Is every $G_delta $, zero-set?












1












$begingroup$


Zero-set means a set of the form:



$Z(f) = { x in X | f(x) = 0 }quadtext{for some } f in C(X)$



$C(X)$ is the ring of continuous function on $X$.



I know that every zero-set is $G_delta $, i.e, a countably intersection of open sets.




Is every $G_delta $, zero-set? if not, can you give me an simple example.







general-topology topological-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    What topological properties does $X$ have?
    $endgroup$
    – i707107
    Jan 13 at 17:36










  • $begingroup$
    For example, an open set is a $G_delta$, and there are likely to be open sets that are not closed, and therefore not zero-sets.
    $endgroup$
    – GEdgar
    Jan 13 at 17:41
















1












$begingroup$


Zero-set means a set of the form:



$Z(f) = { x in X | f(x) = 0 }quadtext{for some } f in C(X)$



$C(X)$ is the ring of continuous function on $X$.



I know that every zero-set is $G_delta $, i.e, a countably intersection of open sets.




Is every $G_delta $, zero-set? if not, can you give me an simple example.







general-topology topological-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    What topological properties does $X$ have?
    $endgroup$
    – i707107
    Jan 13 at 17:36










  • $begingroup$
    For example, an open set is a $G_delta$, and there are likely to be open sets that are not closed, and therefore not zero-sets.
    $endgroup$
    – GEdgar
    Jan 13 at 17:41














1












1








1


0



$begingroup$


Zero-set means a set of the form:



$Z(f) = { x in X | f(x) = 0 }quadtext{for some } f in C(X)$



$C(X)$ is the ring of continuous function on $X$.



I know that every zero-set is $G_delta $, i.e, a countably intersection of open sets.




Is every $G_delta $, zero-set? if not, can you give me an simple example.







general-topology topological-groups







share|cite|improve this question











$endgroup$




Zero-set means a set of the form:



$Z(f) = { x in X | f(x) = 0 }quadtext{for some } f in C(X)$



$C(X)$ is the ring of continuous function on $X$.



I know that every zero-set is $G_delta $, i.e, a countably intersection of open sets.




Is every $G_delta $, zero-set? if not, can you give me an simple example.







general-topology topological-groups




general-topology topological-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 17:43









GEdgar

63.6k269176




63.6k269176










asked Jan 13 at 17:33









adinadin

142




142












  • $begingroup$
    What topological properties does $X$ have?
    $endgroup$
    – i707107
    Jan 13 at 17:36










  • $begingroup$
    For example, an open set is a $G_delta$, and there are likely to be open sets that are not closed, and therefore not zero-sets.
    $endgroup$
    – GEdgar
    Jan 13 at 17:41


















  • $begingroup$
    What topological properties does $X$ have?
    $endgroup$
    – i707107
    Jan 13 at 17:36










  • $begingroup$
    For example, an open set is a $G_delta$, and there are likely to be open sets that are not closed, and therefore not zero-sets.
    $endgroup$
    – GEdgar
    Jan 13 at 17:41
















$begingroup$
What topological properties does $X$ have?
$endgroup$
– i707107
Jan 13 at 17:36




$begingroup$
What topological properties does $X$ have?
$endgroup$
– i707107
Jan 13 at 17:36












$begingroup$
For example, an open set is a $G_delta$, and there are likely to be open sets that are not closed, and therefore not zero-sets.
$endgroup$
– GEdgar
Jan 13 at 17:41




$begingroup$
For example, an open set is a $G_delta$, and there are likely to be open sets that are not closed, and therefore not zero-sets.
$endgroup$
– GEdgar
Jan 13 at 17:41










3 Answers
3






active

oldest

votes


















2












$begingroup$

Trivially not, in general: there is no continuous function on $mathbb{R}$ whose zero set is $(0,1)$, since if $f$ were such a function, then it would satisfy $$f(1) = lim_{xto1}f(x) = lim_{xto1^-}0 = 0,$$ so $1 in Z(f)$, so $Z(f) neq (0,1)$, but as an open set, $(0,1)$ is clearly $G_delta$: it is the intersection of countably many copies of itself.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what you mean about "$f(1) = lim_{xto1}f(x) = lim_{xto1^-}0 = 0,$?
    $endgroup$
    – adin
    Jan 13 at 18:41










  • $begingroup$
    For a continuous function $f$, the value of $f$ at any point is equal to its two-sided limit at that point, which is equal to each of the one-sided limits there. Applying this with the left-handed limit at $1$ for any continuous function such that $Z(f) supseteq (0,1)$ gives that $f(1) = 0$.
    $endgroup$
    – user3482749
    Jan 13 at 18:47



















1












$begingroup$

$Z(f)$ has to be closed, so any $G_delta$ set which is not closed will be a counterexample.



