clarification of G?












1












$begingroup$




Let $X$ be a metric space and $fcolon X rightarrow mathbb R$ be a continuous function. Let $G = { (x , f(x) ) : x in X }$ be the graph of $f$. Then which one is true?






  1. $G$ is homeomorphic to $X$


  2. $G$ is homeomorphic to $mathbb R$


  3. $G$ is homeomorphic to $X times mathbb R$


  4. $G$ is homeomorphic to $mathbb R times X$



My attempt : My answer None of the option is correct



For option $1)$ Consider $X=(0,1)$ and $f(x)=x$ then $G={(x,x):xin (0,1)}$ which is a closed set but $(0,1)$ is not



option $2)$ $X=[0,1]$ which is compact and $f(x)=x$.Its graph is compact but $mathbb R$ is not.



option $3)$ $f(x)=0$ .Then graph of $f={(x,0):xin mathbb R}$ i.e. the $x$ axis .Now remove the point $(0,0)$ from the graph of $f$ ,it becomes disconnected but $mathbb R^2setminus {(0,0)}$ is not.



option $4)$.Consider $f(x)=x$ .Then graph of $f={(0,y):xin mathbb R}$ i.e. the $y$ axis .Now remove the point $(0,0)$ from the graph of $f$ ,it becomes disconnected but $mathbb R^2setminus {(0,0)}$ is not.



Is my answer is correct or not correct ?



Pliz tell me






general-topology







share|cite|improve this question









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  • 1




    $begingroup$
    For (1): $G$ doesn't look very closed to me (in particular: its complement doesn't include a neighbourhood of $(0,0)$, but does include $(0,0)$). The others are fine.
    $endgroup$
    – user3482749
    Dec 4 '18 at 14:04










  • $begingroup$
    thanks u @user3482749.
    $endgroup$
    – jasmine
    Dec 4 '18 at 14:23






  • 1




    $begingroup$
    3. and 4. are the "same" because $X times Y$ and $Y times X$ are homeomorphic for any two spaces $X,Y$.
    $endgroup$
    – Paul Frost
    Dec 4 '18 at 18:28
















1












$begingroup$




Let $X$ be a metric space and $fcolon X rightarrow mathbb R$ be a continuous function. Let $G = { (x , f(x) ) : x in X }$ be the graph of $f$. Then which one is true?






  1. $G$ is homeomorphic to $X$


  2. $G$ is homeomorphic to $mathbb R$


  3. $G$ is homeomorphic to $X times mathbb R$


  4. $G$ is homeomorphic to $mathbb R times X$



My attempt : My answer None of the option is correct



For option $1)$ Consider $X=(0,1)$ and $f(x)=x$ then $G={(x,x):xin (0,1)}$ which is a closed set but $(0,1)$ is not



option $2)$ $X=[0,1]$ which is compact and $f(x)=x$.Its graph is compact but $mathbb R$ is not.



option $3)$ $f(x)=0$ .Then graph of $f={(x,0):xin mathbb R}$ i.e. the $x$ axis .Now remove the point $(0,0)$ from the graph of $f$ ,it becomes disconnected but $mathbb R^2setminus {(0,0)}$ is not.



option $4)$.Consider $f(x)=x$ .Then graph of $f={(0,y):xin mathbb R}$ i.e. the $y$ axis .Now remove the point $(0,0)$ from the graph of $f$ ,it becomes disconnected but $mathbb R^2setminus {(0,0)}$ is not.



Is my answer is correct or not correct ?



Pliz tell me






general-topology







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    For (1): $G$ doesn't look very closed to me (in particular: its complement doesn't include a neighbourhood of $(0,0)$, but does include $(0,0)$). The others are fine.
    $endgroup$
    – user3482749
    Dec 4 '18 at 14:04










  • $begingroup$
    thanks u @user3482749.
    $endgroup$
    – jasmine
    Dec 4 '18 at 14:23






  • 1




    $begingroup$
    3. and 4. are the "same" because $X times Y$ and $Y times X$ are homeomorphic for any two spaces $X,Y$.
    $endgroup$
    – Paul Frost
    Dec 4 '18 at 18:28














1












1








1





$begingroup$




Let $X$ be a metric space and $fcolon X rightarrow mathbb R$ be a continuous function. Let $G = { (x , f(x) ) : x in X }$ be the graph of $f$. Then which one is true?






