G is the Cantor set, $X$ is a countable set of G hence GX is dense
$begingroup$
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
asked Jan 25 at 20:12
Pedro GomesPedro Gomes
1,9132721
$endgroup$
add a comment |
$begingroup$
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
asked Jan 25 at 20:12
Pedro GomesPedro Gomes
1,9132721
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 25 at 20:12
Pedro GomesPedro Gomes
1,9132721
$endgroup$
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
asked Jan 25 at 20:12
Pedro GomesPedro Gomes
1,9132721
asked Jan 25 at 20:12
Pedro GomesPedro Gomes
1,9132721
asked Jan 25 at 20:12
Pedro GomesPedro Gomes
1,9132721
asked Jan 25 at 20:12
Pedro GomesPedro Gomes
1,9132721
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
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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.
answered Jan 25 at 21:06
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
add a comment |
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1 Answer
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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.
answered Jan 25 at 21:06
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 25 at 21:06
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 25 at 21:06
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
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.
answered Jan 25 at 21:06
Henno BrandsmaHenno Brandsma
112k348121
edited Jan 25 at 22:32
answered Jan 25 at 21:06
Henno BrandsmaHenno Brandsma
112k348121
answered Jan 25 at 21:06
Henno BrandsmaHenno Brandsma
112k348121
answered Jan 25 at 21:06
Henno BrandsmaHenno Brandsma
112k348121
112k348121
add a comment |
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$begingroup$
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