Proving $[0,1]capmathbb{Q}$ is not compact in $mathbb{Q}$ [duplicate]
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This question already has an answer here:
An open covering of $mathbb{Q} cap [0,1]$ that does not contain any finite subcovering
1 answer
Why is $[0, 1] cap mathbb{Q}$ not compact in $mathbb{Q}$? [duplicate]
6 answers
Is $[0,1]capmathbb{Q}$ a compact subset of $mathbb{Q}$.
No it is not. Since $mathbb{Q}$ is dense in $mathbb{R}$ for any open set in [0,1] there is a rational that belongs to that set. Since there are infinite open sets in [0,1] by the standard topology there is not a sub covering that would be finite, since it be composed of infinite number of rationals.
Question:
Is this proof right?
Thanks in advance!
real-analysis general-topology
asked Nov 28 '18 at 18:20
Pedro GomesPedro Gomes
1,7452721
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marked as duplicate by Brahadeesh, Stahl, Watson, Alexander Gruber♦ Nov 30 '18 at 3:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
An open covering of $mathbb{Q} cap [0,1]$ that does not contain any finite subcovering
1 answer
Why is $[0, 1] cap mathbb{Q}$ not compact in $mathbb{Q}$? [duplicate]
6 answers
Is $[0,1]capmathbb{Q}$ a compact subset of $mathbb{Q}$.
No it is not. Since $mathbb{Q}$ is dense in $mathbb{R}$ for any open set in [0,1] there is a rational that belongs to that set. Since there are infinite open sets in [0,1] by the standard topology there is not a sub covering that would be finite, since it be composed of infinite number of rationals.
Question:
Is this proof right?
Thanks in advance!
real-analysis general-topology
asked Nov 28 '18 at 18:20
Pedro GomesPedro Gomes
1,7452721
$endgroup$
marked as duplicate by Brahadeesh, Stahl, Watson, Alexander Gruber♦ Nov 30 '18 at 3:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
No, its not correct. The same argumentation is valid for $[0,1]$ itself which is compact. In particular the conclusion that there is no finite sub covering came from nowhere. The main argument should be that there is an irrational that is a limit of rationals.
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– freakish
Nov 28 '18 at 18:24
$begingroup$
But $left[ 0, 1 right]$ is compact in $mathbb{R}$ with the usual topology. Now, the set $left[ 0, 1 right] cap mathbb{Q}$ is a closed and bounded subset of a compact set. Therefore it is compact in $mathbb{R}$. Also, it will be compact in the relativised topology on $mathbb{Q}$.
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– Aniruddha Deshmukh
Nov 28 '18 at 18:25
3
$begingroup$
@AniruddhaDeshmukh $[0,1] cap mathbb Q$ is not closed in $mathbb R$, therefore not compact in $mathbb R$. If it were compact in $mathbb Q$, it would be compact in $mathbb R$.
$endgroup$
– Robert Israel
Nov 28 '18 at 18:40
2
$begingroup$
Yes. I was wrong there!
$endgroup$
– Aniruddha Deshmukh
Nov 28 '18 at 18:42
$begingroup$
"Since there are infinite open sets in [0,1] by the standard topology there is not a sub covering that would be finite". That would imply that no infinite set could possibly be compact. Which you know is false. And the logic seems to be if you have an infinite set then you can't have finite subsets. That isn't at all valid.
$endgroup$
– fleablood
Nov 28 '18 at 18:58
add a comment |
$begingroup$
This question already has an answer here:
An open covering of $mathbb{Q} cap [0,1]$ that does not contain any finite subcovering
1 answer
Why is $[0, 1] cap mathbb{Q}$ not compact in $mathbb{Q}$? [duplicate]
6 answers
Is $[0,1]capmathbb{Q}$ a compact subset of $mathbb{Q}$.
No it is not. Since $mathbb{Q}$ is dense in $mathbb{R}$ for any open set in [0,1] there is a rational that belongs to that set. Since there are infinite open sets in [0,1] by the standard topology there is not a sub covering that would be finite, since it be composed of infinite number of rationals.
Question:
Is this proof right?
Thanks in advance!
real-analysis general-topology
asked Nov 28 '18 at 18:20
Pedro GomesPedro Gomes
1,7452721
$endgroup$
This question already has an answer here:
An open covering of $mathbb{Q} cap [0,1]$ that does not contain any finite subcovering
1 answer
Why is $[0, 1] cap mathbb{Q}$ not compact in $mathbb{Q}$? [duplicate]
6 answers
Is $[0,1]capmathbb{Q}$ a compact subset of $mathbb{Q}$.
