Is $mathbb{Q} cap (a,b)$ for irrationals $a,b$ is not compact in the $mathbb{Q}$ with $d(x,y)=|x-y|$?
$begingroup$
In my exam of analysis we ask to proof $mathbb{Q} cap (sqrt{2},sqrt{3})$ is not compact . i have few questions about this problem . i have a solution for the problem but beside my problem one of my friend said we can use Heine–Borel theorem to solve this problem but i think there should be a flaw in his solution basicly i want to know is my solution correct and is there any flaw in my friedns solution . i will put them below . I would really appreciate if some one help me to find out .
Thanks .
My solution :
Define $f:mathbb{Q} cap (a,b) to mathbb{R}$ , $f(x)= frac{1}{b-x}$ . Since ${f(x)}^{'}= frac{1}{{(b-x)}^{2}} > 0$ it is strictly increasing. Now to show this set is a compact set we create a Cover in such way it does not contain any finite sub covering for mentioned set. Set $h(q) = frac{1}{2^{f(q)}}$ thus $N_{h(q)}(q)$ is a coverin for our set. suppose $q in mathbb{Q} cap (a,b)$ thus $N_{h(q)}(q)=(q- frac{1}{2^{f(q)}},q+ frac{1}{2^{f(q)}})$. now again we have $N_{q+ frac{1}{2^{f(q)}}}(q+ frac{1}{2^{f(q)}})=(q+ frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}},q+ frac{1}{2^{f(q)}}+{frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}) $. now it is suffices to show that :
$$q<q+ frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$$
which is equivalent to :
$$0< frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$$
which is always true because $f(q)$ is strictly increasing thus $2^{f(q)}<2^{f(q+ frac{1}{2^{f(q)}})}$ so $frac{1}{2^{f(q)}} > {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$ or $0< frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$. Thus for any rational number $q$ there is an other rational number which does not contain in it's conver so we can not eliminate any of covers thus the mentioned set is not compact .
It is my solution , but my friends solution seems to have flaw. By searching mathexchange i findout his solution to be true but not in the form he says. I mean they way he writed his solution and his reasoning are wrong. By reading This solution i got the reason completly.
My friends solution:
Since $mathbb{Q} cap (sqrt{2},sqrt{3}) subset mathbb{R} $ and since $mathbb{Q} cap (sqrt{2},sqrt{3})$ is not closed in $mathbb{R}$ thus by Heine–Borel theorem it is not compact in $R$ and since $mathbb{Q} subset mathbb{R}$ thus it is not compact in $mathbb{Q}$.
general-topology analysis proof-verification metric-spaces compactness
edited Dec 14 '18 at 13:43
José Carlos Santos
173k23133242
asked Dec 14 '18 at 13:31
alex ivanovalex ivanov
12
$endgroup$
add a comment |
$begingroup$
In my exam of analysis we ask to proof $mathbb{Q} cap (sqrt{2},sqrt{3})$ is not compact . i have few questions about this problem . i have a solution for the problem but beside my problem one of my friend said we can use Heine–Borel theorem to solve this problem but i think there should be a flaw in his solution basicly i want to know is my solution correct and is there any flaw in my friedns solution . i will put them below . I would really appreciate if some one help me to find out .
Thanks .
