If the metric space is $mathbb Q$, is it connected?
2
$begingroup$
My idea:
I first observe its component, for example $dfrac{3}{2}$. Then, I consider in a metric space $mathbb{Q}$. I know that $left{dfrac{3}{2}right}$ is relatively closed to $mathbb{Q}$. However, the closure of $mathbb{Q}setminusleft{dfrac{3}{2}right}$ is $mathbb{Q}$. It would become $left{dfrac{3}{2}right}capmathbb{Q}$ instead of $emptyset$.
I could not find out two closed set to disconnect the metric space $mathbb{Q}$. Where am I wrong since intuitively $mathbb{Q}$ is not connected in $mathbb{R}$?
general-topology
edited Dec 24 '18 at 14:07
Saad
19.9k92352
asked Dec 24 '18 at 12:26
AndrewAndrew
243
$endgroup$
|
show 1 more comment
2
$begingroup$
My idea:
I first observe its component, for example $dfrac{3}{2}$. Then, I consider in a metric space $mathbb{Q}$. I know that $left{dfrac{3}{2}right}$ is relatively closed to $mathbb{Q}$. However, the closure of $mathbb{Q}setminusleft{dfrac{3}{2}right}$ is $mathbb{Q}$. It would become $left{dfrac{3}{2}right}capmathbb{Q}$ instead of $emptyset$.
I could not find out two closed set to disconnect the metric space $mathbb{Q}$. Where am I wrong since intuitively $mathbb{Q}$ is not connected in $mathbb{R}$?
general-topology
edited Dec 24 '18 at 14:07
Saad
19.9k92352
asked Dec 24 '18 at 12:26
AndrewAndrew
243
$endgroup$
$begingroup$
"intuitively Q isn't connected in R?" Why do you have that intuition? If you can exploit that more directly, it might be easier.
$endgroup$
– Mark S.
Dec 24 '18 at 12:28
$begingroup$
You are right $mathbb Qsetminus{3/2}$ is not closed, so that won't quite work. Try to think using irrational numbers.
$endgroup$
– Wojowu
Dec 24 '18 at 12:29
$begingroup$
I have used irrational number in R. but I think that I can't use irrational number in metric space Q
$endgroup$
– Andrew
Dec 24 '18 at 12:35
$begingroup$
You don't have to use an irrational number in the metric space Q. But you can nonetheless use it to define subsets in Q.
$endgroup$
– Lee Mosher
Dec 24 '18 at 12:38
$begingroup$
Think of it this way: I can define the following subsets of $mathbb Z$: $$A = {n in mathbb Z mid n >frac{1}{2}} = {1,2,3,...}, qquad B = {n in mathbb Z mid n < frac{1}{2}} = {...,-2,-1,0}$$ It makes perfect sense to write those sets and work with them, even though $frac{1}{2} notin mathbb Z$. The reason this works is because $mathbb Z$ and $frac{1}{2}$ are both contained in a larger set where inequality is defined, namely $mathbb R$.
$endgroup$
– Lee Mosher
Dec 24 '18 at 12:40
|
show 1 more comment
2
2
2
$begingroup$
My idea:
I first observe its component, for example $dfrac{3}{2}$. Then, I consider in a metric space $mathbb{Q}$. I know that $left{dfrac{3}{2}right}$ is relatively closed to $mathbb{Q}$. However, the closure of $mathbb{Q}setminusleft{dfrac{3}{2}right}$ is $mathbb{Q}$. It would become $left{dfrac{3}{2}right}capmathbb{Q}$ instead of $emptyset$.
I could not find out two closed set to disconnect the metric space $mathbb{Q}$. Where am I wrong since intuitively $mathbb{Q}$ is not connected in $mathbb{R}$?
general-topology
edited Dec 24 '18 at 14:07
Saad
19.9k92352
asked Dec 24 '18 at 12:26
AndrewAndrew
243
$endgroup$
My idea:
I first observe its component, for example $dfrac{3}{2}$. Then, I consider in a metric space $mathbb{Q}$. I know that $left{dfrac{3}{2}right}$ is relatively closed to $mathbb{Q}$. However, the closure of $mathbb{Q}setminusleft{dfrac{3}{2}right}$ is $mathbb{Q}$. It would become $left{dfrac{3}{2}right}capmathbb{Q}$ instead of $emptyset$.