No it is not. Since $mathbb{Q}$ is dense in $mathbb{R}$ for any open set in [0,1] there is a rational that belongs to that set. Since there are infinite open sets in [0,1] by the standard topology there is not a sub covering that would be finite, since it be composed of infinite number of rationals.
Question:
Is this proof right?
Thanks in advance!
This question already has an answer here:
An open covering of $mathbb{Q} cap [0,1]$ that does not contain any finite subcovering
1 answer
Why is $[0, 1] cap mathbb{Q}$ not compact in $mathbb{Q}$? [duplicate]
6 answers
real-analysis general-topology
real-analysis general-topology
asked Nov 28 '18 at 18:20
Pedro GomesPedro Gomes
1,7452721
asked Nov 28 '18 at 18:20
Pedro GomesPedro Gomes
1,7452721
asked Nov 28 '18 at 18:20
Pedro GomesPedro Gomes
1,7452721
asked Nov 28 '18 at 18:20
Pedro GomesPedro Gomes
1,7452721
asked Nov 28 '18 at 18:20
Pedro GomesPedro Gomes
1,7452721
1,7452721
marked as duplicate by Brahadeesh, Stahl, Watson, Alexander Gruber♦ Nov 30 '18 at 3:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Brahadeesh, Stahl, Watson, Alexander Gruber♦ Nov 30 '18 at 3:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
No, its not correct. The same argumentation is valid for $[0,1]$ itself which is compact. In particular the conclusion that there is no finite sub covering came from nowhere. The main argument should be that there is an irrational that is a limit of rationals.
$endgroup$
– freakish
Nov 28 '18 at 18:24
$begingroup$
But $left[ 0, 1 right]$ is compact in $mathbb{R}$ with the usual topology. Now, the set $left[ 0, 1 right] cap mathbb{Q}$ is a closed and bounded subset of a compact set. Therefore it is compact in $mathbb{R}$. Also, it will be compact in the relativised topology on $mathbb{Q}$.
$endgroup$
– Aniruddha Deshmukh
Nov 28 '18 at 18:25
3
$begingroup$
@AniruddhaDeshmukh $[0,1] cap mathbb Q$ is not closed in $mathbb R$, therefore not compact in $mathbb R$. If it were compact in $mathbb Q$, it would be compact in $mathbb R$.
$endgroup$
– Robert Israel
Nov 28 '18 at 18:40
2
$begingroup$
Yes. I was wrong there!
$endgroup$
– Aniruddha Deshmukh
Nov 28 '18 at 18:42
$begingroup$
"Since there are infinite open sets in [0,1] by the standard topology there is not a sub covering that would be finite". That would imply that no infinite set could possibly be compact. Which you know is false. And the logic seems to be if you have an infinite set then you can't have finite subsets. That isn't at all valid.
$endgroup$
– fleablood
Nov 28 '18 at 18:58
add a comment |
2
$begingroup$
No, its not correct. The same argumentation is valid for $[0,1]$ itself which is compact. In particular the conclusion that there is no finite sub covering came from nowhere. The main argument should be that there is an irrational that is a limit of rationals.
$endgroup$
– freakish
Nov 28 '18 at 18:24
$begingroup$
But $left[ 0, 1 right]$ is compact in $mathbb{R}$ with the usual topology. Now, the set $left[ 0, 1 right] cap mathbb{Q}$ is a closed and bounded subset of a compact set. Therefore it is compact in $mathbb{R}$. Also, it will be compact in the relativised topology on $mathbb{Q}$.
$endgroup$
– Aniruddha Deshmukh
Nov 28 '18 at 18:25
3
$begingroup$
@AniruddhaDeshmukh $[0,1] cap mathbb Q$ is not closed in $mathbb R$, therefore not compact in $mathbb R$. If it were compact in $mathbb Q$, it would be compact in $mathbb R$.
$endgroup$
– Robert Israel
Nov 28 '18 at 18:40
2
$begingroup$
Yes. I was wrong there!
$endgroup$
– Aniruddha Deshmukh
Nov 28 '18 at 18:42
$begingroup$
"Since there are infinite open sets in [0,1] by the standard topology there is not a sub covering that would be finite". That would imply that no infinite set could possibly be compact. Which you know is false. And the logic seems to be if you have an infinite set then you can't have finite subsets. That isn't at all valid.
