G is the Cantor set, $X$ is a countable set of G hence GX is dense












0












$begingroup$



Exercise: Let $G$ be a Cantor set.



a) Prove that $G$ has a countable dense subset.



b)If $X$ is a countable subset of $G$ then $Gsetminus X$ is dense in $G$




a)
It is know that the Cantor set for example the ternary Cantor set has the same cardinality of $mathbb{R}$ despite its length is $0$. Since $mathbb{Q}$ are dense set in $mathbb{R}$ and $G$ is a subset of $mathbb{R}$ with uncountable elements hence $Gcapmathbb{Q}$ must be a countable dense subset of $G$.



b) I have no concrete idea on how to taclkle the question of countability and density in an abstract set $X$.



Questions:



1) Is my proof insofar right? If not. Why not?



2) How is b) solved?



Thanks in advance!






general-topology proof-verification proof-writing cantor-set







share|cite|improve this question









$endgroup$












  • $begingroup$
    For the first question: the cantor space $2^{omega}$ is clearly a Polish space, so you conclude. To see that it is Polish, consider that it is the countable product of the space ${0,1}$ with the discrete topology and this is a Polish space, so it is Polish.
    $endgroup$
    – Lei Feima
    Jan 25 at 20:52












  • $begingroup$
    for the second question: $X$ countable implies that $2^{omega}setminus X$ is uncountable.Then it is dense if and only if, for every open set $A$, $Acap (2^{omega}setminus X)neq emptyset$ , but then an argument from infinite combinatorics show you that it cannot be empty.
    $endgroup$
    – Lei Feima
    Jan 25 at 21:10












  • $begingroup$
    Not every Cantor set in the reals contains any rationals. The srtandard one contains plenty of them, and they do form a dense set. It does require some reasoning. Just compactness of $C$ implies it's separable.
    $endgroup$
    – Henno Brandsma
    Jan 25 at 22:34










  • $begingroup$
    Let $H$ be the set of all irrational numbers. Since $mathbb Q$ is a dense set in $mathbb R$ and $H$ is an uncountable subset of $mathbb R$, can you conclude that $Hcapmathbb Q$ is a countable dense subset of $G$?
    $endgroup$
    – bof
    Jan 26 at 0:55
















0












$begingroup$



Exercise: Let $G$ be a Cantor set.



a) Prove that $G$ has a countable dense subset.



b)If $X$ is a countable subset of $G$ then $Gsetminus X$ is dense in $G$




a)
It is know that the Cantor set for example the ternary Cantor set has the same cardinality of $mathbb{R}$ despite its length is $0$. Since $mathbb{Q}$ are dense set in $mathbb{R}$ and $G$ is a subset of $mathbb{R}$ with uncountable elements hence $Gcapmathbb{Q}$ must be a countable dense subset of $G$.



b) I have no concrete idea on how to taclkle the question of countability and density in an abstract set $X$.



Questions:



1) Is my proof insofar right? If not. Why not?



2) How is b) solved?



Thanks in advance!






general-topology proof-verification proof-writing cantor-set







share|cite|improve this question









$endgroup$












  • $begingroup$
    For the first question: the cantor space $2^{omega}$ is clearly a Polish space, so you conclude. To see that it is Polish, consider that it is the countable product of the space ${0,1}$ with the discrete topology and this is a Polish space, so it is Polish.
    $endgroup$
    – Lei Feima
    Jan 25 at 20:52












  • $begingroup$
    for the second question: $X$ countable implies that $2^{omega}setminus X$ is uncountable.Then it is dense if and only if, for every open set $A$, $Acap (2^{omega}setminus X)neq emptyset$ , but then an argument from infinite combinatorics show you that it cannot be empty.
    $endgroup$
    – Lei Feima
    Jan 25 at 21:10












  • $begingroup$
    Not every Cantor set in the reals contains any rationals. The srtandard one contains plenty of them, and they do form a dense set. It does require some reasoning. Just compactness of $C$ implies it's separable.
    $endgroup$
    – Henno Brandsma
    Jan 25 at 22:34










  • $begingroup$
    Let $H$ be the set of all irrational numbers. Since $mathbb Q$ is a dense set in $mathbb R$ and $H$ is an uncountable subset of $mathbb R$, can you conclude that $Hcapmathbb Q$ is a countable dense subset of $G$?
    $endgroup$
    – bof
    Jan 26 at 0:55














0












0








0





$begingroup$



Exercise: Let $G$ be a Cantor set.



a) Prove that $G$ has a countable dense subset.



b)If $X$ is a countable subset of $G$ then $Gsetminus X$ is dense in $G$




a)
It is know that the Cantor set for example the ternary Cantor set has the same cardinality of $mathbb{R}$ despite its length is $0$. Since $mathbb{Q}$ are dense set in $mathbb{R}$ and $G$ is a subset of $mathbb{R}$ with uncountable elements hence $Gcapmathbb{Q}$ must be a countable dense subset of $G$.



b) I have no concrete idea on how to taclkle the question of countability and density in an abstract set $X$.



