Open sets and intersections
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Suppose $G$ is an open subset of the real number that is not upper bounded. Is there a real number $x > 0$ such that the set of all integer multiples of $x$ intersects $G$ at infinitely many points?
That is, is it true that $exists x in mathbb{R}$ such that ${mxmid min mathbb{Z}}bigcap G$ is infinite?
My intuition tells me yes, since the fact that $G$ is not upper bounded seems to be a major factor here, but I can’t seem to prove it.
real-analysis
edited 14 hours ago
AdditIdent
39319
asked 15 hours ago
IUissopretty
354
|
show 5 more comments
up vote
4
down vote
favorite
Suppose $G$ is an open subset of the real number that is not upper bounded. Is there a real number $x > 0$ such that the set of all integer multiples of $x$ intersects $G$ at infinitely many points?
That is, is it true that $exists x in mathbb{R}$ such that ${mxmid min mathbb{Z}}bigcap G$ is infinite?
My intuition tells me yes, since the fact that $G$ is not upper bounded seems to be a major factor here, but I can’t seem to prove it.
real-analysis
edited 14 hours ago
AdditIdent
39319
asked 15 hours ago
IUissopretty
354
Is $G$ a subset of $mathbb R$?
– 5xum
15 hours ago
Could you explain more?
– Dadrahm
15 hours ago
Yes, I made an edit to the question I asked. It should be slightly clearer now.
– IUissopretty
15 hours ago
i think its, $G subset mathbb{R}$ does $exists~ (x in mathbb{R}) > 0 ~s.t~ A = {ax | a in mathbb{Z}}$ with $A cap G$ being infinitely large.
– Vaas
15 hours ago
What is your proof
– Dadrahm
15 hours ago
|
show 5 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Suppose $G$ is an open subset of the real number that is not upper bounded. Is there a real number $x > 0$ such that the set of all integer multiples of $x$ intersects $G$ at infinitely many points?
That is, is it true that $exists x in mathbb{R}$ such that ${mxmid min mathbb{Z}}bigcap G$ is infinite?
My intuition tells me yes, since the fact that $G$ is not upper bounded seems to be a major factor here, but I can’t seem to prove it.
real-analysis
edited 14 hours ago
AdditIdent
39319
asked 15 hours ago
IUissopretty
354
Suppose $G$ is an open subset of the real number that is not upper bounded. Is there a real number $x > 0$ such that the set of all integer multiples of $x$ intersects $G$ at infinitely many points?
That is, is it true that $exists x in mathbb{R}$ such that ${mxmid min mathbb{Z}}bigcap G$ is infinite?
My intuition tells me yes, since the fact that $G$ is not upper bounded seems to be a major factor here, but I can’t seem to prove it.
real-analysis
real-analysis
edited 14 hours ago
AdditIdent
39319
asked 15 hours ago
IUissopretty
354
edited 14 hours ago
AdditIdent
39319
asked 15 hours ago
IUissopretty
354
edited 14 hours ago
AdditIdent
39319
edited 14 hours ago
AdditIdent
39319
edited 14 hours ago
AdditIdent
39319
39319
asked 15 hours ago
IUissopretty
354
asked 15 hours ago
IUissopretty
354
asked 15 hours ago
IUissopretty
354
354
Is $G$ a subset of $mathbb R$?
– 5xum
15 hours ago
Could you explain more?
– Dadrahm
15 hours ago
Yes, I made an edit to the question I asked. It should be slightly clearer now.
– IUissopretty
15 hours ago
i think its, $G subset mathbb{R}$ does $exists~ (x in mathbb{R}) > 0 ~s.t~ A = {ax | a in mathbb{Z}}$ with $A cap G$ being infinitely large.
– Vaas
15 hours ago
What is your proof
– Dadrahm
15 hours ago
|
show 5 more comments
Is $G$ a subset of $mathbb R$?
– 5xum
15 hours ago
Could you explain more?
– Dadrahm
15 hours ago
Yes, I made an edit to the question I asked. It should be slightly clearer now.
– IUissopretty
15 hours ago
i think its, $G subset mathbb{R}$ does $exists~ (x in mathbb{R}) > 0 ~s.t~ A = {ax | a in mathbb{Z}}$ with $A cap G$ being infinitely large.
– Vaas
15 hours ago
What is your proof
– Dadrahm
15 hours ago
Is $G$ a subset of $mathbb R$?
