Mixing
How do I prove $B_{q} neq emptyset$.
$begingroup$
Suppose that $mathbb P$ is the uniform distribution on $[0,1)$. Partition the interval $[0,1)$ into an equivalence class such that $xsim y$ ($x$ is equivalent to $y$) if $x-yinmathbb Q$, the set of rational numbers.
(a) Show that $sim$ is an equivalence relation.
(b) Show that $mathbb{Q}cap[0,1)$ is an equivalence class.
(c) Find $(pi/10 + 5/6) pmod 1$ up to $10$ decimal places.
(d) Given a subset of $A$ of $[0,1)$ and $xin[0,1)$, define $A_x = x + A = {(x+a) pmod{1} ;|;ain A}$. Then $A_xsubset[0,1)$.
(i) Show that ${piover10} + mathbb{Q}cap[0,1)$ is an equivalence class.
(ii) Show that $x+mathbb{Q}cap[0,1)$ is an equivalence class for any $xin[0,1)setminusmathbb{Q}$.
Ive already answered problem 1 a-d. Now i must prove 2 part a) I'm stuck.
Given $1$, by the Axiom of Choice, there exists a nonempty set $Bsubset[0,1)$ such that $B$ contains exactly one member of each equivalence class. Prove each of the following:
(a) Suppose that $qinmathbb{Q}cap[0,1)$. Show that $B_qnevarnothing$.
probability-theory elementary-set-theory equivalence-relations
edited Jan 30 at 18:53
Decaf-Math
3,422926
asked Jan 30 at 18:02
H.BH.B
195
$endgroup$
add a comment |
$begingroup$
Suppose that $mathbb P$ is the uniform distribution on $[0,1)$. Partition the interval $[0,1)$ into an equivalence class such that $xsim y$ ($x$ is equivalent to $y$) if $x-yinmathbb Q$, the set of rational numbers.
(a) Show that $sim$ is an equivalence relation.
(b) Show that $mathbb{Q}cap[0,1)$ is an equivalence class.
(c) Find $(pi/10 + 5/6) pmod 1$ up to $10$ decimal places.
(d) Given a subset of $A$ of $[0,1)$ and $xin[0,1)$, define $A_x = x + A = {(x+a) pmod{1} ;|;ain A}$. Then $A_xsubset[0,1)$.
(i) Show that ${piover10} + mathbb{Q}cap[0,1)$ is an equivalence class.
(ii) Show that $x+mathbb{Q}cap[0,1)$ is an equivalence class for any $xin[0,1)setminusmathbb{Q}$.
Ive already answered problem 1 a-d. Now i must prove 2 part a) I'm stuck.
Given $1$, by the Axiom of Choice, there exists a nonempty set $Bsubset[0,1)$ such that $B$ contains exactly one member of each equivalence class. Prove each of the following:
(a) Suppose that $qinmathbb{Q}cap[0,1)$. Show that $B_qnevarnothing$.
probability-theory elementary-set-theory equivalence-relations
edited Jan 30 at 18:53
Decaf-Math
3,422926
asked Jan 30 at 18:02
H.BH.B
195
$endgroup$
add a comment |
$begingroup$
Suppose that $mathbb P$ is the uniform distribution on $[0,1)$. Partition the interval $[0,1)$ into an equivalence class such that $xsim y$ ($x$ is equivalent to $y$) if $x-yinmathbb Q$, the set of rational numbers.
(a) Show that $sim$ is an equivalence relation.
(b) Show that $mathbb{Q}cap[0,1)$ is an equivalence class.
(c) Find $(pi/10 + 5/6) pmod 1$ up to $10$ decimal places.
(d) Given a subset of $A$ of $[0,1)$ and $xin[0,1)$, define $A_x = x + A = {(x+a) pmod{1} ;|;ain A}$. Then $A_xsubset[0,1)$.
(i) Show that ${piover10} + mathbb{Q}cap[0,1)$ is an equivalence class.
(ii) Show that $x+mathbb{Q}cap[0,1)$ is an equivalence class for any $xin[0,1)setminusmathbb{Q}$.
Ive already answered problem 1 a-d. Now i must prove 2 part a) I'm stuck.