$endgroup$
– fleablood
Nov 28 '18 at 18:58
2
2
$begingroup$
No, its not correct. The same argumentation is valid for $[0,1]$ itself which is compact. In particular the conclusion that there is no finite sub covering came from nowhere. The main argument should be that there is an irrational that is a limit of rationals.
$endgroup$
– freakish
Nov 28 '18 at 18:24
$begingroup$
No, its not correct. The same argumentation is valid for $[0,1]$ itself which is compact. In particular the conclusion that there is no finite sub covering came from nowhere. The main argument should be that there is an irrational that is a limit of rationals.
$endgroup$
– freakish
Nov 28 '18 at 18:24
$begingroup$
But $left[ 0, 1 right]$ is compact in $mathbb{R}$ with the usual topology. Now, the set $left[ 0, 1 right] cap mathbb{Q}$ is a closed and bounded subset of a compact set. Therefore it is compact in $mathbb{R}$. Also, it will be compact in the relativised topology on $mathbb{Q}$.
$endgroup$
– Aniruddha Deshmukh
Nov 28 '18 at 18:25
$begingroup$
But $left[ 0, 1 right]$ is compact in $mathbb{R}$ with the usual topology. Now, the set $left[ 0, 1 right] cap mathbb{Q}$ is a closed and bounded subset of a compact set. Therefore it is compact in $mathbb{R}$. Also, it will be compact in the relativised topology on $mathbb{Q}$.
$endgroup$
– Aniruddha Deshmukh
Nov 28 '18 at 18:25
3
3
$begingroup$
@AniruddhaDeshmukh $[0,1] cap mathbb Q$ is not closed in $mathbb R$, therefore not compact in $mathbb R$. If it were compact in $mathbb Q$, it would be compact in $mathbb R$.
$endgroup$
– Robert Israel
Nov 28 '18 at 18:40
$begingroup$
@AniruddhaDeshmukh $[0,1] cap mathbb Q$ is not closed in $mathbb R$, therefore not compact in $mathbb R$. If it were compact in $mathbb Q$, it would be compact in $mathbb R$.
$endgroup$
– Robert Israel
Nov 28 '18 at 18:40
2
2
$begingroup$
Yes. I was wrong there!
$endgroup$
– Aniruddha Deshmukh
Nov 28 '18 at 18:42
$begingroup$
Yes. I was wrong there!
$endgroup$
– Aniruddha Deshmukh
Nov 28 '18 at 18:42
$begingroup$
"Since there are infinite open sets in [0,1] by the standard topology there is not a sub covering that would be finite". That would imply that no infinite set could possibly be compact. Which you know is false. And the logic seems to be if you have an infinite set then you can't have finite subsets. That isn't at all valid.
$endgroup$
– fleablood
Nov 28 '18 at 18:58
$begingroup$
"Since there are infinite open sets in [0,1] by the standard topology there is not a sub covering that would be finite". That would imply that no infinite set could possibly be compact. Which you know is false. And the logic seems to be if you have an infinite set then you can't have finite subsets. That isn't at all valid.
$endgroup$
– fleablood
Nov 28 '18 at 18:58
add a comment |
4 Answers
4
active
oldest
votes
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We know not-closed sets are not compact. Why? Because if the set is not closed there is a limit point not in it. And we can have infinite number of open sets that cover everything "up to" the limit point and there can't be a finite subcover as there will be a measurable quantifiable "gap" between the "largest" interval and the limit point that will not be covered.
So let's take some limit point in $[0,1]cap mathbb Q$ that is not in $[0,1]cap Q$; in other words any irrational number between $0$ and $1$. Let $x; 0 < x < 1$ and $x$ is irrational.
We can take the open sets $U_i= (-1, x-frac 1i)cup (x+frac 1i,2)cap mathbb Q$. (As $(-1, x-frac 1i)cup (x+frac 1i,2)$ is open in $mathbb R$ we know $(-1, x-frac 1i)cup (x+frac 1i,2)cap mathbb Q$ is open in $mathbb Q$. And $cup U_i$ covers $[0,1]cap mathbb Q = ([0,x)cup(x,1])cap mathbb Q subset( (-1,x)cup(x, 2))cap mathbb Q$.
And ${U_i}$ has no finite subcover.
After that brainstorming we can put it in simpler terms:
For any irrational $x: 0 < x < 1$ then $[0,x) cup (x, 1]$ is not compact in $mathbb R$. An open cover that has no subcover would be ${V_i|V_i = (-1,x-frac 1i)cup (x+frac 1i, 2); i in mathbb N}$.
