Mixing
Writing the set of Liouville numbers as an intersection of dense open subsets of $mathbb{R}$
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On the Wikipedia page they write the set of all Liouville numbers as $displaystyle{L = bigcap _ { n = 1 } ^ { infty } U _ { n }}$, and it's stated that each $U_n$ is an open dense subsets of $mathbb{R}$, where:
$$U _ { n } = bigcup _ { q = 2 } ^ { infty } bigcup _ { p = - infty } ^ { infty } left{ x in mathbb { R } : 0 < left| x - frac { p } { q } right| < frac { 1 } { q ^ { n } } right} = bigcup _ { q = 2 } ^ { infty } bigcup _ { p = - infty } ^ { infty } left( frac { p } { q } - frac { 1 } { q ^ { n } } , frac { p } { q } + frac { 1 } { q ^ { n } } right) backslash left{ frac { p } { q } right}$$
My question: why is each $U_n$ an open dense subset of $mathbb{R}$?
real-analysis number-theory liouville-numbers
asked Jan 28 at 14:45
Matheus AndradeMatheus Andrade
1,340418
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add a comment |
$begingroup$
On the Wikipedia page they write the set of all Liouville numbers as $displaystyle{L = bigcap _ { n = 1 } ^ { infty } U _ { n }}$, and it's stated that each $U_n$ is an open dense subsets of $mathbb{R}$, where:
$$U _ { n } = bigcup _ { q = 2 } ^ { infty } bigcup _ { p = - infty } ^ { infty } left{ x in mathbb { R } : 0 < left| x - frac { p } { q } right| < frac { 1 } { q ^ { n } } right} = bigcup _ { q = 2 } ^ { infty } bigcup _ { p = - infty } ^ { infty } left( frac { p } { q } - frac { 1 } { q ^ { n } } , frac { p } { q } + frac { 1 } { q ^ { n } } right) backslash left{ frac { p } { q } right}$$
My question: why is each $U_n$ an open dense subset of $mathbb{R}$?
real-analysis number-theory liouville-numbers
asked Jan 28 at 14:45
Matheus AndradeMatheus Andrade
1,340418
$endgroup$
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Open is obvious: union of open sets. Dense takes a little more work. Note that this implies that the Liouville numbers are dense in the reals, by Baire category theorem.
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– Math_QED
Jan 28 at 14:51
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Ah, I see now how obvious it is that they are open. But how would one show that they're dense in the reals?
$endgroup$
– Matheus Andrade
Jan 28 at 14:59
add a comment |
$begingroup$
On the Wikipedia page they write the set of all Liouville numbers as $displaystyle{L = bigcap _ { n = 1 } ^ { infty } U _ { n }}$, and it's stated that each $U_n$ is an open dense subsets of $mathbb{R}$, where:
$$U _ { n } = bigcup _ { q = 2 } ^ { infty } bigcup _ { p = - infty } ^ { infty } left{ x in mathbb { R } : 0 < left| x - frac { p } { q } right| < frac { 1 } { q ^ { n } } right} = bigcup _ { q = 2 } ^ { infty } bigcup _ { p = - infty } ^ { infty } left( frac { p } { q } - frac { 1 } { q ^ { n } } , frac { p } { q } + frac { 1 } { q ^ { n } } right) backslash left{ frac { p } { q } right}$$
My question: why is each $U_n$ an open dense subset of $mathbb{R}$?
real-analysis number-theory liouville-numbers
asked Jan 28 at 14:45
Matheus AndradeMatheus Andrade
1,340418
$endgroup$
On the Wikipedia page they write the set of all Liouville numbers as $displaystyle{L = bigcap _ { n = 1 } ^ { infty } U _ { n }}$, and it's stated that each $U_n$ is an open dense subsets of $mathbb{R}$, where:
$$U _ { n } = bigcup _ { q = 2 } ^ { infty } bigcup _ { p = - infty } ^ { infty } left{ x in mathbb { R } : 0 < left| x - frac { p } { q } right| < frac { 1 } { q ^ { n } } right} = bigcup _ { q = 2 } ^ { infty } bigcup _ { p = - infty } ^ { infty } left( frac { p } { q } - frac { 1 } { q ^ { n } } , frac { p } { q } + frac { 1 } { q ^ { n } } right) backslash left{ frac { p } { q } right}$$
My question: why is each $U_n$ an open dense subset of $mathbb{R}$?
