Given that $X$ is zero dimensional, does this imply that $C_p(X)$ is zero dimensional?












1












$begingroup$


A zero dimensional space is a space which has a basis consisting of clopen sets.



$C_p(X)$ is the space of continuous real valued functions with the topology of pointwise convergence. (This is equivalent to taking it to be a subspace of $mathbb{R}^X$, the family of all functions $X to mathbb{R}$, with the usual product topology.)



My question is:




Given that $X$ is zero dimensional, does this imply that $C_p(X)$ is zero dimensional?
If not, are there any common conditions which imply the zero dimensionality of $C_p(X)$?




Thank you!






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    (Presumably you are aware of the fact that constant functions are always continuous.)
    $endgroup$
    – Mariano Suárez-Álvarez
    Feb 4 '14 at 15:51










  • $begingroup$
    The answer is then that $C_p(X)$ is never zero dimensional, simply because it is a connected space and with more than one point.
    $endgroup$
    – Mariano Suárez-Álvarez
    Feb 4 '14 at 16:14


















1












$begingroup$


A zero dimensional space is a space which has a basis consisting of clopen sets.



$C_p(X)$ is the space of continuous real valued functions with the topology of pointwise convergence. (This is equivalent to taking it to be a subspace of $mathbb{R}^X$, the family of all functions $X to mathbb{R}$, with the usual product topology.)



My question is:




Given that $X$ is zero dimensional, does this imply that $C_p(X)$ is zero dimensional?
If not, are there any common conditions which imply the zero dimensionality of $C_p(X)$?




Thank you!






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    (Presumably you are aware of the fact that constant functions are always continuous.)
    $endgroup$
    – Mariano Suárez-Álvarez
    Feb 4 '14 at 15:51










  • $begingroup$
    The answer is then that $C_p(X)$ is never zero dimensional, simply because it is a connected space and with more than one point.
    $endgroup$
    – Mariano Suárez-Álvarez
    Feb 4 '14 at 16:14
















1












1








1





$begingroup$


A zero dimensional space is a space which has a basis consisting of clopen sets.



$C_p(X)$ is the space of continuous real valued functions with the topology of pointwise convergence. (This is equivalent to taking it to be a subspace of $mathbb{R}^X$, the family of all functions $X to mathbb{R}$, with the usual product topology.)



My question is:




Given that $X$ is zero dimensional, does this imply that $C_p(X)$ is zero dimensional?
If not, are there any common conditions which imply the zero dimensionality of $C_p(X)$?




Thank you!






general-topology







share|cite|improve this question











$endgroup$




A zero dimensional space is a space which has a basis consisting of clopen sets.



$C_p(X)$ is the space of continuous real valued functions with the topology of pointwise convergence. (This is equivalent to taking it to be a subspace of $mathbb{R}^X$, the family of all functions $X to mathbb{R}$, with the usual product topology.)



My question is:




Given that $X$ is zero dimensional, does this imply that $C_p(X)$ is zero dimensional?
If not, are there any common conditions which imply the zero dimensionality of $C_p(X)$?




Thank you!






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 Feb 4 '14 at 16:39









user642796

44.9k565119




44.9k565119










asked Feb 4 '14 at 15:39









topsitopsi

2,112720




2,112720












  • $begingroup$
    (Presumably you are aware of the fact that constant functions are always continuous.)
    $endgroup$
    – Mariano Suárez-Álvarez
    Feb 4 '14 at 15:51










  • $begingroup$
    The answer is then that $C_p(X)$ is never zero dimensional, simply because it is a connected space and with more than one point.
    $endgroup$
    – Mariano Suárez-Álvarez
    Feb 4 '14 at 16:14




















  • $begingroup$
    (Presumably you are aware of the fact that constant functions are always continuous.)
    $endgroup$
    – Mariano Suárez-Álvarez
    Feb 4 '14 at 15:51










  • $begingroup$
    The answer is then that $C_p(X)$ is never zero dimensional, simply because it is a connected space and with more than one point.
    $endgroup$
    – Mariano Suárez-Álvarez
    Feb 4 '14 at 16:14


















$begingroup$
(Presumably you are aware of the fact that constant functions are always continuous.)
$endgroup$
– Mariano Suárez-Álvarez
Feb 4 '14 at 15:51




$begingroup$
(Presumably you are aware of the fact that constant functions are always continuous.)
$endgroup$
– Mariano Suárez-Álvarez
Feb 4 '14 at 15:51












