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Prove Inf A = 0 where set $A = m + nomega: m + nomega > 0, m, n in mathbbZ$, $omega$ is a positive irrational number.
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I found a solution of this question but the solution seemingly showed that there is only $0$ in set $A$ and this is obviously impossible. Where I am wrong? And in the last part of this solution, it says $1 = m_0alpha$. Is it because we can set $n = 0$, and $m = 1$, then $1$ is in $A$ and every member of $A$ is equal to some multiplication of $alpha$? 
calculus
edited Aug 6 at 16:11
Mason
1,2401224
asked Aug 6 at 15:57
Cathy
1207
add a comment |Â
up vote
3
down vote
favorite
I found a solution of this question but the solution seemingly showed that there is only $0$ in set $A$ and this is obviously impossible. Where I am wrong? And in the last part of this solution, it says $1 = m_0alpha$. Is it because we can set $n = 0$, and $m = 1$, then $1$ is in $A$ and every member of $A$ is equal to some multiplication of $alpha$?
calculus
edited Aug 6 at 16:11
Mason
1,2401224
asked Aug 6 at 15:57
Cathy
1207
1
The first sentence "inf A exists and is positive, i.e., $inf A geq 0$" is problematic, as "positive" is not the same as "greater than or equal to zero." It just shows the writer is comfortable with awkward wordings.
– Michael
Aug 6 at 16:00
(1) Where is the problem from? (2) Is it easy enough to replace this image with the text itself? (3) Can these tags be improved upon?
– Mason
Aug 6 at 16:16
Actually in every other language than English, positive means superior or equal to 0.
– Pjonin
Aug 6 at 16:19
$forall$ languages $neq$ English? That seems like hyperbole... and also a good reason to use another word in this forum. "Non Negative" is the unambiguous way English speaking mathematicians have resolved this.
– Mason
Aug 6 at 16:25
1
Nice proof in your post. The conclusions in the proof are based on assumption $alpha>0$ so is not true that $A$ consists of $0$ alone.
– Paramanand Singh
Aug 7 at 5:24
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I found a solution of this question but the solution seemingly showed that there is only $0$ in set $A$ and this is obviously impossible. Where I am wrong? And in the last part of this solution, it says $1 = m_0alpha$. Is it because we can set $n = 0$, and $m = 1$, then $1$ is in $A$ and every member of $A$ is equal to some multiplication of $alpha$?
calculus
edited Aug 6 at 16:11
Mason
1,2401224
asked Aug 6 at 15:57
Cathy
1207
I found a solution of this question but the solution seemingly showed that there is only $0$ in set $A$ and this is obviously impossible. Where I am wrong? And in the last part of this solution, it says $1 = m_0alpha$. Is it because we can set $n = 0$, and $m = 1$, then $1$ is in $A$ and every member of $A$ is equal to some multiplication of $alpha$?
calculus
edited Aug 6 at 16:11
Mason
1,2401224
asked Aug 6 at 15:57
Cathy
1207
edited Aug 6 at 16:11
Mason
1,2401224
edited Aug 6 at 16:11
Mason
1,2401224
edited Aug 6 at 16:11
Mason
1,2401224
1,2401224
asked Aug 6 at 15:57
Cathy
1207
asked Aug 6 at 15:57
Cathy
1207
asked Aug 6 at 15:57
Cathy
1207
1207
1
The first sentence "inf A exists and is positive, i.e., $inf A geq 0$" is problematic, as "positive" is not the same as "greater than or equal to zero." It just shows the writer is comfortable with awkward wordings.
– Michael
Aug 6 at 16:00
(1) Where is the problem from? (2) Is it easy enough to replace this image with the text itself? (3) Can these tags be improved upon?
– Mason
Aug 6 at 16:16
Actually in every other language than English, positive means superior or equal to 0.
– Pjonin
Aug 6 at 16:19
$forall$ languages $neq$ English? That seems like hyperbole... and also a good reason to use another word in this forum. "Non Negative" is the unambiguous way English speaking mathematicians have resolved this.
– Mason
Aug 6 at 16:25
1
Nice proof in your post. The conclusions in the proof are based on assumption $alpha>0$ so is not true that $A$ consists of $0$ alone.
– Paramanand Singh
Aug 7 at 5:24
add a comment |Â
1
The first sentence "inf A exists and is positive, i.e., $inf A geq 0$" is problematic, as "positive" is not the same as "greater than or equal to zero." It just shows the writer is comfortable with awkward wordings.