My solution :
Define $f:mathbb{Q} cap (a,b) to mathbb{R}$ , $f(x)= frac{1}{b-x}$ . Since ${f(x)}^{'}= frac{1}{{(b-x)}^{2}} > 0$ it is strictly increasing. Now to show this set is a compact set we create a Cover in such way it does not contain any finite sub covering for mentioned set. Set $h(q) = frac{1}{2^{f(q)}}$ thus $N_{h(q)}(q)$ is a coverin for our set. suppose $q in mathbb{Q} cap (a,b)$ thus $N_{h(q)}(q)=(q- frac{1}{2^{f(q)}},q+ frac{1}{2^{f(q)}})$. now again we have $N_{q+ frac{1}{2^{f(q)}}}(q+ frac{1}{2^{f(q)}})=(q+ frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}},q+ frac{1}{2^{f(q)}}+{frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}) $. now it is suffices to show that :
$$q<q+ frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$$
which is equivalent to :
$$0< frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$$
which is always true because $f(q)$ is strictly increasing thus $2^{f(q)}<2^{f(q+ frac{1}{2^{f(q)}})}$ so $frac{1}{2^{f(q)}} > {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$ or $0< frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$. Thus for any rational number $q$ there is an other rational number which does not contain in it's conver so we can not eliminate any of covers thus the mentioned set is not compact .
It is my solution , but my friends solution seems to have flaw. By searching mathexchange i findout his solution to be true but not in the form he says. I mean they way he writed his solution and his reasoning are wrong. By reading This solution i got the reason completly.
My friends solution:
Since $mathbb{Q} cap (sqrt{2},sqrt{3}) subset mathbb{R} $ and since $mathbb{Q} cap (sqrt{2},sqrt{3})$ is not closed in $mathbb{R}$ thus by Heine–Borel theorem it is not compact in $R$ and since $mathbb{Q} subset mathbb{R}$ thus it is not compact in $mathbb{Q}$.
general-topology analysis proof-verification metric-spaces compactness
edited Dec 14 '18 at 13:43
José Carlos Santos
173k23133242
asked Dec 14 '18 at 13:31
alex ivanovalex ivanov
12
$endgroup$
$begingroup$
Honestly, your friend's solution looks fine as long as the Heine-Borel theorem has already been taught.
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 13:38
$begingroup$
@YuiTo Cheng oh, thanks . i am not take the general-topology course but my friend already finished the course. so i am newbie and do not know much about Hausdorff superspace .
$endgroup$
– alex ivanov
Dec 14 '18 at 13:41
$begingroup$
Isn't the Heine-Borel theorem part of an analysis course? I'm quite curious.
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 13:47
$begingroup$
@YuiTo Cheng ofcourse, but in our book it is at the middle of chapter and the exam did not contain those pages. So we are limited to definition only !
$endgroup$
– alex ivanov
Dec 14 '18 at 13:48
add a comment |
$begingroup$
In my exam of analysis we ask to proof $mathbb{Q} cap (sqrt{2},sqrt{3})$ is not compact . i have few questions about this problem . i have a solution for the problem but beside my problem one of my friend said we can use Heine–Borel theorem to solve this problem but i think there should be a flaw in his solution basicly i want to know is my solution correct and is there any flaw in my friedns solution . i will put them below . I would really appreciate if some one help me to find out .
Thanks .
My solution :
Define $f:mathbb{Q} cap (a,b) to mathbb{R}$ , $f(x)= frac{1}{b-x}$ . Since ${f(x)}^{'}= frac{1}{{(b-x)}^{2}} > 0$ it is strictly increasing. Now to show this set is a compact set we create a Cover in such way it does not contain any finite sub covering for mentioned set. Set $h(q) = frac{1}{2^{f(q)}}$ thus $N_{h(q)}(q)$ is a coverin for our set. suppose $q in mathbb{Q} cap (a,b)$ thus $N_{h(q)}(q)=(q- frac{1}{2^{f(q)}},q+ frac{1}{2^{f(q)}})$. now again we have $N_{q+ frac{1}{2^{f(q)}}}(q+ frac{1}{2^{f(q)}})=(q+ frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}},q+ frac{1}{2^{f(q)}}+{frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}) $. now it is suffices to show that :
$$q<q+ frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$$
which is equivalent to :
$$0< frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$$
which is always true because $f(q)$ is strictly increasing thus $2^{f(q)}<2^{f(q+ frac{1}{2^{f(q)}})}$ so $frac{1}{2^{f(q)}} > {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$ or $0< frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$. Thus for any rational number $q$ there is an other rational number which does not contain in it's conver so we can not eliminate any of covers thus the mentioned set is not compact .