$begingroup$
The answer is then that $C_p(X)$ is never zero dimensional, simply because it is a connected space and with more than one point.
$endgroup$
– Mariano Suárez-Álvarez
Feb 4 '14 at 16:14






$begingroup$
The answer is then that $C_p(X)$ is never zero dimensional, simply because it is a connected space and with more than one point.
$endgroup$
– Mariano Suárez-Álvarez
Feb 4 '14 at 16:14












2 Answers
2






active

oldest

votes


















0












$begingroup$

For $X=pt$, we have $C_p(X)cong Bbb R$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what is the space $pt$? the space I had in mind was a completely regular general topological space $X$. If $X$ is zero dimentionas, does it imply that $C_p(X)$ is alos?
    $endgroup$
    – topsi
    Feb 4 '14 at 15:58










  • $begingroup$
    $pt$ is the topological space of a point.
    $endgroup$
    – PVAL-inactive
    Feb 4 '14 at 16:14












  • $begingroup$
    Interesting. I guess we can say that this space satisfy all seperation axioms? (since we don't really have more then one point)
    $endgroup$
    – topsi
    Feb 4 '14 at 16:20










  • $begingroup$
    What about, for instance, a space with 3 points with the discrete topology? can we say that $C_p(X) cong Bbb R$
    $endgroup$
    – topsi
    Feb 4 '14 at 16:23






  • 1




    $begingroup$
    @Shir: If $X$ is discrete, then all functions $X to mathbb{R}$ are continuous, so $C_p (X) = mathbb{R}^X$.
    $endgroup$
    – user642796
    Feb 4 '14 at 16:35



















0












$begingroup$

$C_p(X)$ is never zero-dimensional when $X ne emptyset$.



Let $varphi : X to Y$ be any map (= continuous function) between toplogical spaces $X,Y$. Define
$$varphi^* : C_p(Y) to C_p(X), varphi^*(f) = f circ varphi .$$
Let us verify that $varphi^*$ is continuous. A subbase for the topology of pointwise convergence on $C_p(Z)$ is given by the sets $M(z,U) = { f in C_p(Z) mid f(z) in U }$, where $z in Z$ and $U subset mathbb{R}$ is open. It therefore suffices to show that all $(varphi^*)^{-1}(M(x,U))$ are open. But
$$(varphi^*)^{-1}(M(x,U)) = { f in C_p(Y) mid varphi^*(f) = f circ varphi in M(x,U) } = { f in C_p(Y) mid (f circ varphi)(x) = f(varphi(x)) in U } = M(varphi(x),U) .$$
If $psi : Y to Z$ is another map, then we have $(psi circ varphi)^* = varphi^* circ psi^*$. Moreover, $id_X^* = id_{C_p(X)}$.



Let $T$ denote a space containing only one point. Then $C_p(T) approx mathbb{R}$.
For $X ne emptyset$ let $i : T to X$ be any map and $r : X to T$ the unique map. Then $r circ i = id_T$ and we conclude $i^* circ r^* = id_{C_p(T)}$. Hence $r^* : C_p(T) to C_p(X)$ is an embedding (i.e. a establishes a homeomorphism between $C_p(T)$ and $r^*(C_p(T)) subset C_p(X)$).



This means that $C_p(X)$ contains a subspace homeomorphic to $mathbb{R}$.



If $C_p(X)$ were zero-dimensional, then so would be all subspaces and in particular $mathbb{R}$ would be zero-dimensional, which is absurd.






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    For $X=pt$, we have $C_p(X)cong Bbb R$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      what is the space $pt$? the space I had in mind was a completely regular general topological space $X$. If $X$ is zero dimentionas, does it imply that $C_p(X)$ is alos?
      $endgroup$
      – topsi
      Feb 4 '14 at 15:58










    • $begingroup$
      $pt$ is the topological space of a point.
      $endgroup$
      – PVAL-inactive
      Feb 4 '14 at 16:14












    • $begingroup$
      Interesting. I guess we can say that this space satisfy all seperation axioms? (since we don't really have more then one point)
      $endgroup$
      – topsi
      Feb 4 '14 at 16:20










    • $begingroup$
      What about, for instance, a space with 3 points with the discrete topology? can we say that $C_p(X) cong Bbb R$
      $endgroup$
      – topsi
      Feb 4 '14 at 16:23






    • 1




      $begingroup$
      @Shir: If $X$ is discrete, then all functions $X to mathbb{R}$ are continuous, so $C_p (X) = mathbb{R}^X$.
      $endgroup$
      – user642796
      Feb 4 '14 at 16:35
