I could not find out two closed set to disconnect the metric space $mathbb{Q}$. Where am I wrong since intuitively $mathbb{Q}$ is not connected in $mathbb{R}$?
general-topology
general-topology
edited Dec 24 '18 at 14:07
Saad
19.9k92352
asked Dec 24 '18 at 12:26
AndrewAndrew
243
edited Dec 24 '18 at 14:07
Saad
19.9k92352
asked Dec 24 '18 at 12:26
AndrewAndrew
243
edited Dec 24 '18 at 14:07
Saad
19.9k92352
edited Dec 24 '18 at 14:07
Saad
19.9k92352
edited Dec 24 '18 at 14:07
Saad
19.9k92352
19.9k92352
asked Dec 24 '18 at 12:26
AndrewAndrew
243
asked Dec 24 '18 at 12:26
AndrewAndrew
243
asked Dec 24 '18 at 12:26
AndrewAndrew
243
243
$begingroup$
"intuitively Q isn't connected in R?" Why do you have that intuition? If you can exploit that more directly, it might be easier.
$endgroup$
– Mark S.
Dec 24 '18 at 12:28
$begingroup$
You are right $mathbb Qsetminus{3/2}$ is not closed, so that won't quite work. Try to think using irrational numbers.
$endgroup$
– Wojowu
Dec 24 '18 at 12:29
$begingroup$
I have used irrational number in R. but I think that I can't use irrational number in metric space Q
$endgroup$
– Andrew
Dec 24 '18 at 12:35
$begingroup$
You don't have to use an irrational number in the metric space Q. But you can nonetheless use it to define subsets in Q.
$endgroup$
– Lee Mosher
Dec 24 '18 at 12:38
$begingroup$
Think of it this way: I can define the following subsets of $mathbb Z$: $$A = {n in mathbb Z mid n >frac{1}{2}} = {1,2,3,...}, qquad B = {n in mathbb Z mid n < frac{1}{2}} = {...,-2,-1,0}$$ It makes perfect sense to write those sets and work with them, even though $frac{1}{2} notin mathbb Z$. The reason this works is because $mathbb Z$ and $frac{1}{2}$ are both contained in a larger set where inequality is defined, namely $mathbb R$.
$endgroup$
– Lee Mosher
Dec 24 '18 at 12:40
|
show 1 more comment
$begingroup$
"intuitively Q isn't connected in R?" Why do you have that intuition? If you can exploit that more directly, it might be easier.
$endgroup$
– Mark S.
Dec 24 '18 at 12:28
$begingroup$
You are right $mathbb Qsetminus{3/2}$ is not closed, so that won't quite work. Try to think using irrational numbers.
$endgroup$
– Wojowu
Dec 24 '18 at 12:29
$begingroup$
I have used irrational number in R. but I think that I can't use irrational number in metric space Q
$endgroup$
– Andrew
Dec 24 '18 at 12:35
$begingroup$
You don't have to use an irrational number in the metric space Q. But you can nonetheless use it to define subsets in Q.
$endgroup$
– Lee Mosher
Dec 24 '18 at 12:38
$begingroup$
Think of it this way: I can define the following subsets of $mathbb Z$: $$A = {n in mathbb Z mid n >frac{1}{2}} = {1,2,3,...}, qquad B = {n in mathbb Z mid n < frac{1}{2}} = {...,-2,-1,0}$$ It makes perfect sense to write those sets and work with them, even though $frac{1}{2} notin mathbb Z$. The reason this works is because $mathbb Z$ and $frac{1}{2}$ are both contained in a larger set where inequality is defined, namely $mathbb R$.
$endgroup$
– Lee Mosher
Dec 24 '18 at 12:40
$begingroup$
"intuitively Q isn't connected in R?" Why do you have that intuition? If you can exploit that more directly, it might be easier.
$endgroup$
– Mark S.
Dec 24 '18 at 12:28
$begingroup$
"intuitively Q isn't connected in R?" Why do you have that intuition? If you can exploit that more directly, it might be easier.
$endgroup$
– Mark S.
Dec 24 '18 at 12:28
$begingroup$
You are right $mathbb Qsetminus{3/2}$ is not closed, so that won't quite work. Try to think using irrational numbers.
$endgroup$
– Wojowu
Dec 24 '18 at 12:29
$begingroup$
You are right $mathbb Qsetminus{3/2}$ is not closed, so that won't quite work. Try to think using irrational numbers.
$endgroup$
– Wojowu
Dec 24 '18 at 12:29
$begingroup$
I have used irrational number in R. but I think that I can't use irrational number in metric space Q
$endgroup$
– Andrew
Dec 24 '18 at 12:35
$begingroup$
I have used irrational number in R. but I think that I can't use irrational number in metric space Q
$endgroup$
– Andrew
Dec 24 '18 at 12:35
$begingroup$
You don't have to use an irrational number in the metric space Q. But you can nonetheless use it to define subsets in Q.
$endgroup$
– Lee Mosher
Dec 24 '18 at 12:38
$begingroup$
You don't have to use an irrational number in the metric space Q. But you can nonetheless use it to define subsets in Q.