– Michael
Aug 6 at 16:00
(1) Where is the problem from? (2) Is it easy enough to replace this image with the text itself? (3) Can these tags be improved upon?
– Mason
Aug 6 at 16:16
Actually in every other language than English, positive means superior or equal to 0.
– Pjonin
Aug 6 at 16:19
$forall$ languages $neq$ English? That seems like hyperbole... and also a good reason to use another word in this forum. "Non Negative" is the unambiguous way English speaking mathematicians have resolved this.
– Mason
Aug 6 at 16:25
1
Nice proof in your post. The conclusions in the proof are based on assumption $alpha>0$ so is not true that $A$ consists of $0$ alone.
– Paramanand Singh
Aug 7 at 5:24
1
1
The first sentence "inf A exists and is positive, i.e., $inf A geq 0$" is problematic, as "positive" is not the same as "greater than or equal to zero." It just shows the writer is comfortable with awkward wordings.
– Michael
Aug 6 at 16:00
The first sentence "inf A exists and is positive, i.e., $inf A geq 0$" is problematic, as "positive" is not the same as "greater than or equal to zero." It just shows the writer is comfortable with awkward wordings.
– Michael
Aug 6 at 16:00
(1) Where is the problem from? (2) Is it easy enough to replace this image with the text itself? (3) Can these tags be improved upon?
– Mason
Aug 6 at 16:16
(1) Where is the problem from? (2) Is it easy enough to replace this image with the text itself? (3) Can these tags be improved upon?
– Mason
Aug 6 at 16:16
Actually in every other language than English, positive means superior or equal to 0.
– Pjonin
Aug 6 at 16:19
Actually in every other language than English, positive means superior or equal to 0.
– Pjonin
Aug 6 at 16:19
$forall$ languages $neq$ English? That seems like hyperbole... and also a good reason to use another word in this forum. "Non Negative" is the unambiguous way English speaking mathematicians have resolved this.
– Mason
Aug 6 at 16:25
$forall$ languages $neq$ English? That seems like hyperbole... and also a good reason to use another word in this forum. "Non Negative" is the unambiguous way English speaking mathematicians have resolved this.
– Mason
Aug 6 at 16:25
1
1
Nice proof in your post. The conclusions in the proof are based on assumption $alpha>0$ so is not true that $A$ consists of $0$ alone.
– Paramanand Singh
Aug 7 at 5:24
Nice proof in your post. The conclusions in the proof are based on assumption $alpha>0$ so is not true that $A$ consists of $0$ alone.
– Paramanand Singh
Aug 7 at 5:24
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Given $xinmathbbR$, we denote the largest integer that is not
greater than $x$ by $[x]$. That is, $[x]=maxninmathbbZmid nleq x$.
Define a function $theta:mathbbRrightarrow[0,1)$ by $theta(x)=x-[x]$.
For each $ninmathbbN$, define $x_n=theta(nomega)$. Firstly,
observe that $x_mneq x_n$ whenever $mneq n$. (For, suppose
the contrary that there exist $mneq n$ such that $x_m=x_n$.
Then $momega=N_1+x_m$ and $nomega=N_2+x_n$ for some integers
$N_1$and $N_2$. Now we have: $momega-N_1=x_m=x_n=nomega-N_2$
which implies $omega=fracN_1-N_2m-ninmathbbQ$, which
is a contradiction)
Let $varepsilon>0$ be given. Since the sequence $(x_n)_n$ is
bounded, by Bolzano-Weierstrass Theorem, it has a convergent subsequence.
In particular, there exist distinct $n_1,n_2inmathbbN$ such
that $0<|x_n_1-x_n_2|<varepsilon$. Without loss of generality,
we may assume that $x_n_1>x_n_2$ (otherwise, swap $x_n_1$
and $x_n_2$). Note that $x_n_1=n_1omega-N_1$ and $x_n_2=n_2omega-N_2$
for some integers $N_1$, $N_2$. Hence $x_n_1-x_n_2=(n_1-n_2)omega-(N_1-N_2)in A$.
This shows that $inf A=0$.
answered Aug 6 at 16:50
Danny Pak-Keung Chan
1,87628
Since $A$ is a subset of $[0,infty)$, obviously $inf A$ exists and $inf Ageq 0$. The difficult part is to show that $inf A=0$.