    0












    $begingroup$

    For $X=pt$, we have $C_p(X)cong Bbb R$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      what is the space $pt$? the space I had in mind was a completely regular general topological space $X$. If $X$ is zero dimentionas, does it imply that $C_p(X)$ is alos?
      $endgroup$
      – topsi
      Feb 4 '14 at 15:58










    • $begingroup$
      $pt$ is the topological space of a point.
      $endgroup$
      – PVAL-inactive
      Feb 4 '14 at 16:14












    • $begingroup$
      Interesting. I guess we can say that this space satisfy all seperation axioms? (since we don't really have more then one point)
      $endgroup$
      – topsi
      Feb 4 '14 at 16:20










    • $begingroup$
      What about, for instance, a space with 3 points with the discrete topology? can we say that $C_p(X) cong Bbb R$
      $endgroup$
      – topsi
      Feb 4 '14 at 16:23






    • 1




      $begingroup$
      @Shir: If $X$ is discrete, then all functions $X to mathbb{R}$ are continuous, so $C_p (X) = mathbb{R}^X$.
      $endgroup$
      – user642796
      Feb 4 '14 at 16:35














    0












    0








    0





    $begingroup$

    For $X=pt$, we have $C_p(X)cong Bbb R$.






    share|cite|improve this answer









    $endgroup$



    For $X=pt$, we have $C_p(X)cong Bbb R$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Feb 4 '14 at 15:50









    PVAL-inactivePVAL-inactive

    7,08011843




    7,08011843












    • $begingroup$
      what is the space $pt$? the space I had in mind was a completely regular general topological space $X$. If $X$ is zero dimentionas, does it imply that $C_p(X)$ is alos?
      $endgroup$
      – topsi
      Feb 4 '14 at 15:58










    • $begingroup$
      $pt$ is the topological space of a point.
      $endgroup$
      – PVAL-inactive
      Feb 4 '14 at 16:14












    • $begingroup$
      Interesting. I guess we can say that this space satisfy all seperation axioms? (since we don't really have more then one point)
      $endgroup$
      – topsi
      Feb 4 '14 at 16:20










    • $begingroup$
      What about, for instance, a space with 3 points with the discrete topology? can we say that $C_p(X) cong Bbb R$
      $endgroup$
      – topsi
      Feb 4 '14 at 16:23






    • 1




      $begingroup$
      @Shir: If $X$ is discrete, then all functions $X to mathbb{R}$ are continuous, so $C_p (X) = mathbb{R}^X$.
      $endgroup$
      – user642796
      Feb 4 '14 at 16:35


















    • $begingroup$
      what is the space $pt$? the space I had in mind was a completely regular general topological space $X$. If $X$ is zero dimentionas, does it imply that $C_p(X)$ is alos?
      $endgroup$
      – topsi
      Feb 4 '14 at 15:58










    • $begingroup$
      $pt$ is the topological space of a point.
      $endgroup$
      – PVAL-inactive
      Feb 4 '14 at 16:14












    • $begingroup$
      Interesting. I guess we can say that this space satisfy all seperation axioms? (since we don't really have more then one point)
      $endgroup$
      – topsi
      Feb 4 '14 at 16:20










    • $begingroup$
      What about, for instance, a space with 3 points with the discrete topology? can we say that $C_p(X) cong Bbb R$
      $endgroup$
      – topsi
      Feb 4 '14 at 16:23






    • 1




      $begingroup$
      @Shir: If $X$ is discrete, then all functions $X to mathbb{R}$ are continuous, so $C_p (X) = mathbb{R}^X$.
      $endgroup$
      – user642796
      Feb 4 '14 at 16:35
















    $begingroup$
    what is the space $pt$? the space I had in mind was a completely regular general topological space $X$. If $X$ is zero dimentionas, does it imply that $C_p(X)$ is alos?
    $endgroup$
    – topsi
    Feb 4 '14 at 15:58




    $begingroup$
    what is the space $pt$? the space I had in mind was a completely regular general topological space $X$. If $X$ is zero dimentionas, does it imply that $C_p(X)$ is alos?
    $endgroup$
    – topsi
    Feb 4 '14 at 15:58












    $begingroup$
    $pt$ is the topological space of a point.
    $endgroup$
    – PVAL-inactive
    Feb 4 '14 at 16:14






    $begingroup$
    $pt$ is the topological space of a point.
    $endgroup$
    – PVAL-inactive
    Feb 4 '14 at 16:14