  1. $G$ is homeomorphic to $X$


  2. $G$ is homeomorphic to $mathbb R$


  3. $G$ is homeomorphic to $X times mathbb R$


  4. $G$ is homeomorphic to $mathbb R times X$



My attempt : My answer None of the option is correct



For option $1)$ Consider $X=(0,1)$ and $f(x)=x$ then $G={(x,x):xin (0,1)}$ which is a closed set but $(0,1)$ is not



option $2)$ $X=[0,1]$ which is compact and $f(x)=x$.Its graph is compact but $mathbb R$ is not.



option $3)$ $f(x)=0$ .Then graph of $f={(x,0):xin mathbb R}$ i.e. the $x$ axis .Now remove the point $(0,0)$ from the graph of $f$ ,it becomes disconnected but $mathbb R^2setminus {(0,0)}$ is not.



option $4)$.Consider $f(x)=x$ .Then graph of $f={(0,y):xin mathbb R}$ i.e. the $y$ axis .Now remove the point $(0,0)$ from the graph of $f$ ,it becomes disconnected but $mathbb R^2setminus {(0,0)}$ is not.



Is my answer is correct or not correct ?



Pliz tell me






general-topology







share|cite|improve this question









$endgroup$






Let $X$ be a metric space and $fcolon X rightarrow mathbb R$ be a continuous function. Let $G = { (x , f(x) ) : x in X }$ be the graph of $f$. Then which one is true?






  1. $G$ is homeomorphic to $X$


  2. $G$ is homeomorphic to $mathbb R$


  3. $G$ is homeomorphic to $X times mathbb R$


  4. $G$ is homeomorphic to $mathbb R times X$



My attempt : My answer None of the option is correct



For option $1)$ Consider $X=(0,1)$ and $f(x)=x$ then $G={(x,x):xin (0,1)}$ which is a closed set but $(0,1)$ is not



option $2)$ $X=[0,1]$ which is compact and $f(x)=x$.Its graph is compact but $mathbb R$ is not.



option $3)$ $f(x)=0$ .Then graph of $f={(x,0):xin mathbb R}$ i.e. the $x$ axis .Now remove the point $(0,0)$ from the graph of $f$ ,it becomes disconnected but $mathbb R^2setminus {(0,0)}$ is not.



option $4)$.Consider $f(x)=x$ .Then graph of $f={(0,y):xin mathbb R}$ i.e. the $y$ axis .Now remove the point $(0,0)$ from the graph of $f$ ,it becomes disconnected but $mathbb R^2setminus {(0,0)}$ is not.



Is my answer is correct or not correct ?



Pliz tell me






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 Dec 4 '18 at 14:01









jasminejasmine

1,791418




1,791418








  • 1




    $begingroup$
    For (1): $G$ doesn't look very closed to me (in particular: its complement doesn't include a neighbourhood of $(0,0)$, but does include $(0,0)$). The others are fine.
    $endgroup$
    – user3482749
    Dec 4 '18 at 14:04










  • $begingroup$
    thanks u @user3482749.
    $endgroup$
    – jasmine
    Dec 4 '18 at 14:23






  • 1




    $begingroup$
    3. and 4. are the "same" because $X times Y$ and $Y times X$ are homeomorphic for any two spaces $X,Y$.
    $endgroup$
    – Paul Frost
    Dec 4 '18 at 18:28














  • 1




    $begingroup$
    For (1): $G$ doesn't look very closed to me (in particular: its complement doesn't include a neighbourhood of $(0,0)$, but does include $(0,0)$). The others are fine.
    $endgroup$
    – user3482749
    Dec 4 '18 at 14:04










  • $begingroup$
    thanks u @user3482749.
    $endgroup$
    – jasmine
    Dec 4 '18 at 14:23






  • 1




    $begingroup$
    3. and 4. are the "same" because $X times Y$ and $Y times X$ are homeomorphic for any two spaces $X,Y$.
    $endgroup$
    – Paul Frost
    Dec 4 '18 at 18:28








1




1




$begingroup$
For (1): $G$ doesn't look very closed to me (in particular: its complement doesn't include a neighbourhood of $(0,0)$, but does include $(0,0)$). The others are fine.
$endgroup$
– user3482749
Dec 4 '18 at 14:04