– Danny Pak-Keung Chan
Aug 6 at 16:56
Thanks! Why there exist distinct $n_1, n_2 in mathbbN$ such that $ 0 < |x_n1 -x_n2| < varepsilon$?
– Cathy
Aug 6 at 18:22
1
Choose a convergent subsequence $(x_n_k)_k$. Then there exists $K$ such that $|x_n_k-x_n_l|<varepsilon$ whenever $k,lgeq K$. Choose distinct $k,lgeq K$. Then $n_k$ and $n_l$ will serve as the integers $n_1$ and $n_2$ in my original proof.
– Danny Pak-Keung Chan
Aug 6 at 19:19
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Given $xinmathbbR$, we denote the largest integer that is not
greater than $x$ by $[x]$. That is, $[x]=maxninmathbbZmid nleq x$.
Define a function $theta:mathbbRrightarrow[0,1)$ by $theta(x)=x-[x]$.
For each $ninmathbbN$, define $x_n=theta(nomega)$. Firstly,
observe that $x_mneq x_n$ whenever $mneq n$. (For, suppose
the contrary that there exist $mneq n$ such that $x_m=x_n$.
Then $momega=N_1+x_m$ and $nomega=N_2+x_n$ for some integers
$N_1$and $N_2$. Now we have: $momega-N_1=x_m=x_n=nomega-N_2$
which implies $omega=fracN_1-N_2m-ninmathbbQ$, which
is a contradiction)
Let $varepsilon>0$ be given. Since the sequence $(x_n)_n$ is
bounded, by Bolzano-Weierstrass Theorem, it has a convergent subsequence.
In particular, there exist distinct $n_1,n_2inmathbbN$ such
that $0<|x_n_1-x_n_2|<varepsilon$. Without loss of generality,
we may assume that $x_n_1>x_n_2$ (otherwise, swap $x_n_1$
and $x_n_2$). Note that $x_n_1=n_1omega-N_1$ and $x_n_2=n_2omega-N_2$
for some integers $N_1$, $N_2$. Hence $x_n_1-x_n_2=(n_1-n_2)omega-(N_1-N_2)in A$.
This shows that $inf A=0$.
answered Aug 6 at 16:50
Danny Pak-Keung Chan
1,87628
Since $A$ is a subset of $[0,infty)$, obviously $inf A$ exists and $inf Ageq 0$. The difficult part is to show that $inf A=0$.
– Danny Pak-Keung Chan
Aug 6 at 16:56
Thanks! Why there exist distinct $n_1, n_2 in mathbbN$ such that $ 0 < |x_n1 -x_n2| < varepsilon$?
– Cathy
Aug 6 at 18:22
1
Choose a convergent subsequence $(x_n_k)_k$. Then there exists $K$ such that $|x_n_k-x_n_l|<varepsilon$ whenever $k,lgeq K$. Choose distinct $k,lgeq K$. Then $n_k$ and $n_l$ will serve as the integers $n_1$ and $n_2$ in my original proof.
– Danny Pak-Keung Chan
Aug 6 at 19:19
add a comment |Â
up vote
1
down vote
Given $xinmathbbR$, we denote the largest integer that is not
greater than $x$ by $[x]$. That is, $[x]=maxninmathbbZmid nleq x$.
Define a function $theta:mathbbRrightarrow[0,1)$ by $theta(x)=x-[x]$.
For each $ninmathbbN$, define $x_n=theta(nomega)$. Firstly,
observe that $x_mneq x_n$ whenever $mneq n$. (For, suppose
the contrary that there exist $mneq n$ such that $x_m=x_n$.
Then $momega=N_1+x_m$ and $nomega=N_2+x_n$ for some integers
$N_1$and $N_2$. Now we have: $momega-N_1=x_m=x_n=nomega-N_2$
which implies $omega=fracN_1-N_2m-ninmathbbQ$, which
is a contradiction)
Let $varepsilon>0$ be given. Since the sequence $(x_n)_n$ is
bounded, by Bolzano-Weierstrass Theorem, it has a convergent subsequence.
In particular, there exist distinct $n_1,n_2inmathbbN$ such
that $0<|x_n_1-x_n_2|<varepsilon$. Without loss of generality,
we may assume that $x_n_1>x_n_2$ (otherwise, swap $x_n_1$
and $x_n_2$). Note that $x_n_1=n_1omega-N_1$ and $x_n_2=n_2omega-N_2$
for some integers $N_1$, $N_2$. Hence $x_n_1-x_n_2=(n_1-n_2)omega-(N_1-N_2)in A$.