As a very specific example, in $mathbb{R}$, the set $(0,1)$ is $G_delta$ (indeed it's open) but it cannot be the zero set of any continuous function.



It's also not true in general that every closed $G_delta$ set is a zero set, not even if $X$ is Hausdorff. See Is a closed $G_delta$ set in a Hausdorff space always a zero set?. (It is true if $X$ is $T_6$, and maybe some weaker condition would also suffice.)






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    A zero-set of $X$ is always a closed $G_delta$ by construction, in any $X$.
    A closed $G_delta$ set is a zero-set when $X$ is metric or more generally $T_4$.



    This need not hold in all spaces, though. You'll meet counterexamples naturally later in the study of rings of continuous functions.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Trivially not, in general: there is no continuous function on $mathbb{R}$ whose zero set is $(0,1)$, since if $f$ were such a function, then it would satisfy $$f(1) = lim_{xto1}f(x) = lim_{xto1^-}0 = 0,$$ so $1 in Z(f)$, so $Z(f) neq (0,1)$, but as an open set, $(0,1)$ is clearly $G_delta$: it is the intersection of countably many copies of itself.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        what you mean about "$f(1) = lim_{xto1}f(x) = lim_{xto1^-}0 = 0,$?
        $endgroup$
        – adin
        Jan 13 at 18:41










      • $begingroup$
        For a continuous function $f$, the value of $f$ at any point is equal to its two-sided limit at that point, which is equal to each of the one-sided limits there. Applying this with the left-handed limit at $1$ for any continuous function such that $Z(f) supseteq (0,1)$ gives that $f(1) = 0$.
        $endgroup$
        – user3482749
        Jan 13 at 18:47
















      2












      $begingroup$

      Trivially not, in general: there is no continuous function on $mathbb{R}$ whose zero set is $(0,1)$, since if $f$ were such a function, then it would satisfy $$f(1) = lim_{xto1}f(x) = lim_{xto1^-}0 = 0,$$ so $1 in Z(f)$, so $Z(f) neq (0,1)$, but as an open set, $(0,1)$ is clearly $G_delta$: it is the intersection of countably many copies of itself.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        what you mean about "$f(1) = lim_{xto1}f(x) = lim_{xto1^-}0 = 0,$?
        $endgroup$
        – adin
        Jan 13 at 18:41










      • $begingroup$
        For a continuous function $f$, the value of $f$ at any point is equal to its two-sided limit at that point, which is equal to each of the one-sided limits there. Applying this with the left-handed limit at $1$ for any continuous function such that $Z(f) supseteq (0,1)$ gives that $f(1) = 0$.
        $endgroup$
        – user3482749
        Jan 13 at 18:47














      2












      2








      2





      $begingroup$

      Trivially not, in general: there is no continuous function on $mathbb{R}$ whose zero set is $(0,1)$, since if $f$ were such a function, then it would satisfy $$f(1) = lim_{xto1}f(x) = lim_{xto1^-}0 = 0,$$ so $1 in Z(f)$, so $Z(f) neq (0,1)$, but as an open set, $(0,1)$ is clearly $G_delta$: it is the intersection of countably many copies of itself.






      share|cite|improve this answer









      $endgroup$



      Trivially not, in general: there is no continuous function on $mathbb{R}$ whose zero set is $(0,1)$, since if $f$ were such a function, then it would satisfy $$f(1) = lim_{xto1}f(x) = lim_{xto1^-}0 = 0,$$ so $1 in Z(f)$, so $Z(f) neq (0,1)$, but as an open set, $(0,1)$ is clearly $G_delta$: it is the intersection of countably many copies of itself.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 13 at 17:40









      user3482749user3482749

      4,3291119




      4,3291119












      • $begingroup$
        what you mean about "$f(1) = lim_{xto1}f(x) = lim_{xto1^-}0 = 0,$?
        $endgroup$
        – adin
        Jan 13 at 18:41










      • $begingroup$
        For a continuous function $f$, the value of $f$ at any point is equal to its two-sided limit at that point, which is equal to each of the one-sided limits there. Applying this with the left-handed limit at $1$ for any continuous function such that $Z(f) supseteq (0,1)$ gives that $f(1) = 0$.
        $endgroup$
        – user3482749
        Jan 13 at 18:47


















      • $begingroup$
        what you mean about "$f(1) = lim_{xto1}f(x) = lim_{xto1^-}0 = 0,$?
        $endgroup$
        – adin
        Jan 13 at 18:41










      • $begingroup$
        For a continuous function $f$, the value of $f$ at any point is equal to its two-sided limit at that point, which is equal to each of the one-sided limits there. Applying this with the left-handed limit at $1$ for any continuous function such that $Z(f) supseteq (0,1)$ gives that $f(1) = 0$.
        $endgroup$
        – user3482749
        Jan 13 at 18:47
