    $begingroup$
    Interesting. I guess we can say that this space satisfy all seperation axioms? (since we don't really have more then one point)
    $endgroup$
    – topsi
    Feb 4 '14 at 16:20




    $begingroup$
    Interesting. I guess we can say that this space satisfy all seperation axioms? (since we don't really have more then one point)
    $endgroup$
    – topsi
    Feb 4 '14 at 16:20












    $begingroup$
    What about, for instance, a space with 3 points with the discrete topology? can we say that $C_p(X) cong Bbb R$
    $endgroup$
    – topsi
    Feb 4 '14 at 16:23




    $begingroup$
    What about, for instance, a space with 3 points with the discrete topology? can we say that $C_p(X) cong Bbb R$
    $endgroup$
    – topsi
    Feb 4 '14 at 16:23




    1




    1




    $begingroup$
    @Shir: If $X$ is discrete, then all functions $X to mathbb{R}$ are continuous, so $C_p (X) = mathbb{R}^X$.
    $endgroup$
    – user642796
    Feb 4 '14 at 16:35




    $begingroup$
    @Shir: If $X$ is discrete, then all functions $X to mathbb{R}$ are continuous, so $C_p (X) = mathbb{R}^X$.
    $endgroup$
    – user642796
    Feb 4 '14 at 16:35











    0












    $begingroup$

    $C_p(X)$ is never zero-dimensional when $X ne emptyset$.



    Let $varphi : X to Y$ be any map (= continuous function) between toplogical spaces $X,Y$. Define
    $$varphi^* : C_p(Y) to C_p(X), varphi^*(f) = f circ varphi .$$
    Let us verify that $varphi^*$ is continuous. A subbase for the topology of pointwise convergence on $C_p(Z)$ is given by the sets $M(z,U) = { f in C_p(Z) mid f(z) in U }$, where $z in Z$ and $U subset mathbb{R}$ is open. It therefore suffices to show that all $(varphi^*)^{-1}(M(x,U))$ are open. But
    $$(varphi^*)^{-1}(M(x,U)) = { f in C_p(Y) mid varphi^*(f) = f circ varphi in M(x,U) } = { f in C_p(Y) mid (f circ varphi)(x) = f(varphi(x)) in U } = M(varphi(x),U) .$$
    If $psi : Y to Z$ is another map, then we have $(psi circ varphi)^* = varphi^* circ psi^*$. Moreover, $id_X^* = id_{C_p(X)}$.



    Let $T$ denote a space containing only one point. Then $C_p(T) approx mathbb{R}$.
    For $X ne emptyset$ let $i : T to X$ be any map and $r : X to T$ the unique map. Then $r circ i = id_T$ and we conclude $i^* circ r^* = id_{C_p(T)}$. Hence $r^* : C_p(T) to C_p(X)$ is an embedding (i.e. a establishes a homeomorphism between $C_p(T)$ and $r^*(C_p(T)) subset C_p(X)$).



    This means that $C_p(X)$ contains a subspace homeomorphic to $mathbb{R}$.



    If $C_p(X)$ were zero-dimensional, then so would be all subspaces and in particular $mathbb{R}$ would be zero-dimensional, which is absurd.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $C_p(X)$ is never zero-dimensional when $X ne emptyset$.



      Let $varphi : X to Y$ be any map (= continuous function) between toplogical spaces $X,Y$. Define
      $$varphi^* : C_p(Y) to C_p(X), varphi^*(f) = f circ varphi .$$
      Let us verify that $varphi^*$ is continuous. A subbase for the topology of pointwise convergence on $C_p(Z)$ is given by the sets $M(z,U) = { f in C_p(Z) mid f(z) in U }$, where $z in Z$ and $U subset mathbb{R}$ is open. It therefore suffices to show that all $(varphi^*)^{-1}(M(x,U))$ are open. But
      $$(varphi^*)^{-1}(M(x,U)) = { f in C_p(Y) mid varphi^*(f) = f circ varphi in M(x,U) } = { f in C_p(Y) mid (f circ varphi)(x) = f(varphi(x)) in U } = M(varphi(x),U) .$$
      If $psi : Y to Z$ is another map, then we have $(psi circ varphi)^* = varphi^* circ psi^*$. Moreover, $id_X^* = id_{C_p(X)}$.