$endgroup$
– Lee Mosher
Dec 24 '18 at 12:38
$begingroup$
Think of it this way: I can define the following subsets of $mathbb Z$: $$A = {n in mathbb Z mid n >frac{1}{2}} = {1,2,3,...}, qquad B = {n in mathbb Z mid n < frac{1}{2}} = {...,-2,-1,0}$$ It makes perfect sense to write those sets and work with them, even though $frac{1}{2} notin mathbb Z$. The reason this works is because $mathbb Z$ and $frac{1}{2}$ are both contained in a larger set where inequality is defined, namely $mathbb R$.
$endgroup$
– Lee Mosher
Dec 24 '18 at 12:40
$begingroup$
Think of it this way: I can define the following subsets of $mathbb Z$: $$A = {n in mathbb Z mid n >frac{1}{2}} = {1,2,3,...}, qquad B = {n in mathbb Z mid n < frac{1}{2}} = {...,-2,-1,0}$$ It makes perfect sense to write those sets and work with them, even though $frac{1}{2} notin mathbb Z$. The reason this works is because $mathbb Z$ and $frac{1}{2}$ are both contained in a larger set where inequality is defined, namely $mathbb R$.
$endgroup$
– Lee Mosher
Dec 24 '18 at 12:40
|
show 1 more comment
1 Answer
1
active
oldest
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1
$begingroup$
$mathbb Q cap (-infty, sqrt 2)$ and $mathbb Q cap ( sqrt 2,infty)$ are closed and disjoint subsets of $mathbb Q$ whose union is $mathbb Q$.
answered Dec 24 '18 at 12:31
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
$endgroup$
$begingroup$
but if you are in metric space Q, it won't have irrational number.
$endgroup$
– Andrew
Dec 24 '18 at 12:34
$begingroup$
@Andrew Set of all rational numbers less than $sqrt 2$ (which is set of all negative rationals, zero, and those positive one's whose squares are less than $ 2$ is awell defined subset of $mathbb Q$.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 12:39
$begingroup$
I think I probably understand your thought. Both of them are closed in Q, but they are disjoint.
$endgroup$
– Andrew
Dec 24 '18 at 12:43
$begingroup$
Write $A={ q in Bbb{Q} mid q^2 < 2 }$ to avoid using irrational numbers to define your clopen subsets.
$endgroup$
– dbx
Dec 24 '18 at 13:22
add a comment |
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1 Answer
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1 Answer
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oldest
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1
$begingroup$
$mathbb Q cap (-infty, sqrt 2)$ and $mathbb Q cap ( sqrt 2,infty)$ are closed and disjoint subsets of $mathbb Q$ whose union is $mathbb Q$.
answered Dec 24 '18 at 12:31
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
$endgroup$
$begingroup$
but if you are in metric space Q, it won't have irrational number.
$endgroup$
– Andrew
Dec 24 '18 at 12:34
$begingroup$
@Andrew Set of all rational numbers less than $sqrt 2$ (which is set of all negative rationals, zero, and those positive one's whose squares are less than $ 2$ is awell defined subset of $mathbb Q$.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 12:39
$begingroup$
I think I probably understand your thought. Both of them are closed in Q, but they are disjoint.
$endgroup$
– Andrew
Dec 24 '18 at 12:43
$begingroup$
Write $A={ q in Bbb{Q} mid q^2 < 2 }$ to avoid using irrational numbers to define your clopen subsets.
$endgroup$
– dbx
Dec 24 '18 at 13:22
add a comment |
1
$begingroup$
$mathbb Q cap (-infty, sqrt 2)$ and $mathbb Q cap ( sqrt 2,infty)$ are closed and disjoint subsets of $mathbb Q$ whose union is $mathbb Q$.
answered Dec 24 '18 at 12:31
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
$endgroup$
$begingroup$
but if you are in metric space Q, it won't have irrational number.
$endgroup$
– Andrew
Dec 24 '18 at 12:34
$begingroup$
@Andrew Set of all rational numbers less than $sqrt 2$ (which is set of all negative rationals, zero, and those positive one's whose squares are less than $ 2$ is awell defined subset of $mathbb Q$.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 12:39
$begingroup$
I think I probably understand your thought. Both of them are closed in Q, but they are disjoint.
$endgroup$
– Andrew
Dec 24 '18 at 12:43
$begingroup$
Write $A={ q in Bbb{Q} mid q^2 < 2 }$ to avoid using irrational numbers to define your clopen subsets.