It is my solution , but my friends solution seems to have flaw. By searching mathexchange i findout his solution to be true but not in the form he says. I mean they way he writed his solution and his reasoning are wrong. By reading This solution i got the reason completly.
My friends solution:
Since $mathbb{Q} cap (sqrt{2},sqrt{3}) subset mathbb{R} $ and since $mathbb{Q} cap (sqrt{2},sqrt{3})$ is not closed in $mathbb{R}$ thus by Heine–Borel theorem it is not compact in $R$ and since $mathbb{Q} subset mathbb{R}$ thus it is not compact in $mathbb{Q}$.
general-topology analysis proof-verification metric-spaces compactness
edited Dec 14 '18 at 13:43
José Carlos Santos
173k23133242
asked Dec 14 '18 at 13:31
alex ivanovalex ivanov
12
$endgroup$
In my exam of analysis we ask to proof $mathbb{Q} cap (sqrt{2},sqrt{3})$ is not compact . i have few questions about this problem . i have a solution for the problem but beside my problem one of my friend said we can use Heine–Borel theorem to solve this problem but i think there should be a flaw in his solution basicly i want to know is my solution correct and is there any flaw in my friedns solution . i will put them below . I would really appreciate if some one help me to find out .
Thanks .
My solution :
Define $f:mathbb{Q} cap (a,b) to mathbb{R}$ , $f(x)= frac{1}{b-x}$ . Since ${f(x)}^{'}= frac{1}{{(b-x)}^{2}} > 0$ it is strictly increasing. Now to show this set is a compact set we create a Cover in such way it does not contain any finite sub covering for mentioned set. Set $h(q) = frac{1}{2^{f(q)}}$ thus $N_{h(q)}(q)$ is a coverin for our set. suppose $q in mathbb{Q} cap (a,b)$ thus $N_{h(q)}(q)=(q- frac{1}{2^{f(q)}},q+ frac{1}{2^{f(q)}})$. now again we have $N_{q+ frac{1}{2^{f(q)}}}(q+ frac{1}{2^{f(q)}})=(q+ frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}},q+ frac{1}{2^{f(q)}}+{frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}) $. now it is suffices to show that :
$$q<q+ frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$$
which is equivalent to :
$$0< frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$$
which is always true because $f(q)$ is strictly increasing thus $2^{f(q)}<2^{f(q+ frac{1}{2^{f(q)}})}$ so $frac{1}{2^{f(q)}} > {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$ or $0< frac{1}{2^{f(q)}}- {frac{1}{2^{f(q+ frac{1}{2^{f(q)}})}}}$. Thus for any rational number $q$ there is an other rational number which does not contain in it's conver so we can not eliminate any of covers thus the mentioned set is not compact .
It is my solution , but my friends solution seems to have flaw. By searching mathexchange i findout his solution to be true but not in the form he says. I mean they way he writed his solution and his reasoning are wrong. By reading This solution i got the reason completly.
My friends solution:
Since $mathbb{Q} cap (sqrt{2},sqrt{3}) subset mathbb{R} $ and since $mathbb{Q} cap (sqrt{2},sqrt{3})$ is not closed in $mathbb{R}$ thus by Heine–Borel theorem it is not compact in $R$ and since $mathbb{Q} subset mathbb{R}$ thus it is not compact in $mathbb{Q}$.
general-topology analysis proof-verification metric-spaces compactness
general-topology analysis proof-verification metric-spaces compactness
edited Dec 14 '18 at 13:43
José Carlos Santos
173k23133242
asked Dec 14 '18 at 13:31
alex ivanovalex ivanov
12
edited Dec 14 '18 at 13:43
José Carlos Santos
173k23133242
asked Dec 14 '18 at 13:31
alex ivanovalex ivanov
12
edited Dec 14 '18 at 13:43
José Carlos Santos
173k23133242
edited Dec 14 '18 at 13:43
José Carlos Santos
173k23133242
edited Dec 14 '18 at 13:43
José Carlos Santos
173k23133242
173k23133242
asked Dec 14 '18 at 13:31
alex ivanovalex ivanov
12
asked Dec 14 '18 at 13:31
alex ivanovalex ivanov
12
asked Dec 14 '18 at 13:31
alex ivanovalex ivanov
12
12
$begingroup$
Honestly, your friend's solution looks fine as long as the Heine-Borel theorem has already been taught.