If we restrict that to $mathbb Q$ then ${U_i| U_i = V_icap mathbb Q}$ is an open (in $mathbb Q$) cover of $([0,x) cup (x, 1])cap mathbb Q$. But $([0,x) cup (x, 1])cap mathbb Q= [0,1]cap mathbb Q$ (because $xnot in mathbb Q$).
So $[0,1]cap mathbb Q$ is not compact.
answered Nov 28 '18 at 19:38
fleabloodfleablood
68.7k22685
$endgroup$
$begingroup$
I appreciate your answer but I have one doubt that is preventing me from understanding this example. How do you know or prove that ${U_i}$ has no finite subcover? Thanks in advance!
$endgroup$
– Pedro Gomes
Nov 29 '18 at 12:24
$begingroup$
Because if you take a finite subcover then there is a max value of $i$. Call it $n$. And the interval $(x-frac 1n, x + frac 1n) $ is not covered. This is a very standard tool in a bag of tricks mathematicians develop.
$endgroup$
– fleablood
Nov 29 '18 at 17:18
add a comment |
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A useful property of compactness is that it is intrinsical. In our case $mathbb{Q}cap [0,1]$ being compact in $mathbb{Q}$ is equivalent to it being compact in $mathbb{R}$. Now the compact subsets of $mathbb{R}$ are the closed bounded sets, and clearly our set is not closed, since its closure is $[0,1]$.
answered Nov 28 '18 at 18:58
Olivier MoschettaOlivier Moschetta
2,8111411
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If you want to show that $[0,1]capmathbb{Q}$ does have an infinite covering which cannot be refined to a finite one consider this:
Let $a_n=sqrt{2}/2-1/n$. Note that each $a_n$ is irrational. Now consider $I_n=(a_n,a_{n+1})$ together with $[0,a_1)$ and $(sqrt{2}/2, 1]$. All those sets together form an open covering of $[0,1]capmathbb{Q}$ but there is no finite refinement. In fact every rational belongs to exactly one element from that set meaning there is no refinement (even infinite) at all.
answered Nov 28 '18 at 19:04
freakishfreakish
11.8k1629
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add a comment |
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Consider, for $varepsilon>0$,
$$
A_varepsilon=bigl((-varepsilon,sqrt{2}/2-varepsilon)cup(sqrt{2}/2+varepsilon,1+varepsilon)bigr)capmathbb{Q}
$$
for $n>0$. Then
$$
bigcup_{varepsilon>0} A_varepsilonsupseteq[0,1]capmathbb{Q}
$$
Is there a finite subcover?
More conceptually, a compact subset of a Hausdorff space is closed.
answered Nov 28 '18 at 18:51
egregegreg
180k1485202
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add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We know not-closed sets are not compact. Why? Because if the set is not closed there is a limit point not in it. And we can have infinite number of open sets that cover everything "up to" the limit point and there can't be a finite subcover as there will be a measurable quantifiable "gap" between the "largest" interval and the limit point that will not be covered.
So let's take some limit point in $[0,1]cap mathbb Q$ that is not in $[0,1]cap Q$; in other words any irrational number between $0$ and $1$. Let $x; 0 < x < 1$ and $x$ is irrational.
We can take the open sets $U_i= (-1, x-frac 1i)cup (x+frac 1i,2)cap mathbb Q$. (As $(-1, x-frac 1i)cup (x+frac 1i,2)$ is open in $mathbb R$ we know $(-1, x-frac 1i)cup (x+frac 1i,2)cap mathbb Q$ is open in $mathbb Q$. And $cup U_i$ covers $[0,1]cap mathbb Q = ([0,x)cup(x,1])cap mathbb Q subset( (-1,x)cup(x, 2))cap mathbb Q$.
And ${U_i}$ has no finite subcover.
After that brainstorming we can put it in simpler terms:
For any irrational $x: 0 < x < 1$ then $[0,x) cup (x, 1]$ is not compact in $mathbb R$. An open cover that has no subcover would be ${V_i|V_i = (-1,x-frac 1i)cup (x+frac 1i, 2); i in mathbb N}$.
If we restrict that to $mathbb Q$ then ${U_i| U_i = V_icap mathbb Q}$ is an open (in $mathbb Q$) cover of $([0,x) cup (x, 1])cap mathbb Q$. But $([0,x) cup (x, 1])cap mathbb Q= [0,1]cap mathbb Q$ (because $xnot in mathbb Q$).
So $[0,1]cap mathbb Q$ is not compact.
answered Nov 28 '18 at 19:38
fleabloodfleablood
68.7k22685
$endgroup$
$begingroup$
I appreciate your answer but I have one doubt that is preventing me from understanding this example. How do you know or prove that ${U_i}$ has no finite subcover? Thanks in advance!