      $begingroup$
      what you mean about "$f(1) = lim_{xto1}f(x) = lim_{xto1^-}0 = 0,$?
      $endgroup$
      – adin
      Jan 13 at 18:41




      $begingroup$
      what you mean about "$f(1) = lim_{xto1}f(x) = lim_{xto1^-}0 = 0,$?
      $endgroup$
      – adin
      Jan 13 at 18:41












      $begingroup$
      For a continuous function $f$, the value of $f$ at any point is equal to its two-sided limit at that point, which is equal to each of the one-sided limits there. Applying this with the left-handed limit at $1$ for any continuous function such that $Z(f) supseteq (0,1)$ gives that $f(1) = 0$.
      $endgroup$
      – user3482749
      Jan 13 at 18:47




      $begingroup$
      For a continuous function $f$, the value of $f$ at any point is equal to its two-sided limit at that point, which is equal to each of the one-sided limits there. Applying this with the left-handed limit at $1$ for any continuous function such that $Z(f) supseteq (0,1)$ gives that $f(1) = 0$.
      $endgroup$
      – user3482749
      Jan 13 at 18:47











      1












      $begingroup$

      $Z(f)$ has to be closed, so any $G_delta$ set which is not closed will be a counterexample.



      As a very specific example, in $mathbb{R}$, the set $(0,1)$ is $G_delta$ (indeed it's open) but it cannot be the zero set of any continuous function.



      It's also not true in general that every closed $G_delta$ set is a zero set, not even if $X$ is Hausdorff. See Is a closed $G_delta$ set in a Hausdorff space always a zero set?. (It is true if $X$ is $T_6$, and maybe some weaker condition would also suffice.)






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        $Z(f)$ has to be closed, so any $G_delta$ set which is not closed will be a counterexample.



        As a very specific example, in $mathbb{R}$, the set $(0,1)$ is $G_delta$ (indeed it's open) but it cannot be the zero set of any continuous function.



        It's also not true in general that every closed $G_delta$ set is a zero set, not even if $X$ is Hausdorff. See Is a closed $G_delta$ set in a Hausdorff space always a zero set?. (It is true if $X$ is $T_6$, and maybe some weaker condition would also suffice.)






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          $Z(f)$ has to be closed, so any $G_delta$ set which is not closed will be a counterexample.



          As a very specific example, in $mathbb{R}$, the set $(0,1)$ is $G_delta$ (indeed it's open) but it cannot be the zero set of any continuous function.



          It's also not true in general that every closed $G_delta$ set is a zero set, not even if $X$ is Hausdorff. See Is a closed $G_delta$ set in a Hausdorff space always a zero set?. (It is true if $X$ is $T_6$, and maybe some weaker condition would also suffice.)






          share|cite|improve this answer











          $endgroup$



          $Z(f)$ has to be closed, so any $G_delta$ set which is not closed will be a counterexample.



          As a very specific example, in $mathbb{R}$, the set $(0,1)$ is $G_delta$ (indeed it's open) but it cannot be the zero set of any continuous function.



          It's also not true in general that every closed $G_delta$ set is a zero set, not even if $X$ is Hausdorff. See Is a closed $G_delta$ set in a Hausdorff space always a zero set?. (It is true if $X$ is $T_6$, and maybe some weaker condition would also suffice.)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 13 at 17:50

























          answered Jan 13 at 17:40









          Nate EldredgeNate Eldredge

          64.6k682174




          64.6k682174























              1












              $begingroup$

              A zero-set of $X$ is always a closed $G_delta$ by construction, in any $X$.
              A closed $G_delta$ set is a zero-set when $X$ is metric or more generally $T_4$.



              This need not hold in all spaces, though. You'll meet counterexamples naturally later in the study of rings of continuous functions.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                A zero-set of $X$ is always a closed $G_delta$ by construction, in any $X$.
                A closed $G_delta$ set is a zero-set when $X$ is metric or more generally $T_4$.



                This need not hold in all spaces, though. You'll meet counterexamples naturally later in the study of rings of continuous functions.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  A zero-set of $X$ is always a closed $G_delta$ by construction, in any $X$.
                  A closed $G_delta$ set is a zero-set when $X$ is metric or more generally $T_4$.



                  This need not hold in all spaces, though. You'll meet counterexamples naturally later in the study of rings of continuous functions.






                  share|cite|improve this answer









                  $endgroup$



                  A zero-set of $X$ is always a closed $G_delta$ by construction, in any $X$.
                  A closed $G_delta$ set is a zero-set when $X$ is metric or more generally $T_4$.



                  This need not hold in all spaces, though. You'll meet counterexamples naturally later in the study of rings of continuous functions.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 13 at 22:57









                  Henno BrandsmaHenno Brandsma

                  117k349127




                  117k349127






























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