real-analysis number-theory liouville-numbers
real-analysis number-theory liouville-numbers
asked Jan 28 at 14:45
Matheus AndradeMatheus Andrade
1,340418
asked Jan 28 at 14:45
Matheus AndradeMatheus Andrade
1,340418
asked Jan 28 at 14:45
Matheus AndradeMatheus Andrade
1,340418
asked Jan 28 at 14:45
Matheus AndradeMatheus Andrade
1,340418
asked Jan 28 at 14:45
Matheus AndradeMatheus Andrade
1,340418
1,340418
$begingroup$
Open is obvious: union of open sets. Dense takes a little more work. Note that this implies that the Liouville numbers are dense in the reals, by Baire category theorem.
$endgroup$
– Math_QED
Jan 28 at 14:51
$begingroup$
Ah, I see now how obvious it is that they are open. But how would one show that they're dense in the reals?
$endgroup$
– Matheus Andrade
Jan 28 at 14:59
add a comment |
$begingroup$
Open is obvious: union of open sets. Dense takes a little more work. Note that this implies that the Liouville numbers are dense in the reals, by Baire category theorem.
$endgroup$
– Math_QED
Jan 28 at 14:51
$begingroup$
Ah, I see now how obvious it is that they are open. But how would one show that they're dense in the reals?
$endgroup$
– Matheus Andrade
Jan 28 at 14:59
$begingroup$
Open is obvious: union of open sets. Dense takes a little more work. Note that this implies that the Liouville numbers are dense in the reals, by Baire category theorem.
$endgroup$
– Math_QED
Jan 28 at 14:51
$begingroup$
Open is obvious: union of open sets. Dense takes a little more work. Note that this implies that the Liouville numbers are dense in the reals, by Baire category theorem.
$endgroup$
– Math_QED
Jan 28 at 14:51
$begingroup$
Ah, I see now how obvious it is that they are open. But how would one show that they're dense in the reals?
$endgroup$
– Matheus Andrade
Jan 28 at 14:59
$begingroup$
Ah, I see now how obvious it is that they are open. But how would one show that they're dense in the reals?
$endgroup$
– Matheus Andrade
Jan 28 at 14:59
add a comment |
1 Answer
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They're dense because $overline{U_n} supset mathbb{Q}$, so $overline{overline{U_n}} = overline{U_n} supset overline{mathbb{Q}} = mathbb{R}$.
answered Jan 28 at 15:11
Matheus AndradeMatheus Andrade
1,340418
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add a comment |
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1 Answer
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$begingroup$
They're dense because $overline{U_n} supset mathbb{Q}$, so $overline{overline{U_n}} = overline{U_n} supset overline{mathbb{Q}} = mathbb{R}$.
answered Jan 28 at 15:11
Matheus AndradeMatheus Andrade
1,340418
$endgroup$
add a comment |
$begingroup$
They're dense because $overline{U_n} supset mathbb{Q}$, so $overline{overline{U_n}} = overline{U_n} supset overline{mathbb{Q}} = mathbb{R}$.
answered Jan 28 at 15:11
Matheus AndradeMatheus Andrade
1,340418
$endgroup$
add a comment |
$begingroup$
They're dense because $overline{U_n} supset mathbb{Q}$, so $overline{overline{U_n}} = overline{U_n} supset overline{mathbb{Q}} = mathbb{R}$.
answered Jan 28 at 15:11
Matheus AndradeMatheus Andrade
1,340418
$endgroup$
They're dense because $overline{U_n} supset mathbb{Q}$, so $overline{overline{U_n}} = overline{U_n} supset overline{mathbb{Q}} = mathbb{R}$.
answered Jan 28 at 15:11
Matheus AndradeMatheus Andrade
1,340418
edited Feb 3 at 21:47
answered Jan 28 at 15:11
Matheus AndradeMatheus Andrade
1,340418
answered Jan 28 at 15:11
Matheus AndradeMatheus Andrade
1,340418
answered Jan 28 at 15:11
Matheus AndradeMatheus Andrade
1,340418
1,340418
add a comment |
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$begingroup$
Open is obvious: union of open sets. Dense takes a little more work. Note that this implies that the Liouville numbers are dense in the reals, by Baire category theorem.
$endgroup$
– Math_QED
Jan 28 at 14:51
$begingroup$
Ah, I see now how obvious it is that they are open. But how would one show that they're dense in the reals?
$endgroup$
– Matheus Andrade
Jan 28 at 14:59