$endgroup$
– Pedro Gomes
Nov 29 '18 at 12:24
$begingroup$
Because if you take a finite subcover then there is a max value of $i$. Call it $n$. And the interval $(x-frac 1n, x + frac 1n) $ is not covered. This is a very standard tool in a bag of tricks mathematicians develop.
$endgroup$
– fleablood
Nov 29 '18 at 17:18
add a comment |
$begingroup$
We know not-closed sets are not compact. Why? Because if the set is not closed there is a limit point not in it. And we can have infinite number of open sets that cover everything "up to" the limit point and there can't be a finite subcover as there will be a measurable quantifiable "gap" between the "largest" interval and the limit point that will not be covered.
So let's take some limit point in $[0,1]cap mathbb Q$ that is not in $[0,1]cap Q$; in other words any irrational number between $0$ and $1$. Let $x; 0 < x < 1$ and $x$ is irrational.
We can take the open sets $U_i= (-1, x-frac 1i)cup (x+frac 1i,2)cap mathbb Q$. (As $(-1, x-frac 1i)cup (x+frac 1i,2)$ is open in $mathbb R$ we know $(-1, x-frac 1i)cup (x+frac 1i,2)cap mathbb Q$ is open in $mathbb Q$. And $cup U_i$ covers $[0,1]cap mathbb Q = ([0,x)cup(x,1])cap mathbb Q subset( (-1,x)cup(x, 2))cap mathbb Q$.
And ${U_i}$ has no finite subcover.
After that brainstorming we can put it in simpler terms:
For any irrational $x: 0 < x < 1$ then $[0,x) cup (x, 1]$ is not compact in $mathbb R$. An open cover that has no subcover would be ${V_i|V_i = (-1,x-frac 1i)cup (x+frac 1i, 2); i in mathbb N}$.
If we restrict that to $mathbb Q$ then ${U_i| U_i = V_icap mathbb Q}$ is an open (in $mathbb Q$) cover of $([0,x) cup (x, 1])cap mathbb Q$. But $([0,x) cup (x, 1])cap mathbb Q= [0,1]cap mathbb Q$ (because $xnot in mathbb Q$).
So $[0,1]cap mathbb Q$ is not compact.
answered Nov 28 '18 at 19:38
fleabloodfleablood
68.7k22685
$endgroup$
$begingroup$
I appreciate your answer but I have one doubt that is preventing me from understanding this example. How do you know or prove that ${U_i}$ has no finite subcover? Thanks in advance!
$endgroup$
– Pedro Gomes
Nov 29 '18 at 12:24
$begingroup$
Because if you take a finite subcover then there is a max value of $i$. Call it $n$. And the interval $(x-frac 1n, x + frac 1n) $ is not covered. This is a very standard tool in a bag of tricks mathematicians develop.
$endgroup$
– fleablood
Nov 29 '18 at 17:18
add a comment |
$begingroup$
We know not-closed sets are not compact. Why? Because if the set is not closed there is a limit point not in it. And we can have infinite number of open sets that cover everything "up to" the limit point and there can't be a finite subcover as there will be a measurable quantifiable "gap" between the "largest" interval and the limit point that will not be covered.
So let's take some limit point in $[0,1]cap mathbb Q$ that is not in $[0,1]cap Q$; in other words any irrational number between $0$ and $1$. Let $x; 0 < x < 1$ and $x$ is irrational.
We can take the open sets $U_i= (-1, x-frac 1i)cup (x+frac 1i,2)cap mathbb Q$. (As $(-1, x-frac 1i)cup (x+frac 1i,2)$ is open in $mathbb R$ we know $(-1, x-frac 1i)cup (x+frac 1i,2)cap mathbb Q$ is open in $mathbb Q$. And $cup U_i$ covers $[0,1]cap mathbb Q = ([0,x)cup(x,1])cap mathbb Q subset( (-1,x)cup(x, 2))cap mathbb Q$.
And ${U_i}$ has no finite subcover.
After that brainstorming we can put it in simpler terms:
For any irrational $x: 0 < x < 1$ then $[0,x) cup (x, 1]$ is not compact in $mathbb R$. An open cover that has no subcover would be ${V_i|V_i = (-1,x-frac 1i)cup (x+frac 1i, 2); i in mathbb N}$.