This shows that $inf A=0$.
answered Aug 6 at 16:50
Danny Pak-Keung Chan
1,87628
Since $A$ is a subset of $[0,infty)$, obviously $inf A$ exists and $inf Ageq 0$. The difficult part is to show that $inf A=0$.
– Danny Pak-Keung Chan
Aug 6 at 16:56
Thanks! Why there exist distinct $n_1, n_2 in mathbbN$ such that $ 0 < |x_n1 -x_n2| < varepsilon$?
– Cathy
Aug 6 at 18:22
1
Choose a convergent subsequence $(x_n_k)_k$. Then there exists $K$ such that $|x_n_k-x_n_l|<varepsilon$ whenever $k,lgeq K$. Choose distinct $k,lgeq K$. Then $n_k$ and $n_l$ will serve as the integers $n_1$ and $n_2$ in my original proof.
– Danny Pak-Keung Chan
Aug 6 at 19:19
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Given $xinmathbbR$, we denote the largest integer that is not
greater than $x$ by $[x]$. That is, $[x]=maxninmathbbZmid nleq x$.
Define a function $theta:mathbbRrightarrow[0,1)$ by $theta(x)=x-[x]$.
For each $ninmathbbN$, define $x_n=theta(nomega)$. Firstly,
observe that $x_mneq x_n$ whenever $mneq n$. (For, suppose
the contrary that there exist $mneq n$ such that $x_m=x_n$.
Then $momega=N_1+x_m$ and $nomega=N_2+x_n$ for some integers
$N_1$and $N_2$. Now we have: $momega-N_1=x_m=x_n=nomega-N_2$
which implies $omega=fracN_1-N_2m-ninmathbbQ$, which
is a contradiction)
Let $varepsilon>0$ be given. Since the sequence $(x_n)_n$ is
bounded, by Bolzano-Weierstrass Theorem, it has a convergent subsequence.
In particular, there exist distinct $n_1,n_2inmathbbN$ such
that $0<|x_n_1-x_n_2|<varepsilon$. Without loss of generality,
we may assume that $x_n_1>x_n_2$ (otherwise, swap $x_n_1$
and $x_n_2$). Note that $x_n_1=n_1omega-N_1$ and $x_n_2=n_2omega-N_2$
for some integers $N_1$, $N_2$. Hence $x_n_1-x_n_2=(n_1-n_2)omega-(N_1-N_2)in A$.
This shows that $inf A=0$.
answered Aug 6 at 16:50
Danny Pak-Keung Chan
1,87628
Given $xinmathbbR$, we denote the largest integer that is not
greater than $x$ by $[x]$. That is, $[x]=maxninmathbbZmid nleq x$.
Define a function $theta:mathbbRrightarrow[0,1)$ by $theta(x)=x-[x]$.
For each $ninmathbbN$, define $x_n=theta(nomega)$. Firstly,
observe that $x_mneq x_n$ whenever $mneq n$. (For, suppose
the contrary that there exist $mneq n$ such that $x_m=x_n$.
Then $momega=N_1+x_m$ and $nomega=N_2+x_n$ for some integers
$N_1$and $N_2$. Now we have: $momega-N_1=x_m=x_n=nomega-N_2$
which implies $omega=fracN_1-N_2m-ninmathbbQ$, which
is a contradiction)
Let $varepsilon>0$ be given. Since the sequence $(x_n)_n$ is
bounded, by Bolzano-Weierstrass Theorem, it has a convergent subsequence.
In particular, there exist distinct $n_1,n_2inmathbbN$ such
that $0<|x_n_1-x_n_2|<varepsilon$. Without loss of generality,
we may assume that $x_n_1>x_n_2$ (otherwise, swap $x_n_1$
and $x_n_2$). Note that $x_n_1=n_1omega-N_1$ and $x_n_2=n_2omega-N_2$
for some integers $N_1$, $N_2$. Hence $x_n_1-x_n_2=(n_1-n_2)omega-(N_1-N_2)in A$.