$begingroup$
For (1): $G$ doesn't look very closed to me (in particular: its complement doesn't include a neighbourhood of $(0,0)$, but does include $(0,0)$). The others are fine.
$endgroup$
– user3482749
Dec 4 '18 at 14:04












$begingroup$
thanks u @user3482749.
$endgroup$
– jasmine
Dec 4 '18 at 14:23




$begingroup$
thanks u @user3482749.
$endgroup$
– jasmine
Dec 4 '18 at 14:23




1




1




$begingroup$
3. and 4. are the "same" because $X times Y$ and $Y times X$ are homeomorphic for any two spaces $X,Y$.
$endgroup$
– Paul Frost
Dec 4 '18 at 18:28




$begingroup$
3. and 4. are the "same" because $X times Y$ and $Y times X$ are homeomorphic for any two spaces $X,Y$.
$endgroup$
– Paul Frost
Dec 4 '18 at 18:28










1 Answer
1






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oldest

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1












$begingroup$

Your counter example for (1) doesn't work for two reasons: because ${(x,x)mid x in (0,1)} $ is not closed in the plane (it does not contain the limit points $(0,0)$ and $(1,1)$), and moreover, being open or closed are weak conditions to look at, since every space is both open and closed with respect to itself.



Item (1) is actually true, with homeomorphism $$Xni xmapsto (x,f(x))in G $$and inverse $$Gni (x,y)mapsto x in X. $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks Ivo terek
    $endgroup$
    – jasmine
    Dec 4 '18 at 14:23











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1 Answer
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1 Answer
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active

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active

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active

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1












$begingroup$

Your counter example for (1) doesn't work for two reasons: because ${(x,x)mid x in (0,1)} $ is not closed in the plane (it does not contain the limit points $(0,0)$ and $(1,1)$), and moreover, being open or closed are weak conditions to look at, since every space is both open and closed with respect to itself.



Item (1) is actually true, with homeomorphism $$Xni xmapsto (x,f(x))in G $$and inverse $$Gni (x,y)mapsto x in X. $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks Ivo terek
    $endgroup$
    – jasmine
    Dec 4 '18 at 14:23
















1












$begingroup$

Your counter example for (1) doesn't work for two reasons: because ${(x,x)mid x in (0,1)} $ is not closed in the plane (it does not contain the limit points $(0,0)$ and $(1,1)$), and moreover, being open or closed are weak conditions to look at, since every space is both open and closed with respect to itself.



Item (1) is actually true, with homeomorphism $$Xni xmapsto (x,f(x))in G $$and inverse $$Gni (x,y)mapsto x in X. $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks Ivo terek
    $endgroup$
    – jasmine
    Dec 4 '18 at 14:23














1












1








1





$begingroup$

Your counter example for (1) doesn't work for two reasons: because ${(x,x)mid x in (0,1)} $ is not closed in the plane (it does not contain the limit points $(0,0)$ and $(1,1)$), and moreover, being open or closed are weak conditions to look at, since every space is both open and closed with respect to itself.



Item (1) is actually true, with homeomorphism $$Xni xmapsto (x,f(x))in G $$and inverse $$Gni (x,y)mapsto x in X. $$






share|cite|improve this answer









$endgroup$



Your counter example for (1) doesn't work for two reasons: because ${(x,x)mid x in (0,1)} $ is not closed in the plane (it does not contain the limit points $(0,0)$ and $(1,1)$), and moreover, being open or closed are weak conditions to look at, since every space is both open and closed with respect to itself.



Item (1) is actually true, with homeomorphism $$Xni xmapsto (x,f(x))in G $$and inverse $$Gni (x,y)mapsto x in X. $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 14:16









Ivo TerekIvo Terek

46.3k954142




46.3k954142












  • $begingroup$
    thanks Ivo terek
    $endgroup$
    – jasmine
    Dec 4 '18 at 14:23


















  • $begingroup$
    thanks Ivo terek
    $endgroup$
    – jasmine
    Dec 4 '18 at 14:23
















$begingroup$
thanks Ivo terek
$endgroup$
– jasmine
Dec 4 '18 at 14:23




$begingroup$
thanks Ivo terek
$endgroup$
– jasmine
Dec 4 '18 at 14:23


















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