If we restrict that to $mathbb Q$ then ${U_i| U_i = V_icap mathbb Q}$ is an open (in $mathbb Q$) cover of $([0,x) cup (x, 1])cap mathbb Q$. But $([0,x) cup (x, 1])cap mathbb Q= [0,1]cap mathbb Q$ (because $xnot in mathbb Q$).
So $[0,1]cap mathbb Q$ is not compact.
answered Nov 28 '18 at 19:38
fleabloodfleablood
68.7k22685
$endgroup$
We know not-closed sets are not compact. Why? Because if the set is not closed there is a limit point not in it. And we can have infinite number of open sets that cover everything "up to" the limit point and there can't be a finite subcover as there will be a measurable quantifiable "gap" between the "largest" interval and the limit point that will not be covered.
So let's take some limit point in $[0,1]cap mathbb Q$ that is not in $[0,1]cap Q$; in other words any irrational number between $0$ and $1$. Let $x; 0 < x < 1$ and $x$ is irrational.
We can take the open sets $U_i= (-1, x-frac 1i)cup (x+frac 1i,2)cap mathbb Q$. (As $(-1, x-frac 1i)cup (x+frac 1i,2)$ is open in $mathbb R$ we know $(-1, x-frac 1i)cup (x+frac 1i,2)cap mathbb Q$ is open in $mathbb Q$. And $cup U_i$ covers $[0,1]cap mathbb Q = ([0,x)cup(x,1])cap mathbb Q subset( (-1,x)cup(x, 2))cap mathbb Q$.
And ${U_i}$ has no finite subcover.
After that brainstorming we can put it in simpler terms:
For any irrational $x: 0 < x < 1$ then $[0,x) cup (x, 1]$ is not compact in $mathbb R$. An open cover that has no subcover would be ${V_i|V_i = (-1,x-frac 1i)cup (x+frac 1i, 2); i in mathbb N}$.
If we restrict that to $mathbb Q$ then ${U_i| U_i = V_icap mathbb Q}$ is an open (in $mathbb Q$) cover of $([0,x) cup (x, 1])cap mathbb Q$. But $([0,x) cup (x, 1])cap mathbb Q= [0,1]cap mathbb Q$ (because $xnot in mathbb Q$).
So $[0,1]cap mathbb Q$ is not compact.
answered Nov 28 '18 at 19:38
fleabloodfleablood
68.7k22685
answered Nov 28 '18 at 19:38
fleabloodfleablood
68.7k22685
answered Nov 28 '18 at 19:38
fleabloodfleablood
68.7k22685
answered Nov 28 '18 at 19:38
fleabloodfleablood
68.7k22685
68.7k22685
$begingroup$
I appreciate your answer but I have one doubt that is preventing me from understanding this example. How do you know or prove that ${U_i}$ has no finite subcover? Thanks in advance!
$endgroup$
– Pedro Gomes
Nov 29 '18 at 12:24
$begingroup$
Because if you take a finite subcover then there is a max value of $i$. Call it $n$. And the interval $(x-frac 1n, x + frac 1n) $ is not covered. This is a very standard tool in a bag of tricks mathematicians develop.
$endgroup$
– fleablood
Nov 29 '18 at 17:18
add a comment |
$begingroup$
I appreciate your answer but I have one doubt that is preventing me from understanding this example. How do you know or prove that ${U_i}$ has no finite subcover? Thanks in advance!
$endgroup$
– Pedro Gomes
Nov 29 '18 at 12:24
$begingroup$
Because if you take a finite subcover then there is a max value of $i$. Call it $n$. And the interval $(x-frac 1n, x + frac 1n) $ is not covered. This is a very standard tool in a bag of tricks mathematicians develop.
$endgroup$
– fleablood
Nov 29 '18 at 17:18
$begingroup$
I appreciate your answer but I have one doubt that is preventing me from understanding this example. How do you know or prove that ${U_i}$ has no finite subcover? Thanks in advance!
$endgroup$
– Pedro Gomes
Nov 29 '18 at 12:24
$begingroup$
I appreciate your answer but I have one doubt that is preventing me from understanding this example. How do you know or prove that ${U_i}$ has no finite subcover? Thanks in advance!
$endgroup$
– Pedro Gomes
Nov 29 '18 at 12:24
$begingroup$
Because if you take a finite subcover then there is a max value of $i$. Call it $n$. And the interval $(x-frac 1n, x + frac 1n) $ is not covered. This is a very standard tool in a bag of tricks mathematicians develop.
$endgroup$
– fleablood
Nov 29 '18 at 17:18
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Because if you take a finite subcover then there is a max value of $i$. Call it $n$. And the interval $(x-frac 1n, x + frac 1n) $ is not covered. This is a very standard tool in a bag of tricks mathematicians develop.