Given $1$, by the Axiom of Choice, there exists a nonempty set $Bsubset[0,1)$ such that $B$ contains exactly one member of each equivalence class. Prove each of the following:
(a) Suppose that $qinmathbb{Q}cap[0,1)$. Show that $B_qnevarnothing$.
probability-theory elementary-set-theory equivalence-relations
edited Jan 30 at 18:53
Decaf-Math
3,422926
asked Jan 30 at 18:02
H.BH.B
195
$endgroup$
Suppose that $mathbb P$ is the uniform distribution on $[0,1)$. Partition the interval $[0,1)$ into an equivalence class such that $xsim y$ ($x$ is equivalent to $y$) if $x-yinmathbb Q$, the set of rational numbers.
(a) Show that $sim$ is an equivalence relation.
(b) Show that $mathbb{Q}cap[0,1)$ is an equivalence class.
(c) Find $(pi/10 + 5/6) pmod 1$ up to $10$ decimal places.
(d) Given a subset of $A$ of $[0,1)$ and $xin[0,1)$, define $A_x = x + A = {(x+a) pmod{1} ;|;ain A}$. Then $A_xsubset[0,1)$.
(i) Show that ${piover10} + mathbb{Q}cap[0,1)$ is an equivalence class.
(ii) Show that $x+mathbb{Q}cap[0,1)$ is an equivalence class for any $xin[0,1)setminusmathbb{Q}$.
Ive already answered problem 1 a-d. Now i must prove 2 part a) I'm stuck.
Given $1$, by the Axiom of Choice, there exists a nonempty set $Bsubset[0,1)$ such that $B$ contains exactly one member of each equivalence class. Prove each of the following:
(a) Suppose that $qinmathbb{Q}cap[0,1)$. Show that $B_qnevarnothing$.
probability-theory elementary-set-theory equivalence-relations
probability-theory elementary-set-theory equivalence-relations
edited Jan 30 at 18:53
Decaf-Math
3,422926
asked Jan 30 at 18:02
H.BH.B
195
edited Jan 30 at 18:53
Decaf-Math
3,422926
asked Jan 30 at 18:02
H.BH.B
195
edited Jan 30 at 18:53
Decaf-Math
3,422926
edited Jan 30 at 18:53
Decaf-Math
3,422926
edited Jan 30 at 18:53
Decaf-Math
3,422926
3,422926
asked Jan 30 at 18:02
H.BH.B
195
asked Jan 30 at 18:02
H.BH.B
195
asked Jan 30 at 18:02
H.BH.B
195
195
add a comment |
add a comment |
1 Answer
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$begingroup$
$B$ is nonempty, and $B_q$ is just a translation of $B$ by $q$.
Choose $x in B$. Then $x + q mod 1 = x+q - n in B_q$ for some integer $n$.
answered Jan 30 at 18:13
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
$begingroup$
Yes, i understand that it is a translation, but i am having a hard time figuring out how to put that into words to prove that it is nonempty
$endgroup$
– H.B
Jan 30 at 18:16
$begingroup$
If you have an element, and you translate that element, it is still an element. It can't disappear.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:21
$begingroup$
so this would just be a straight forward proof?
$endgroup$
– H.B
Jan 30 at 18:30
$begingroup$
it just follows from since B is not empty then B_q is nonempty since B_q=x+q?
$endgroup$
– H.B
Jan 30 at 18:37
$begingroup$
Yeah. It's really immediate
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:44
add a comment |
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$begingroup$
$B$ is nonempty, and $B_q$ is just a translation of $B$ by $q$.
Choose $x in B$. Then $x + q mod 1 = x+q - n in B_q$ for some integer $n$.
answered Jan 30 at 18:13
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
$begingroup$
Yes, i understand that it is a translation, but i am having a hard time figuring out how to put that into words to prove that it is nonempty
$endgroup$
– H.B
Jan 30 at 18:16
$begingroup$
If you have an element, and you translate that element, it is still an element. It can't disappear.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:21
$begingroup$
so this would just be a straight forward proof?
$endgroup$
– H.B
Jan 30 at 18:30
$begingroup$
it just follows from since B is not empty then B_q is nonempty since B_q=x+q?