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 13:38
$begingroup$
@YuiTo Cheng oh, thanks . i am not take the general-topology course but my friend already finished the course. so i am newbie and do not know much about Hausdorff superspace .
$endgroup$
– alex ivanov
Dec 14 '18 at 13:41
$begingroup$
Isn't the Heine-Borel theorem part of an analysis course? I'm quite curious.
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 13:47
$begingroup$
@YuiTo Cheng ofcourse, but in our book it is at the middle of chapter and the exam did not contain those pages. So we are limited to definition only !
$endgroup$
– alex ivanov
Dec 14 '18 at 13:48
add a comment |
$begingroup$
Honestly, your friend's solution looks fine as long as the Heine-Borel theorem has already been taught.
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 13:38
$begingroup$
@YuiTo Cheng oh, thanks . i am not take the general-topology course but my friend already finished the course. so i am newbie and do not know much about Hausdorff superspace .
$endgroup$
– alex ivanov
Dec 14 '18 at 13:41
$begingroup$
Isn't the Heine-Borel theorem part of an analysis course? I'm quite curious.
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 13:47
$begingroup$
@YuiTo Cheng ofcourse, but in our book it is at the middle of chapter and the exam did not contain those pages. So we are limited to definition only !
$endgroup$
– alex ivanov
Dec 14 '18 at 13:48
$begingroup$
Honestly, your friend's solution looks fine as long as the Heine-Borel theorem has already been taught.
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 13:38
$begingroup$
Honestly, your friend's solution looks fine as long as the Heine-Borel theorem has already been taught.
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 13:38
$begingroup$
@YuiTo Cheng oh, thanks . i am not take the general-topology course but my friend already finished the course. so i am newbie and do not know much about Hausdorff superspace .
$endgroup$
– alex ivanov
Dec 14 '18 at 13:41
$begingroup$
@YuiTo Cheng oh, thanks . i am not take the general-topology course but my friend already finished the course. so i am newbie and do not know much about Hausdorff superspace .
$endgroup$
– alex ivanov
Dec 14 '18 at 13:41
$begingroup$
Isn't the Heine-Borel theorem part of an analysis course? I'm quite curious.
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 13:47
$begingroup$
Isn't the Heine-Borel theorem part of an analysis course? I'm quite curious.
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 13:47
$begingroup$
@YuiTo Cheng ofcourse, but in our book it is at the middle of chapter and the exam did not contain those pages. So we are limited to definition only !
$endgroup$
– alex ivanov
Dec 14 '18 at 13:48
$begingroup$
@YuiTo Cheng ofcourse, but in our book it is at the middle of chapter and the exam did not contain those pages. So we are limited to definition only !
$endgroup$
– alex ivanov
Dec 14 '18 at 13:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your friend's solution is fine. If you and to prove that $mathbb{Q}cap(a,b)$ is not compact using the definition of compact set, simply consider the cover$$left{mathbb{Q}cap(a,b)capleft(-infty,b-frac1nright),middle|,ninmathbb Nright},$$which obviously has no finite subcover.
answered Dec 14 '18 at 13:42
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
$begingroup$
brilliant! . but what about my solution ?
$endgroup$
– alex ivanov
Dec 14 '18 at 13:46
$begingroup$
I don't understand it. In particular, I have no idea about what's the meaning of “for any rational number $q$ there is an other rational number which does not contain in it's conver”.