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– fleablood
Nov 29 '18 at 17:18
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A useful property of compactness is that it is intrinsical. In our case $mathbb{Q}cap [0,1]$ being compact in $mathbb{Q}$ is equivalent to it being compact in $mathbb{R}$. Now the compact subsets of $mathbb{R}$ are the closed bounded sets, and clearly our set is not closed, since its closure is $[0,1]$.
answered Nov 28 '18 at 18:58
Olivier MoschettaOlivier Moschetta
2,8111411
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A useful property of compactness is that it is intrinsical. In our case $mathbb{Q}cap [0,1]$ being compact in $mathbb{Q}$ is equivalent to it being compact in $mathbb{R}$. Now the compact subsets of $mathbb{R}$ are the closed bounded sets, and clearly our set is not closed, since its closure is $[0,1]$.
answered Nov 28 '18 at 18:58
Olivier MoschettaOlivier Moschetta
2,8111411
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add a comment |
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A useful property of compactness is that it is intrinsical. In our case $mathbb{Q}cap [0,1]$ being compact in $mathbb{Q}$ is equivalent to it being compact in $mathbb{R}$. Now the compact subsets of $mathbb{R}$ are the closed bounded sets, and clearly our set is not closed, since its closure is $[0,1]$.
answered Nov 28 '18 at 18:58
Olivier MoschettaOlivier Moschetta
2,8111411
$endgroup$
A useful property of compactness is that it is intrinsical. In our case $mathbb{Q}cap [0,1]$ being compact in $mathbb{Q}$ is equivalent to it being compact in $mathbb{R}$. Now the compact subsets of $mathbb{R}$ are the closed bounded sets, and clearly our set is not closed, since its closure is $[0,1]$.
answered Nov 28 '18 at 18:58
Olivier MoschettaOlivier Moschetta
2,8111411
answered Nov 28 '18 at 18:58
Olivier MoschettaOlivier Moschetta
2,8111411
answered Nov 28 '18 at 18:58
Olivier MoschettaOlivier Moschetta
2,8111411
answered Nov 28 '18 at 18:58
Olivier MoschettaOlivier Moschetta
2,8111411
2,8111411
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If you want to show that $[0,1]capmathbb{Q}$ does have an infinite covering which cannot be refined to a finite one consider this:
Let $a_n=sqrt{2}/2-1/n$. Note that each $a_n$ is irrational. Now consider $I_n=(a_n,a_{n+1})$ together with $[0,a_1)$ and $(sqrt{2}/2, 1]$. All those sets together form an open covering of $[0,1]capmathbb{Q}$ but there is no finite refinement. In fact every rational belongs to exactly one element from that set meaning there is no refinement (even infinite) at all.
answered Nov 28 '18 at 19:04
freakishfreakish
11.8k1629
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If you want to show that $[0,1]capmathbb{Q}$ does have an infinite covering which cannot be refined to a finite one consider this:
Let $a_n=sqrt{2}/2-1/n$. Note that each $a_n$ is irrational. Now consider $I_n=(a_n,a_{n+1})$ together with $[0,a_1)$ and $(sqrt{2}/2, 1]$. All those sets together form an open covering of $[0,1]capmathbb{Q}$ but there is no finite refinement. In fact every rational belongs to exactly one element from that set meaning there is no refinement (even infinite) at all.
answered Nov 28 '18 at 19:04
freakishfreakish
11.8k1629
$endgroup$
add a comment |
$begingroup$
If you want to show that $[0,1]capmathbb{Q}$ does have an infinite covering which cannot be refined to a finite one consider this:
Let $a_n=sqrt{2}/2-1/n$. Note that each $a_n$ is irrational. Now consider $I_n=(a_n,a_{n+1})$ together with $[0,a_1)$ and $(sqrt{2}/2, 1]$. All those sets together form an open covering of $[0,1]capmathbb{Q}$ but there is no finite refinement. In fact every rational belongs to exactly one element from that set meaning there is no refinement (even infinite) at all.
answered Nov 28 '18 at 19:04
freakishfreakish
11.8k1629
$endgroup$
If you want to show that $[0,1]capmathbb{Q}$ does have an infinite covering which cannot be refined to a finite one consider this:
Let $a_n=sqrt{2}/2-1/n$. Note that each $a_n$ is irrational. Now consider $I_n=(a_n,a_{n+1})$ together with $[0,a_1)$ and $(sqrt{2}/2, 1]$. All those sets together form an open covering of $[0,1]capmathbb{Q}$ but there is no finite refinement. In fact every rational belongs to exactly one element from that set meaning there is no refinement (even infinite) at all.
