covered by its singletons. There is but one subcover (the cover itself, as we cannot omit one) and if it's finite so is the compact set.



          $(f) implies (a)$: your argument shows that all convergent sequences in $M$ are eventually constant (or the sequence with its limit is an infinite compact subset). This implies condition $(1)$ above by standard arguments, and thus also $(a)$.



          For $(c) implies (d)$, suppose $M$ has a cluster point $p$. Then, by standard arguments, we can find a sequence $x_n to p$ where all $x_n neq p$. Let $g$ be any bijection that is the identity except that we interchange $p$ and, say, $x_1$. Then the fact that $g$ is a homeomorphism would imply that $x_n to x_1$ also, which is impossible by unicity of limits.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I got it! For (e) implies (f), I've tried using some sequence argument, but using cover gets easier. Thank you!
            $endgroup$
            – Lucas Corrêa
            Dec 21 '18 at 18:59
















          2












          $begingroup$

          Not a mistake but easier: (d), $M$ has no cluster points, can be translated as



          $$forall x in M: exists r_x >0: B(x,r_x) = {x}tag{1}$$



          which says that all singleton sets are open, hence all sets are open. Hence $(e)$.



          $(e) implies (f)$ is easy: we know all singletons are open, so a compact set can be disjointly covered by its singletons. There is but one subcover (the cover itself, as we cannot omit one) and if it's finite so is the compact set.



          $(f) implies (a)$: your argument shows that all convergent sequences in $M$ are eventually constant (or the sequence with its limit is an infinite compact subset). This implies condition $(1)$ above by standard arguments, and thus also $(a)$.



          For $(c) implies (d)$, suppose $M$ has a cluster point $p$. Then, by standard arguments, we can find a sequence $x_n to p$ where all $x_n neq p$. Let $g$ be any bijection that is the identity except that we interchange $p$ and, say, $x_1$. Then the fact that $g$ is a homeomorphism would imply that $x_n to x_1$ also, which is impossible by unicity of limits.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I got it! For (e) implies (f), I've tried using some sequence argument, but using cover gets easier. Thank you!
            $endgroup$
            – Lucas Corrêa
            Dec 21 '18 at 18:59














          2












          2








          2





          $begingroup$

          Not a mistake but easier: (d), $M$ has no cluster points, can be translated as



          $$forall x in M: exists r_x >0: B(x,r_x) = {x}tag{1}$$



          which says that all singleton sets are open, hence all sets are open. Hence $(e)$.



          $(e) implies (f)$ is easy: we know all singletons are open, so a compact set can be disjointly covered by its singletons. There is but one subcover (the cover itself, as we cannot omit one) and if it's finite so is the compact set.



          $(f) implies (a)$: your argument shows that all convergent sequences in $M$ are eventually constant (or the sequence with its limit is an infinite compact subset). This implies condition $(1)$ above by standard arguments, and thus also $(a)$.



          For $(c) implies (d)$, suppose $M$ has a cluster point $p$. Then, by standard arguments, we can find a sequence $x_n to p$ where all $x_n neq p$. Let $g$ be any bijection that is the identity except that we interchange $p$ and, say, $x_1$. Then the fact that $g$ is a homeomorphism would imply that $x_n to x_1$ also, which is impossible by unicity of limits.






          share|cite|improve this answer











          $endgroup$



          Not a mistake but easier: (d), $M$ has no cluster points, can be translated as



          $$forall x in M: exists r_x >0: B(x,r_x) = {x}tag{1}$$



          which says that all singleton sets are open, hence all sets are open. Hence $(e)$.



          $(e) implies (f)$ is easy: we know all singletons are open, so a compact set can be disjointly covered by its singletons. There is but one subcover (the cover itself, as we cannot omit one) and if it's finite so is the compact set.



          $(f) implies (a)$: your argument shows that all convergent sequences in $M$ are eventually constant (or the sequence with its limit is an infinite compact subset). This implies condition $(1)$ above by standard arguments, and thus also $(a)$.



          For $(c) implies (d)$, suppose $M$ has a cluster point $p$. Then, by standard arguments, we can find a sequence $x_n to p$ where all $x_n neq p$. Let $g$ be any bijection that is the identity except that we interchange $p$ and, say, $x_1$. Then the fact that $g$ is a homeomorphism would imply that $x_n to x_1$ also, which is impossible by unicity of limits.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 21 '18 at 18:36

























          answered Dec 21 '18 at 18:28









          Henno BrandsmaHenno Brandsma

          112k348120




          112k348120












          • $begingroup$
            I got it! For (e) implies (f), I've tried using some sequence argument, but using cover gets easier. Thank you!
            $endgroup$
            – Lucas Corrêa
            Dec 21 '18 at 18:59


















          • $begingroup$
            I got it! For (e) implies (f), I've tried using some sequence argument, but using cover gets easier. Thank you!
            $endgroup$
            – Lucas Corrêa
            Dec 21 '18 at 18:59
















          $begingroup$
          I got it! For (e) implies (f), I've tried using some sequence argument, but using cover gets easier. Thank you!
          $endgroup$
          – Lucas Corrêa
          Dec 21 '18 at 18:59




          $begingroup$
          I got it! For (e) implies (f), I've tried using some sequence argument, but using cover gets easier. Thank you!
          $endgroup$
          – Lucas Corrêa
          Dec 21 '18 at 18:59


















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