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:49
$begingroup$
I mean if we consider the interval for the $q$ , there will be an other rational like $t$ which does not contain in the defined interval for $q$.
$endgroup$
– alex ivanov
Dec 14 '18 at 13:52
$begingroup$
How could possibly the number $t$ contain an interval?
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:54
$begingroup$
I think i have to make my solution more cleare ! need some time. Thank you in advance.
$endgroup$
– alex ivanov
Dec 14 '18 at 14:02
add a comment |
$begingroup$
For any real $a,b$ with $a<b$ the space $Bbb Qcap (a,b)$ is not compact. Let $x_0=a$ and let $(x_n)_{ngeq 1}$ be a strictly increasing sequence of irrationals converging to $b, $ with $x_1>a. $ Let $C={Bbb Qcap (x_n,x_{n+1}): ngeq 0}.$ Then $C$ is an infinite open cover of $Bbb Qcap (a,b),$ but $C$ is an irreducible cover: No proper subset of C is a cover.
By some simple modifications to this, you can also show that $Bbb Qcap [a,b)$ and $Bbb Qcap (a,b]$ and $Bbb Qcap [a,b]$ are also not compact.
answered Dec 14 '18 at 15:53
DanielWainfleetDanielWainfleet
35.8k31648
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Your friend's solution is fine. If you and to prove that $mathbb{Q}cap(a,b)$ is not compact using the definition of compact set, simply consider the cover$$left{mathbb{Q}cap(a,b)capleft(-infty,b-frac1nright),middle|,ninmathbb Nright},$$which obviously has no finite subcover.
answered Dec 14 '18 at 13:42
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
$begingroup$
brilliant! . but what about my solution ?
$endgroup$
– alex ivanov
Dec 14 '18 at 13:46
$begingroup$
I don't understand it. In particular, I have no idea about what's the meaning of “for any rational number $q$ there is an other rational number which does not contain in it's conver”.
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:49
$begingroup$
I mean if we consider the interval for the $q$ , there will be an other rational like $t$ which does not contain in the defined interval for $q$.
$endgroup$
– alex ivanov
Dec 14 '18 at 13:52
$begingroup$
How could possibly the number $t$ contain an interval?
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:54
$begingroup$
I think i have to make my solution more cleare ! need some time. Thank you in advance.
$endgroup$
– alex ivanov
Dec 14 '18 at 14:02
add a comment |
$begingroup$
Your friend's solution is fine. If you and to prove that $mathbb{Q}cap(a,b)$ is not compact using the definition of compact set, simply consider the cover$$left{mathbb{Q}cap(a,b)capleft(-infty,b-frac1nright),middle|,ninmathbb Nright},$$which obviously has no finite subcover.
answered Dec 14 '18 at 13:42
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
$begingroup$
brilliant! . but what about my solution ?
$endgroup$
– alex ivanov
Dec 14 '18 at 13:46
$begingroup$
I don't understand it. In particular, I have no idea about what's the meaning of “for any rational number $q$ there is an other rational number which does not contain in it's conver”.
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:49
$begingroup$
I mean if we consider the interval for the $q$ , there will be an other rational like $t$ which does not contain in the defined interval for $q$.
$endgroup$
– alex ivanov
Dec 14 '18 at 13:52
$begingroup$
How could possibly the number $t$ contain an interval?
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:54
$begingroup$
I think i have to make my solution more cleare ! need some time. Thank you in advance.