Questions:



1) Is my proof insofar right? If not. Why not?



2) How is b) solved?



Thanks in advance!






general-topology proof-verification proof-writing cantor-set







share|cite|improve this question









$endgroup$





Exercise: Let $G$ be a Cantor set.



a) Prove that $G$ has a countable dense subset.



b)If $X$ is a countable subset of $G$ then $Gsetminus X$ is dense in $G$




a)
It is know that the Cantor set for example the ternary Cantor set has the same cardinality of $mathbb{R}$ despite its length is $0$. Since $mathbb{Q}$ are dense set in $mathbb{R}$ and $G$ is a subset of $mathbb{R}$ with uncountable elements hence $Gcapmathbb{Q}$ must be a countable dense subset of $G$.



b) I have no concrete idea on how to taclkle the question of countability and density in an abstract set $X$.



Questions:



1) Is my proof insofar right? If not. Why not?



2) How is b) solved?



Thanks in advance!






general-topology proof-verification proof-writing cantor-set




general-topology proof-verification proof-writing cantor-set






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 25 at 20:12









Pedro GomesPedro Gomes

1,9132721




1,9132721












  • $begingroup$
    For the first question: the cantor space $2^{omega}$ is clearly a Polish space, so you conclude. To see that it is Polish, consider that it is the countable product of the space ${0,1}$ with the discrete topology and this is a Polish space, so it is Polish.
    $endgroup$
    – Lei Feima
    Jan 25 at 20:52












  • $begingroup$
    for the second question: $X$ countable implies that $2^{omega}setminus X$ is uncountable.Then it is dense if and only if, for every open set $A$, $Acap (2^{omega}setminus X)neq emptyset$ , but then an argument from infinite combinatorics show you that it cannot be empty.
    $endgroup$
    – Lei Feima
    Jan 25 at 21:10












  • $begingroup$
    Not every Cantor set in the reals contains any rationals. The srtandard one contains plenty of them, and they do form a dense set. It does require some reasoning. Just compactness of $C$ implies it's separable.
    $endgroup$
    – Henno Brandsma
    Jan 25 at 22:34










  • $begingroup$
    Let $H$ be the set of all irrational numbers. Since $mathbb Q$ is a dense set in $mathbb R$ and $H$ is an uncountable subset of $mathbb R$, can you conclude that $Hcapmathbb Q$ is a countable dense subset of $G$?
    $endgroup$
    – bof
    Jan 26 at 0:55


















  • $begingroup$
    For the first question: the cantor space $2^{omega}$ is clearly a Polish space, so you conclude. To see that it is Polish, consider that it is the countable product of the space ${0,1}$ with the discrete topology and this is a Polish space, so it is Polish.
    $endgroup$
    – Lei Feima
    Jan 25 at 20:52












  • $begingroup$
    for the second question: $X$ countable implies that $2^{omega}setminus X$ is uncountable.Then it is dense if and only if, for every open set $A$, $Acap (2^{omega}setminus X)neq emptyset$ , but then an argument from infinite combinatorics show you that it cannot be empty.
    $endgroup$
    – Lei Feima
    Jan 25 at 21:10












  • $begingroup$
    Not every Cantor set in the reals contains any rationals. The srtandard one contains plenty of them, and they do form a dense set. It does require some reasoning. Just compactness of $C$ implies it's separable.
    $endgroup$
    – Henno Brandsma
    Jan 25 at 22:34










  • $begingroup$
    Let $H$ be the set of all irrational numbers. Since $mathbb Q$ is a dense set in $mathbb R$ and $H$ is an uncountable subset of $mathbb R$, can you conclude that $Hcapmathbb Q$ is a countable dense subset of $G$?
    $endgroup$
    – bof
    Jan 26 at 0:55
















$begingroup$
For the first question: the cantor space $2^{omega}$ is clearly a Polish space, so you conclude. To see that it is Polish, consider that it is the countable product of the space ${0,1}$ with the discrete topology and this is a Polish space, so it is Polish.
$endgroup$
– Lei Feima
Jan 25 at 20:52






$begingroup$
For the first question: the cantor space $2^{omega}$ is clearly a Polish space, so you conclude. To see that it is Polish, consider that it is the countable product of the space ${0,1}$ with the discrete topology and this is a Polish space, so it is Polish.
$endgroup$
– Lei Feima
Jan 25 at 20:52














$begingroup$
for the second question: $X$ countable implies that $2^{omega}setminus X$ is uncountable.Then it is dense if and only if, for every open set $A$, $Acap (2^{omega}setminus X)neq emptyset$ , but then an argument from infinite combinatorics show you that it cannot be empty.
$endgroup$
– Lei Feima
Jan 25 at 21:10






$begingroup$
for the second question: $X$ countable implies that $2^{omega}setminus X$ is uncountable.Then it is dense if and only if, for every open set $A$, $Acap (2^{omega}setminus X)neq emptyset$ , but then an argument from infinite combinatorics show you that it cannot be empty.
$endgroup$
– Lei Feima
Jan 25 at 21:10