$endgroup$
– dbx
Dec 24 '18 at 13:22
add a comment |
1
1
1
$begingroup$
$mathbb Q cap (-infty, sqrt 2)$ and $mathbb Q cap ( sqrt 2,infty)$ are closed and disjoint subsets of $mathbb Q$ whose union is $mathbb Q$.
answered Dec 24 '18 at 12:31
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
$endgroup$
$mathbb Q cap (-infty, sqrt 2)$ and $mathbb Q cap ( sqrt 2,infty)$ are closed and disjoint subsets of $mathbb Q$ whose union is $mathbb Q$.
answered Dec 24 '18 at 12:31
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
answered Dec 24 '18 at 12:31
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
answered Dec 24 '18 at 12:31
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
answered Dec 24 '18 at 12:31
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
65.6k42766
$begingroup$
but if you are in metric space Q, it won't have irrational number.
$endgroup$
– Andrew
Dec 24 '18 at 12:34
$begingroup$
@Andrew Set of all rational numbers less than $sqrt 2$ (which is set of all negative rationals, zero, and those positive one's whose squares are less than $ 2$ is awell defined subset of $mathbb Q$.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 12:39
$begingroup$
I think I probably understand your thought. Both of them are closed in Q, but they are disjoint.
$endgroup$
– Andrew
Dec 24 '18 at 12:43
$begingroup$
Write $A={ q in Bbb{Q} mid q^2 < 2 }$ to avoid using irrational numbers to define your clopen subsets.
$endgroup$
– dbx
Dec 24 '18 at 13:22
add a comment |
$begingroup$
but if you are in metric space Q, it won't have irrational number.
$endgroup$
– Andrew
Dec 24 '18 at 12:34
$begingroup$
@Andrew Set of all rational numbers less than $sqrt 2$ (which is set of all negative rationals, zero, and those positive one's whose squares are less than $ 2$ is awell defined subset of $mathbb Q$.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 12:39
$begingroup$
I think I probably understand your thought. Both of them are closed in Q, but they are disjoint.
$endgroup$
– Andrew
Dec 24 '18 at 12:43
$begingroup$
Write $A={ q in Bbb{Q} mid q^2 < 2 }$ to avoid using irrational numbers to define your clopen subsets.
$endgroup$
– dbx
Dec 24 '18 at 13:22
$begingroup$
but if you are in metric space Q, it won't have irrational number.
$endgroup$
– Andrew
Dec 24 '18 at 12:34
$begingroup$
but if you are in metric space Q, it won't have irrational number.
$endgroup$
– Andrew
Dec 24 '18 at 12:34
$begingroup$
@Andrew Set of all rational numbers less than $sqrt 2$ (which is set of all negative rationals, zero, and those positive one's whose squares are less than $ 2$ is awell defined subset of $mathbb Q$.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 12:39
$begingroup$
@Andrew Set of all rational numbers less than $sqrt 2$ (which is set of all negative rationals, zero, and those positive one's whose squares are less than $ 2$ is awell defined subset of $mathbb Q$.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 12:39
$begingroup$
I think I probably understand your thought. Both of them are closed in Q, but they are disjoint.
$endgroup$
– Andrew
Dec 24 '18 at 12:43
$begingroup$
I think I probably understand your thought. Both of them are closed in Q, but they are disjoint.
$endgroup$
– Andrew
Dec 24 '18 at 12:43
$begingroup$
Write $A={ q in Bbb{Q} mid q^2 < 2 }$ to avoid using irrational numbers to define your clopen subsets.
$endgroup$
– dbx
Dec 24 '18 at 13:22
$begingroup$
Write $A={ q in Bbb{Q} mid q^2 < 2 }$ to avoid using irrational numbers to define your clopen subsets.
$endgroup$
– dbx
Dec 24 '18 at 13:22
add a comment |
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$begingroup$
"intuitively Q isn't connected in R?" Why do you have that intuition? If you can exploit that more directly, it might be easier.
$endgroup$
– Mark S.
Dec 24 '18 at 12:28
$begingroup$
You are right $mathbb Qsetminus{3/2}$ is not closed, so that won't quite work. Try to think using irrational numbers.
$endgroup$
– Wojowu
Dec 24 '18 at 12:29
$begingroup$
I have used irrational number in R. but I think that I can't use irrational number in metric space Q
$endgroup$
– Andrew
Dec 24 '18 at 12:35
$begingroup$
You don't have to use an irrational number in the metric space Q. But you can nonetheless use it to define subsets in Q.
$endgroup$
– Lee Mosher
Dec 24 '18 at 12:38
$begingroup$
Think of it this way: I can define the following subsets of $mathbb Z$: $$A = {n in mathbb Z mid n >frac{1}{2}} = {1,2,3,...}, qquad B = {n in mathbb Z mid n < frac{1}{2}} = {...,-2,-1,0}$$ It makes perfect sense to write those sets and work with them, even though $frac{1}{2} notin mathbb Z$. The reason this works is because $mathbb Z$ and $frac{1}{2}$ are both contained in a larger set where inequality is defined, namely $mathbb R$.
$endgroup$
– Lee Mosher
Dec 24 '18 at 12:40