answered Nov 28 '18 at 19:04
freakishfreakish
11.8k1629
edited Nov 28 '18 at 19:33
answered Nov 28 '18 at 19:04
freakishfreakish
11.8k1629
answered Nov 28 '18 at 19:04
freakishfreakish
11.8k1629
answered Nov 28 '18 at 19:04
freakishfreakish
11.8k1629
11.8k1629
add a comment |
add a comment |
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Consider, for $varepsilon>0$,
$$
A_varepsilon=bigl((-varepsilon,sqrt{2}/2-varepsilon)cup(sqrt{2}/2+varepsilon,1+varepsilon)bigr)capmathbb{Q}
$$
for $n>0$. Then
$$
bigcup_{varepsilon>0} A_varepsilonsupseteq[0,1]capmathbb{Q}
$$
Is there a finite subcover?
More conceptually, a compact subset of a Hausdorff space is closed.
answered Nov 28 '18 at 18:51
egregegreg
180k1485202
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Consider, for $varepsilon>0$,
$$
A_varepsilon=bigl((-varepsilon,sqrt{2}/2-varepsilon)cup(sqrt{2}/2+varepsilon,1+varepsilon)bigr)capmathbb{Q}
$$
for $n>0$. Then
$$
bigcup_{varepsilon>0} A_varepsilonsupseteq[0,1]capmathbb{Q}
$$
Is there a finite subcover?
More conceptually, a compact subset of a Hausdorff space is closed.
answered Nov 28 '18 at 18:51
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Consider, for $varepsilon>0$,
$$
A_varepsilon=bigl((-varepsilon,sqrt{2}/2-varepsilon)cup(sqrt{2}/2+varepsilon,1+varepsilon)bigr)capmathbb{Q}
$$
for $n>0$. Then
$$
bigcup_{varepsilon>0} A_varepsilonsupseteq[0,1]capmathbb{Q}
$$
Is there a finite subcover?
More conceptually, a compact subset of a Hausdorff space is closed.
answered Nov 28 '18 at 18:51
egregegreg
180k1485202
$endgroup$
Consider, for $varepsilon>0$,
$$
A_varepsilon=bigl((-varepsilon,sqrt{2}/2-varepsilon)cup(sqrt{2}/2+varepsilon,1+varepsilon)bigr)capmathbb{Q}
$$
for $n>0$. Then
$$
bigcup_{varepsilon>0} A_varepsilonsupseteq[0,1]capmathbb{Q}
$$
Is there a finite subcover?
More conceptually, a compact subset of a Hausdorff space is closed.
answered Nov 28 '18 at 18:51
egregegreg
180k1485202
edited Nov 28 '18 at 20:50
answered Nov 28 '18 at 18:51
egregegreg
180k1485202
answered Nov 28 '18 at 18:51
egregegreg
180k1485202
answered Nov 28 '18 at 18:51
egregegreg
180k1485202
180k1485202
add a comment |
add a comment |
2
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No, its not correct. The same argumentation is valid for $[0,1]$ itself which is compact. In particular the conclusion that there is no finite sub covering came from nowhere. The main argument should be that there is an irrational that is a limit of rationals.
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– freakish
Nov 28 '18 at 18:24
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But $left[ 0, 1 right]$ is compact in $mathbb{R}$ with the usual topology. Now, the set $left[ 0, 1 right] cap mathbb{Q}$ is a closed and bounded subset of a compact set. Therefore it is compact in $mathbb{R}$. Also, it will be compact in the relativised topology on $mathbb{Q}$.
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– Aniruddha Deshmukh
Nov 28 '18 at 18:25
3
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@AniruddhaDeshmukh $[0,1] cap mathbb Q$ is not closed in $mathbb R$, therefore not compact in $mathbb R$. If it were compact in $mathbb Q$, it would be compact in $mathbb R$.
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– Robert Israel
Nov 28 '18 at 18:40
2
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Yes. I was wrong there!
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– Aniruddha Deshmukh
Nov 28 '18 at 18:42
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"Since there are infinite open sets in [0,1] by the standard topology there is not a sub covering that would be finite". That would imply that no infinite set could possibly be compact. Which you know is false. And the logic seems to be if you have an infinite set then you can't have finite subsets. That isn't at all valid.
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– fleablood
Nov 28 '18 at 18:58