$endgroup$
– alex ivanov
Dec 14 '18 at 14:02
add a comment |
$begingroup$
Your friend's solution is fine. If you and to prove that $mathbb{Q}cap(a,b)$ is not compact using the definition of compact set, simply consider the cover$$left{mathbb{Q}cap(a,b)capleft(-infty,b-frac1nright),middle|,ninmathbb Nright},$$which obviously has no finite subcover.
answered Dec 14 '18 at 13:42
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
Your friend's solution is fine. If you and to prove that $mathbb{Q}cap(a,b)$ is not compact using the definition of compact set, simply consider the cover$$left{mathbb{Q}cap(a,b)capleft(-infty,b-frac1nright),middle|,ninmathbb Nright},$$which obviously has no finite subcover.
answered Dec 14 '18 at 13:42
José Carlos SantosJosé Carlos Santos
173k23133242
answered Dec 14 '18 at 13:42
José Carlos SantosJosé Carlos Santos
173k23133242
answered Dec 14 '18 at 13:42
José Carlos SantosJosé Carlos Santos
173k23133242
answered Dec 14 '18 at 13:42
José Carlos SantosJosé Carlos Santos
173k23133242
173k23133242
$begingroup$
brilliant! . but what about my solution ?
$endgroup$
– alex ivanov
Dec 14 '18 at 13:46
$begingroup$
I don't understand it. In particular, I have no idea about what's the meaning of “for any rational number $q$ there is an other rational number which does not contain in it's conver”.
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:49
$begingroup$
I mean if we consider the interval for the $q$ , there will be an other rational like $t$ which does not contain in the defined interval for $q$.
$endgroup$
– alex ivanov
Dec 14 '18 at 13:52
$begingroup$
How could possibly the number $t$ contain an interval?
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:54
$begingroup$
I think i have to make my solution more cleare ! need some time. Thank you in advance.
$endgroup$
– alex ivanov
Dec 14 '18 at 14:02
add a comment |
$begingroup$
brilliant! . but what about my solution ?
$endgroup$
– alex ivanov
Dec 14 '18 at 13:46
$begingroup$
I don't understand it. In particular, I have no idea about what's the meaning of “for any rational number $q$ there is an other rational number which does not contain in it's conver”.
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– José Carlos Santos
Dec 14 '18 at 13:49
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I mean if we consider the interval for the $q$ , there will be an other rational like $t$ which does not contain in the defined interval for $q$.
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– alex ivanov
Dec 14 '18 at 13:52
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How could possibly the number $t$ contain an interval?
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– José Carlos Santos
Dec 14 '18 at 13:54
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I think i have to make my solution more cleare ! need some time. Thank you in advance.
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– alex ivanov
Dec 14 '18 at 14:02
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brilliant! . but what about my solution ?
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– alex ivanov
Dec 14 '18 at 13:46
$begingroup$
brilliant! . but what about my solution ?
$endgroup$
– alex ivanov
Dec 14 '18 at 13:46
$begingroup$
I don't understand it. In particular, I have no idea about what's the meaning of “for any rational number $q$ there is an other rational number which does not contain in it's conver”.
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:49
$begingroup$
I don't understand it. In particular, I have no idea about what's the meaning of “for any rational number $q$ there is an other rational number which does not contain in it's conver”.
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:49
$begingroup$
I mean if we consider the interval for the $q$ , there will be an other rational like $t$ which does not contain in the defined interval for $q$.
$endgroup$
– alex ivanov
Dec 14 '18 at 13:52
$begingroup$
I mean if we consider the interval for the $q$ , there will be an other rational like $t$ which does not contain in the defined interval for $q$.
$endgroup$
– alex ivanov
Dec 14 '18 at 13:52
$begingroup$
How could possibly the number $t$ contain an interval?
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:54
$begingroup$
How could possibly the number $t$ contain an interval?
$endgroup$
– José Carlos Santos
Dec 14 '18 at 13:54
$begingroup$
I think i have to make my solution more cleare ! need some time. Thank you in advance.
$endgroup$
– alex ivanov
Dec 14 '18 at 14:02
$begingroup$
I think i have to make my solution more cleare ! need some time. Thank you in advance.
