Mixing
Open Cover of (0,1) by infinite collection of sets
$begingroup$
text given in"Calculus on manifolds"
In the above text, it's mentioned that no finite collection of open cover of the form $(frac{1}{n},1-frac{1}{n})$, n $in N $ can cover the interval (0,1). But if we take n = 1, it gives the inteval (1,0) which does cover the interval (0,1).
So, is the statement given in the book wrong?
calculus general-topology manifolds
edited Feb 2 at 11:33
Asaf Karagila♦
308k33441775
asked Feb 2 at 11:00
AstikAstik
41
$endgroup$
add a comment |
$begingroup$
text given in"Calculus on manifolds"
In the above text, it's mentioned that no finite collection of open cover of the form $(frac{1}{n},1-frac{1}{n})$, n $in N $ can cover the interval (0,1). But if we take n = 1, it gives the inteval (1,0) which does cover the interval (0,1).
So, is the statement given in the book wrong?
calculus general-topology manifolds
edited Feb 2 at 11:33
Asaf Karagila♦
308k33441775
asked Feb 2 at 11:00
AstikAstik
41
$endgroup$
4
$begingroup$
The link you provide clearly specifies $n>1$.
$endgroup$
– lulu
Feb 2 at 11:02
5
$begingroup$
$(1,0) = {xinmathbb{R}: 1<x<0} = emptyset$
$endgroup$
– kellty
Feb 2 at 11:03
add a comment |
$begingroup$
text given in"Calculus on manifolds"
In the above text, it's mentioned that no finite collection of open cover of the form $(frac{1}{n},1-frac{1}{n})$, n $in N $ can cover the interval (0,1). But if we take n = 1, it gives the inteval (1,0) which does cover the interval (0,1).
So, is the statement given in the book wrong?
calculus general-topology manifolds
edited Feb 2 at 11:33
Asaf Karagila♦
308k33441775
asked Feb 2 at 11:00
AstikAstik
41
$endgroup$
text given in"Calculus on manifolds"
In the above text, it's mentioned that no finite collection of open cover of the form $(frac{1}{n},1-frac{1}{n})$, n $in N $ can cover the interval (0,1). But if we take n = 1, it gives the inteval (1,0) which does cover the interval (0,1).
So, is the statement given in the book wrong?
calculus general-topology manifolds
calculus general-topology manifolds
edited Feb 2 at 11:33
Asaf Karagila♦
308k33441775
asked Feb 2 at 11:00
AstikAstik
41
edited Feb 2 at 11:33
Asaf Karagila♦
308k33441775
asked Feb 2 at 11:00
AstikAstik
41
edited Feb 2 at 11:33
Asaf Karagila♦
308k33441775
edited Feb 2 at 11:33
Asaf Karagila♦
308k33441775
edited Feb 2 at 11:33
Asaf Karagila♦
308k33441775
308k33441775
asked Feb 2 at 11:00
AstikAstik
41
asked Feb 2 at 11:00
AstikAstik
41
asked Feb 2 at 11:00
AstikAstik
41
41
4
$begingroup$
The link you provide clearly specifies $n>1$.
$endgroup$
– lulu
Feb 2 at 11:02
5
$begingroup$
$(1,0) = {xinmathbb{R}: 1<x<0} = emptyset$
$endgroup$
– kellty
Feb 2 at 11:03
add a comment |
4
$begingroup$
The link you provide clearly specifies $n>1$.
$endgroup$
– lulu
Feb 2 at 11:02
5
$begingroup$
$(1,0) = {xinmathbb{R}: 1<x<0} = emptyset$
$endgroup$
– kellty
Feb 2 at 11:03
4
4
$begingroup$
The link you provide clearly specifies $n>1$.
$endgroup$
– lulu
Feb 2 at 11:02
$begingroup$
The link you provide clearly specifies $n>1$.
$endgroup$
– lulu
Feb 2 at 11:02
5
5
$begingroup$
$(1,0) = {xinmathbb{R}: 1<x<0} = emptyset$
$endgroup$
– kellty
Feb 2 at 11:03
$begingroup$
$(1,0) = {xinmathbb{R}: 1<x<0} = emptyset$
$endgroup$
– kellty
Feb 2 at 11:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are wrong, because for $n=1$ you get
$$
left(frac11,1-frac11right)=(1,0)=emptyset.
$$
You have to consider that
$$
(a,b):={xinmathbb R~:~a<x<b}.
$$
Therefore $a>b$ will give $(a,b)=emptyset$.
answered Feb 2 at 11:03
Mundron SchmidtMundron Schmidt
7,4942729
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You are wrong, because for $n=1$ you get
$$
left(frac11,1-frac11right)=(1,0)=emptyset.
$$
You have to consider that
$$
(a,b):={xinmathbb R~:~a<x<b}.
$$
Therefore $a>b$ will give $(a,b)=emptyset$.
answered Feb 2 at 11:03
Mundron SchmidtMundron Schmidt
7,4942729
$endgroup$
add a comment |
$begingroup$
You are wrong, because for $n=1$ you get
$$
left(frac11,1-frac11right)=(1,0)=emptyset.
$$
You have to consider that
$$
(a,b):={xinmathbb R~:~a<x<b}.
$$
Therefore $a>b$ will give $(a,b)=emptyset$.
answered Feb 2 at 11:03
Mundron SchmidtMundron Schmidt
7,4942729
$endgroup$
add a comment |
$begingroup$
You are wrong, because for $n=1$ you get
$$
left(frac11,1-frac11right)=(1,0)=emptyset.
$$
You have to consider that
$$
(a,b):={xinmathbb R~:~a<x<b}.
$$
Therefore $a>b$ will give $(a,b)=emptyset$.
answered Feb 2 at 11:03
Mundron SchmidtMundron Schmidt
7,4942729
$endgroup$
You are wrong, because for $n=1$ you get
$$
left(frac11,1-frac11right)=(1,0)=emptyset.
$$
You have to consider that
$$
(a,b):={xinmathbb R~:~a<x<b}.
$$
Therefore $a>b$ will give $(a,b)=emptyset$.
answered Feb 2 at 11:03
Mundron SchmidtMundron Schmidt
7,4942729
answered Feb 2 at 11:03
Mundron SchmidtMundron Schmidt
7,4942729
answered Feb 2 at 11:03
Mundron SchmidtMundron Schmidt
7,4942729
answered Feb 2 at 11:03
Mundron SchmidtMundron Schmidt
7,4942729
7,4942729
add a comment |
add a comment |
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4
$begingroup$
The link you provide clearly specifies $n>1$.
$endgroup$
– lulu
Feb 2 at 11:02
5
$begingroup$
$(1,0) = {xinmathbb{R}: 1<x<0} = emptyset$
$endgroup$
– kellty
Feb 2 at 11:03