This shows that $inf A=0$.
answered Aug 6 at 16:50
Danny Pak-Keung Chan
1,87628
answered Aug 6 at 16:50
Danny Pak-Keung Chan
1,87628
answered Aug 6 at 16:50
Danny Pak-Keung Chan
1,87628
answered Aug 6 at 16:50
Danny Pak-Keung Chan
1,87628
1,87628
Since $A$ is a subset of $[0,infty)$, obviously $inf A$ exists and $inf Ageq 0$. The difficult part is to show that $inf A=0$.
– Danny Pak-Keung Chan
Aug 6 at 16:56
Thanks! Why there exist distinct $n_1, n_2 in mathbbN$ such that $ 0 < |x_n1 -x_n2| < varepsilon$?
– Cathy
Aug 6 at 18:22
1
Choose a convergent subsequence $(x_n_k)_k$. Then there exists $K$ such that $|x_n_k-x_n_l|<varepsilon$ whenever $k,lgeq K$. Choose distinct $k,lgeq K$. Then $n_k$ and $n_l$ will serve as the integers $n_1$ and $n_2$ in my original proof.
– Danny Pak-Keung Chan
Aug 6 at 19:19
add a comment |Â
Since $A$ is a subset of $[0,infty)$, obviously $inf A$ exists and $inf Ageq 0$. The difficult part is to show that $inf A=0$.
– Danny Pak-Keung Chan
Aug 6 at 16:56
Thanks! Why there exist distinct $n_1, n_2 in mathbbN$ such that $ 0 < |x_n1 -x_n2| < varepsilon$?
– Cathy
Aug 6 at 18:22
1
Choose a convergent subsequence $(x_n_k)_k$. Then there exists $K$ such that $|x_n_k-x_n_l|<varepsilon$ whenever $k,lgeq K$. Choose distinct $k,lgeq K$. Then $n_k$ and $n_l$ will serve as the integers $n_1$ and $n_2$ in my original proof.
– Danny Pak-Keung Chan
Aug 6 at 19:19
Since $A$ is a subset of $[0,infty)$, obviously $inf A$ exists and $inf Ageq 0$. The difficult part is to show that $inf A=0$.
– Danny Pak-Keung Chan
Aug 6 at 16:56
Since $A$ is a subset of $[0,infty)$, obviously $inf A$ exists and $inf Ageq 0$. The difficult part is to show that $inf A=0$.
– Danny Pak-Keung Chan
Aug 6 at 16:56
Thanks! Why there exist distinct $n_1, n_2 in mathbbN$ such that $ 0 < |x_n1 -x_n2| < varepsilon$?
– Cathy
Aug 6 at 18:22
Thanks! Why there exist distinct $n_1, n_2 in mathbbN$ such that $ 0 < |x_n1 -x_n2| < varepsilon$?
– Cathy
Aug 6 at 18:22
1
1
Choose a convergent subsequence $(x_n_k)_k$. Then there exists $K$ such that $|x_n_k-x_n_l|<varepsilon$ whenever $k,lgeq K$. Choose distinct $k,lgeq K$. Then $n_k$ and $n_l$ will serve as the integers $n_1$ and $n_2$ in my original proof.
– Danny Pak-Keung Chan
Aug 6 at 19:19
Choose a convergent subsequence $(x_n_k)_k$. Then there exists $K$ such that $|x_n_k-x_n_l|<varepsilon$ whenever $k,lgeq K$. Choose distinct $k,lgeq K$. Then $n_k$ and $n_l$ will serve as the integers $n_1$ and $n_2$ in my original proof.
– Danny Pak-Keung Chan
Aug 6 at 19:19
add a comment |Â
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1
The first sentence "inf A exists and is positive, i.e., $inf A geq 0$" is problematic, as "positive" is not the same as "greater than or equal to zero." It just shows the writer is comfortable with awkward wordings.
– Michael
Aug 6 at 16:00
(1) Where is the problem from? (2) Is it easy enough to replace this image with the text itself? (3) Can these tags be improved upon?
– Mason
Aug 6 at 16:16
Actually in every other language than English, positive means superior or equal to 0.
– Pjonin
Aug 6 at 16:19
$forall$ languages $neq$ English? That seems like hyperbole... and also a good reason to use another word in this forum. "Non Negative" is the unambiguous way English speaking mathematicians have resolved this.
– Mason
Aug 6 at 16:25
1
Nice proof in your post. The conclusions in the proof are based on assumption $alpha>0$ so is not true that $A$ consists of $0$ alone.
– Paramanand Singh
Aug 7 at 5:24