$endgroup$
– alex ivanov
Dec 14 '18 at 14:02
add a comment |
$begingroup$
For any real $a,b$ with $a<b$ the space $Bbb Qcap (a,b)$ is not compact. Let $x_0=a$ and let $(x_n)_{ngeq 1}$ be a strictly increasing sequence of irrationals converging to $b, $ with $x_1>a. $ Let $C={Bbb Qcap (x_n,x_{n+1}): ngeq 0}.$ Then $C$ is an infinite open cover of $Bbb Qcap (a,b),$ but $C$ is an irreducible cover: No proper subset of C is a cover.
By some simple modifications to this, you can also show that $Bbb Qcap [a,b)$ and $Bbb Qcap (a,b]$ and $Bbb Qcap [a,b]$ are also not compact.
answered Dec 14 '18 at 15:53
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
add a comment |
$begingroup$
For any real $a,b$ with $a<b$ the space $Bbb Qcap (a,b)$ is not compact. Let $x_0=a$ and let $(x_n)_{ngeq 1}$ be a strictly increasing sequence of irrationals converging to $b, $ with $x_1>a. $ Let $C={Bbb Qcap (x_n,x_{n+1}): ngeq 0}.$ Then $C$ is an infinite open cover of $Bbb Qcap (a,b),$ but $C$ is an irreducible cover: No proper subset of C is a cover.
By some simple modifications to this, you can also show that $Bbb Qcap [a,b)$ and $Bbb Qcap (a,b]$ and $Bbb Qcap [a,b]$ are also not compact.
answered Dec 14 '18 at 15:53
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
add a comment |
$begingroup$
For any real $a,b$ with $a<b$ the space $Bbb Qcap (a,b)$ is not compact. Let $x_0=a$ and let $(x_n)_{ngeq 1}$ be a strictly increasing sequence of irrationals converging to $b, $ with $x_1>a. $ Let $C={Bbb Qcap (x_n,x_{n+1}): ngeq 0}.$ Then $C$ is an infinite open cover of $Bbb Qcap (a,b),$ but $C$ is an irreducible cover: No proper subset of C is a cover.
By some simple modifications to this, you can also show that $Bbb Qcap [a,b)$ and $Bbb Qcap (a,b]$ and $Bbb Qcap [a,b]$ are also not compact.
answered Dec 14 '18 at 15:53
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
For any real $a,b$ with $a<b$ the space $Bbb Qcap (a,b)$ is not compact. Let $x_0=a$ and let $(x_n)_{ngeq 1}$ be a strictly increasing sequence of irrationals converging to $b, $ with $x_1>a. $ Let $C={Bbb Qcap (x_n,x_{n+1}): ngeq 0}.$ Then $C$ is an infinite open cover of $Bbb Qcap (a,b),$ but $C$ is an irreducible cover: No proper subset of C is a cover.
By some simple modifications to this, you can also show that $Bbb Qcap [a,b)$ and $Bbb Qcap (a,b]$ and $Bbb Qcap [a,b]$ are also not compact.
answered Dec 14 '18 at 15:53
DanielWainfleetDanielWainfleet
35.8k31648
answered Dec 14 '18 at 15:53
DanielWainfleetDanielWainfleet
35.8k31648
answered Dec 14 '18 at 15:53
DanielWainfleetDanielWainfleet
35.8k31648
answered Dec 14 '18 at 15:53
DanielWainfleetDanielWainfleet
35.8k31648
35.8k31648
add a comment |
add a comment |
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$begingroup$
Honestly, your friend's solution looks fine as long as the Heine-Borel theorem has already been taught.
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 13:38
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@YuiTo Cheng oh, thanks . i am not take the general-topology course but my friend already finished the course. so i am newbie and do not know much about Hausdorff superspace .
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– alex ivanov
Dec 14 '18 at 13:41
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Isn't the Heine-Borel theorem part of an analysis course? I'm quite curious.
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– YuiTo Cheng
Dec 14 '18 at 13:47
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@YuiTo Cheng ofcourse, but in our book it is at the middle of chapter and the exam did not contain those pages. So we are limited to definition only !
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– alex ivanov
Dec 14 '18 at 13:48