$begingroup$
Not every Cantor set in the reals contains any rationals. The srtandard one contains plenty of them, and they do form a dense set. It does require some reasoning. Just compactness of $C$ implies it's separable.
$endgroup$
– Henno Brandsma
Jan 25 at 22:34




$begingroup$
Not every Cantor set in the reals contains any rationals. The srtandard one contains plenty of them, and they do form a dense set. It does require some reasoning. Just compactness of $C$ implies it's separable.
$endgroup$
– Henno Brandsma
Jan 25 at 22:34












$begingroup$
Let $H$ be the set of all irrational numbers. Since $mathbb Q$ is a dense set in $mathbb R$ and $H$ is an uncountable subset of $mathbb R$, can you conclude that $Hcapmathbb Q$ is a countable dense subset of $G$?
$endgroup$
– bof
Jan 26 at 0:55




$begingroup$
Let $H$ be the set of all irrational numbers. Since $mathbb Q$ is a dense set in $mathbb R$ and $H$ is an uncountable subset of $mathbb R$, can you conclude that $Hcapmathbb Q$ is a countable dense subset of $G$?
$endgroup$
– bof
Jan 26 at 0:55










1 Answer
1






active

oldest

votes


















1












$begingroup$

A compact metric space is separable, so the Cantor set is too. For a concrete dense set in the Cantor set: take all the end points of the intervals in the construction: at stage $0$ we have $[0,1]$ and we add $0$ and $1$ to the dense set; at stage $1$ we have $[0,frac13]$ and $[frac23,1]$ and we add $frac13,frac23$ to the dense set; at the next stage we add $frac{1}{9}, frac29, frac79, frac89$ etc. In the end we get all "rationals" in the ternary expansion: all those numbers (that have only $0$ and $2$'s in a ternary expansion) that are eventually periodic.



All non-empty open subsets of $C$ are uncountable (they all contain sets homeomorphic to the whole Cantor set, even), so all of them intersect the complement of any countable set.






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    $begingroup$

    A compact metric space is separable, so the Cantor set is too. For a concrete dense set in the Cantor set: take all the end points of the intervals in the construction: at stage $0$ we have $[0,1]$ and we add $0$ and $1$ to the dense set; at stage $1$ we have $[0,frac13]$ and $[frac23,1]$ and we add $frac13,frac23$ to the dense set; at the next stage we add $frac{1}{9}, frac29, frac79, frac89$ etc. In the end we get all "rationals" in the ternary expansion: all those numbers (that have only $0$ and $2$'s in a ternary expansion) that are eventually periodic.



    All non-empty open subsets of $C$ are uncountable (they all contain sets homeomorphic to the whole Cantor set, even), so all of them intersect the complement of any countable set.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      A compact metric space is separable, so the Cantor set is too. For a concrete dense set in the Cantor set: take all the end points of the intervals in the construction: at stage $0$ we have $[0,1]$ and we add $0$ and $1$ to the dense set; at stage $1$ we have $[0,frac13]$ and $[frac23,1]$ and we add $frac13,frac23$ to the dense set; at the next stage we add $frac{1}{9}, frac29, frac79, frac89$ etc. In the end we get all "rationals" in the ternary expansion: all those numbers (that have only $0$ and $2$'s in a ternary expansion) that are eventually periodic.



      All non-empty open subsets of $C$ are uncountable (they all contain sets homeomorphic to the whole Cantor set, even), so all of them intersect the complement of any countable set.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        A compact metric space is separable, so the Cantor set is too. For a concrete dense set in the Cantor set: take all the end points of the intervals in the construction: at stage $0$ we have $[0,1]$ and we add $0$ and $1$ to the dense set; at stage $1$ we have $[0,frac13]$ and $[frac23,1]$ and we add $frac13,frac23$ to the dense set; at the next stage we add $frac{1}{9}, frac29, frac79, frac89$ etc. In the end we get all "rationals" in the ternary expansion: all those numbers (that have only $0$ and $2$'s in a ternary expansion) that are eventually periodic.



        All non-empty open subsets of $C$ are uncountable (they all contain sets homeomorphic to the whole Cantor set, even), so all of them intersect the complement of any countable set.






        share|cite|improve this answer











        $endgroup$



        A compact metric space is separable, so the Cantor set is too. For a concrete dense set in the Cantor set: take all the end points of the intervals in the construction: at stage $0$ we have $[0,1]$ and we add $0$ and $1$ to the dense set; at stage $1$ we have $[0,frac13]$ and $[frac23,1]$ and we add $frac13,frac23$ to the dense set; at the next stage we add $frac{1}{9}, frac29, frac79, frac89$ etc. In the end we get all "rationals" in the ternary expansion: all those numbers (that have only $0$ and $2$'s in a ternary expansion) that are eventually periodic.



        All non-empty open subsets of $C$ are uncountable (they all contain sets homeomorphic to the whole Cantor set, even), so all of them intersect the complement of any countable set.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 25 at 22:32

























        answered Jan 25 at 21:06









        Henno BrandsmaHenno Brandsma

        112k348121




        112k348121






























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