      Let $T$ denote a space containing only one point. Then $C_p(T) approx mathbb{R}$.
      For $X ne emptyset$ let $i : T to X$ be any map and $r : X to T$ the unique map. Then $r circ i = id_T$ and we conclude $i^* circ r^* = id_{C_p(T)}$. Hence $r^* : C_p(T) to C_p(X)$ is an embedding (i.e. a establishes a homeomorphism between $C_p(T)$ and $r^*(C_p(T)) subset C_p(X)$).



      This means that $C_p(X)$ contains a subspace homeomorphic to $mathbb{R}$.



      If $C_p(X)$ were zero-dimensional, then so would be all subspaces and in particular $mathbb{R}$ would be zero-dimensional, which is absurd.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $C_p(X)$ is never zero-dimensional when $X ne emptyset$.



        Let $varphi : X to Y$ be any map (= continuous function) between toplogical spaces $X,Y$. Define
        $$varphi^* : C_p(Y) to C_p(X), varphi^*(f) = f circ varphi .$$
        Let us verify that $varphi^*$ is continuous. A subbase for the topology of pointwise convergence on $C_p(Z)$ is given by the sets $M(z,U) = { f in C_p(Z) mid f(z) in U }$, where $z in Z$ and $U subset mathbb{R}$ is open. It therefore suffices to show that all $(varphi^*)^{-1}(M(x,U))$ are open. But
        $$(varphi^*)^{-1}(M(x,U)) = { f in C_p(Y) mid varphi^*(f) = f circ varphi in M(x,U) } = { f in C_p(Y) mid (f circ varphi)(x) = f(varphi(x)) in U } = M(varphi(x),U) .$$
        If $psi : Y to Z$ is another map, then we have $(psi circ varphi)^* = varphi^* circ psi^*$. Moreover, $id_X^* = id_{C_p(X)}$.



        Let $T$ denote a space containing only one point. Then $C_p(T) approx mathbb{R}$.
        For $X ne emptyset$ let $i : T to X$ be any map and $r : X to T$ the unique map. Then $r circ i = id_T$ and we conclude $i^* circ r^* = id_{C_p(T)}$. Hence $r^* : C_p(T) to C_p(X)$ is an embedding (i.e. a establishes a homeomorphism between $C_p(T)$ and $r^*(C_p(T)) subset C_p(X)$).



        This means that $C_p(X)$ contains a subspace homeomorphic to $mathbb{R}$.



        If $C_p(X)$ were zero-dimensional, then so would be all subspaces and in particular $mathbb{R}$ would be zero-dimensional, which is absurd.






        share|cite|improve this answer









        $endgroup$



        $C_p(X)$ is never zero-dimensional when $X ne emptyset$.



        Let $varphi : X to Y$ be any map (= continuous function) between toplogical spaces $X,Y$. Define
        $$varphi^* : C_p(Y) to C_p(X), varphi^*(f) = f circ varphi .$$
        Let us verify that $varphi^*$ is continuous. A subbase for the topology of pointwise convergence on $C_p(Z)$ is given by the sets $M(z,U) = { f in C_p(Z) mid f(z) in U }$, where $z in Z$ and $U subset mathbb{R}$ is open. It therefore suffices to show that all $(varphi^*)^{-1}(M(x,U))$ are open. But
        $$(varphi^*)^{-1}(M(x,U)) = { f in C_p(Y) mid varphi^*(f) = f circ varphi in M(x,U) } = { f in C_p(Y) mid (f circ varphi)(x) = f(varphi(x)) in U } = M(varphi(x),U) .$$
        If $psi : Y to Z$ is another map, then we have $(psi circ varphi)^* = varphi^* circ psi^*$. Moreover, $id_X^* = id_{C_p(X)}$.



        Let $T$ denote a space containing only one point. Then $C_p(T) approx mathbb{R}$.
        For $X ne emptyset$ let $i : T to X$ be any map and $r : X to T$ the unique map. Then $r circ i = id_T$ and we conclude $i^* circ r^* = id_{C_p(T)}$. Hence $r^* : C_p(T) to C_p(X)$ is an embedding (i.e. a establishes a homeomorphism between $C_p(T)$ and $r^*(C_p(T)) subset C_p(X)$).



        This means that $C_p(X)$ contains a subspace homeomorphic to $mathbb{R}$.



        If $C_p(X)$ were zero-dimensional, then so would be all subspaces and in particular $mathbb{R}$ would be zero-dimensional, which is absurd.







        share|cite|improve this answer












        share|cite|improve this answer



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        answered Jan 4 at 12:49









        Paul FrostPaul Frost

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