– 5xum
15 hours ago
Is $G$ a subset of $mathbb R$?
– 5xum
15 hours ago
Could you explain more?
– Dadrahm
15 hours ago
Could you explain more?
– Dadrahm
15 hours ago
Yes, I made an edit to the question I asked. It should be slightly clearer now.
– IUissopretty
15 hours ago
Yes, I made an edit to the question I asked. It should be slightly clearer now.
– IUissopretty
15 hours ago
i think its, $G subset mathbb{R}$ does $exists~ (x in mathbb{R}) > 0 ~s.t~ A = {ax | a in mathbb{Z}}$ with $A cap G$ being infinitely large.
– Vaas
15 hours ago
i think its, $G subset mathbb{R}$ does $exists~ (x in mathbb{R}) > 0 ~s.t~ A = {ax | a in mathbb{Z}}$ with $A cap G$ being infinitely large.
– Vaas
15 hours ago
What is your proof
– Dadrahm
15 hours ago
What is your proof
– Dadrahm
15 hours ago
|
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
Yes. Suppose $nexists x$ such that ${nx:ninmathbb{Z}}cap G$ is infinite, then for any $epsilon>0$, ${nepsilon:ninmathbb{Z}}cap G$ is finite.
Let $n_0 = max {n inmathbb{Z}:nepsilon in G}$.
Take $ain G$ such that $a > n_0epsilon$. $[because G$ is unbounded above$]$.
Let $S = {x in G: (a,x) subset G}.$ $Sne phi$ as $G$ is open.
Let $b=sup S$.
So $(a, b) subset G$. Then $|b-a|<epsilon$ also implies $b=a$. We have arrived at a contradiction.
answered 13 hours ago
Offlaw
114
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Would you mind to suggest how does $|b-a|<epsilon$ imply $b=a$?
– AdditIdent
10 hours ago
As $epsilon$ is arbitrarily small.
– Offlaw
9 hours ago
As you choose $n_0$ you implicitly fix an $epsilon$ already, thus your $a$ and things else too depends on $epsilon$.
– AdditIdent
9 hours ago
I understood the problem with $epsilon$. There is a glitch in the proof. I'll try to mend it.
– Offlaw
9 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Yes. Suppose $nexists x$ such that ${nx:ninmathbb{Z}}cap G$ is infinite, then for any $epsilon>0$, ${nepsilon:ninmathbb{Z}}cap G$ is finite.
Let $n_0 = max {n inmathbb{Z}:nepsilon in G}$.
Take $ain G$ such that $a > n_0epsilon$. $[because G$ is unbounded above$]$.
Let $S = {x in G: (a,x) subset G}.$ $Sne phi$ as $G$ is open.
Let $b=sup S$.
So $(a, b) subset G$. Then $|b-a|<epsilon$ also implies $b=a$. We have arrived at a contradiction.
answered 13 hours ago
Offlaw
114
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Would you mind to suggest how does $|b-a|<epsilon$ imply $b=a$?
– AdditIdent
10 hours ago
As $epsilon$ is arbitrarily small.
– Offlaw
9 hours ago
As you choose $n_0$ you implicitly fix an $epsilon$ already, thus your $a$ and things else too depends on $epsilon$.
– AdditIdent
9 hours ago
I understood the problem with $epsilon$. There is a glitch in the proof. I'll try to mend it.
– Offlaw
9 hours ago
add a comment |
up vote
0
down vote
Yes. Suppose $nexists x$ such that ${nx:ninmathbb{Z}}cap G$ is infinite, then for any $epsilon>0$, ${nepsilon:ninmathbb{Z}}cap G$ is finite.
Let $n_0 = max {n inmathbb{Z}:nepsilon in G}$.
Take $ain G$ such that $a > n_0epsilon$. $[because G$ is unbounded above$]$.
Let $S = {x in G: (a,x) subset G}.$ $Sne phi$ as $G$ is open.
Let $b=sup S$.
So $(a, b) subset G$. Then $|b-a|<epsilon$ also implies $b=a$. We have arrived at a contradiction.
answered 13 hours ago
Offlaw
114
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Would you mind to suggest how does $|b-a|<epsilon$ imply $b=a$?
– AdditIdent
10 hours ago
As $epsilon$ is arbitrarily small.