$endgroup$
– H.B
Jan 30 at 18:37
$begingroup$
Yeah. It's really immediate
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:44
add a comment |
$begingroup$
$B$ is nonempty, and $B_q$ is just a translation of $B$ by $q$.
Choose $x in B$. Then $x + q mod 1 = x+q - n in B_q$ for some integer $n$.
answered Jan 30 at 18:13
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
$begingroup$
Yes, i understand that it is a translation, but i am having a hard time figuring out how to put that into words to prove that it is nonempty
$endgroup$
– H.B
Jan 30 at 18:16
$begingroup$
If you have an element, and you translate that element, it is still an element. It can't disappear.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:21
$begingroup$
so this would just be a straight forward proof?
$endgroup$
– H.B
Jan 30 at 18:30
$begingroup$
it just follows from since B is not empty then B_q is nonempty since B_q=x+q?
$endgroup$
– H.B
Jan 30 at 18:37
$begingroup$
Yeah. It's really immediate
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:44
add a comment |
$begingroup$
$B$ is nonempty, and $B_q$ is just a translation of $B$ by $q$.
Choose $x in B$. Then $x + q mod 1 = x+q - n in B_q$ for some integer $n$.
answered Jan 30 at 18:13
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
$B$ is nonempty, and $B_q$ is just a translation of $B$ by $q$.
Choose $x in B$. Then $x + q mod 1 = x+q - n in B_q$ for some integer $n$.
answered Jan 30 at 18:13
Nathaniel MayerNathaniel Mayer
1,863516
edited Jan 30 at 18:18
answered Jan 30 at 18:13
Nathaniel MayerNathaniel Mayer
1,863516
answered Jan 30 at 18:13
Nathaniel MayerNathaniel Mayer
1,863516
answered Jan 30 at 18:13
Nathaniel MayerNathaniel Mayer
1,863516
1,863516
$begingroup$
Yes, i understand that it is a translation, but i am having a hard time figuring out how to put that into words to prove that it is nonempty
$endgroup$
– H.B
Jan 30 at 18:16
$begingroup$
If you have an element, and you translate that element, it is still an element. It can't disappear.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:21
$begingroup$
so this would just be a straight forward proof?
$endgroup$
– H.B
Jan 30 at 18:30
$begingroup$
it just follows from since B is not empty then B_q is nonempty since B_q=x+q?
$endgroup$
– H.B
Jan 30 at 18:37
$begingroup$
Yeah. It's really immediate
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:44
add a comment |
$begingroup$
Yes, i understand that it is a translation, but i am having a hard time figuring out how to put that into words to prove that it is nonempty
$endgroup$
– H.B
Jan 30 at 18:16
$begingroup$
If you have an element, and you translate that element, it is still an element. It can't disappear.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:21
$begingroup$
so this would just be a straight forward proof?
$endgroup$
– H.B
Jan 30 at 18:30
$begingroup$
it just follows from since B is not empty then B_q is nonempty since B_q=x+q?
$endgroup$
– H.B
Jan 30 at 18:37
$begingroup$
Yeah. It's really immediate
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:44
$begingroup$
Yes, i understand that it is a translation, but i am having a hard time figuring out how to put that into words to prove that it is nonempty
$endgroup$
– H.B
Jan 30 at 18:16
$begingroup$
Yes, i understand that it is a translation, but i am having a hard time figuring out how to put that into words to prove that it is nonempty
$endgroup$
– H.B
Jan 30 at 18:16
$begingroup$
If you have an element, and you translate that element, it is still an element. It can't disappear.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:21
$begingroup$
If you have an element, and you translate that element, it is still an element. It can't disappear.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:21
$begingroup$
so this would just be a straight forward proof?
$endgroup$
– H.B
Jan 30 at 18:30
$begingroup$
so this would just be a straight forward proof?
$endgroup$
– H.B
Jan 30 at 18:30
$begingroup$
it just follows from since B is not empty then B_q is nonempty since B_q=x+q?
$endgroup$
– H.B
Jan 30 at 18:37
$begingroup$
it just follows from since B is not empty then B_q is nonempty since B_q=x+q?
$endgroup$
– H.B
Jan 30 at 18:37
$begingroup$
Yeah. It's really immediate
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:44
$begingroup$
Yeah. It's really immediate
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:44
add a comment |
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