– Offlaw
9 hours ago
As you choose $n_0$ you implicitly fix an $epsilon$ already, thus your $a$ and things else too depends on $epsilon$.
– AdditIdent
9 hours ago
I understood the problem with $epsilon$. There is a glitch in the proof. I'll try to mend it.
– Offlaw
9 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
Yes. Suppose $nexists x$ such that ${nx:ninmathbb{Z}}cap G$ is infinite, then for any $epsilon>0$, ${nepsilon:ninmathbb{Z}}cap G$ is finite.
Let $n_0 = max {n inmathbb{Z}:nepsilon in G}$.
Take $ain G$ such that $a > n_0epsilon$. $[because G$ is unbounded above$]$.
Let $S = {x in G: (a,x) subset G}.$ $Sne phi$ as $G$ is open.
Let $b=sup S$.
So $(a, b) subset G$. Then $|b-a|<epsilon$ also implies $b=a$. We have arrived at a contradiction.
answered 13 hours ago
Offlaw
114
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Yes. Suppose $nexists x$ such that ${nx:ninmathbb{Z}}cap G$ is infinite, then for any $epsilon>0$, ${nepsilon:ninmathbb{Z}}cap G$ is finite.
Let $n_0 = max {n inmathbb{Z}:nepsilon in G}$.
Take $ain G$ such that $a > n_0epsilon$. $[because G$ is unbounded above$]$.
Let $S = {x in G: (a,x) subset G}.$ $Sne phi$ as $G$ is open.
Let $b=sup S$.
So $(a, b) subset G$. Then $|b-a|<epsilon$ also implies $b=a$. We have arrived at a contradiction.
answered 13 hours ago
Offlaw
114
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 9 hours ago
answered 13 hours ago
Offlaw
114
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 13 hours ago
Offlaw
114
answered 13 hours ago
Offlaw
114
114
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Offlaw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Would you mind to suggest how does $|b-a|<epsilon$ imply $b=a$?
– AdditIdent
10 hours ago
As $epsilon$ is arbitrarily small.
– Offlaw
9 hours ago
As you choose $n_0$ you implicitly fix an $epsilon$ already, thus your $a$ and things else too depends on $epsilon$.
– AdditIdent
9 hours ago
I understood the problem with $epsilon$. There is a glitch in the proof. I'll try to mend it.
– Offlaw
9 hours ago
add a comment |
Would you mind to suggest how does $|b-a|<epsilon$ imply $b=a$?
– AdditIdent
10 hours ago
As $epsilon$ is arbitrarily small.
– Offlaw
9 hours ago
As you choose $n_0$ you implicitly fix an $epsilon$ already, thus your $a$ and things else too depends on $epsilon$.
– AdditIdent
9 hours ago
I understood the problem with $epsilon$. There is a glitch in the proof. I'll try to mend it.
– Offlaw
9 hours ago
Would you mind to suggest how does $|b-a|<epsilon$ imply $b=a$?
– AdditIdent
10 hours ago
Would you mind to suggest how does $|b-a|<epsilon$ imply $b=a$?
– AdditIdent
10 hours ago
As $epsilon$ is arbitrarily small.
– Offlaw
9 hours ago
As $epsilon$ is arbitrarily small.
– Offlaw
9 hours ago
As you choose $n_0$ you implicitly fix an $epsilon$ already, thus your $a$ and things else too depends on $epsilon$.
– AdditIdent
9 hours ago
As you choose $n_0$ you implicitly fix an $epsilon$ already, thus your $a$ and things else too depends on $epsilon$.
– AdditIdent
9 hours ago
I understood the problem with $epsilon$. There is a glitch in the proof. I'll try to mend it.
– Offlaw
9 hours ago
I understood the problem with $epsilon$. There is a glitch in the proof. I'll try to mend it.
– Offlaw
9 hours ago
add a comment |
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Is $G$ a subset of $mathbb R$?
– 5xum
15 hours ago
Could you explain more?
– Dadrahm
15 hours ago
Yes, I made an edit to the question I asked. It should be slightly clearer now.
– IUissopretty
15 hours ago
i think its, $G subset mathbb{R}$ does $exists~ (x in mathbb{R}) > 0 ~s.t~ A = {ax | a in mathbb{Z}}$ with $A cap G$ being infinitely large.
– Vaas
15 hours ago
What is your proof
– Dadrahm
15 hours ago