Mixing
$[0,1]$ is compact regarding the basis $mathscr{B}={{q}:qinmathbb{Q}cap[0,1]}cup{]a,b[:0leqslant...
$begingroup$
Consider $mathscr{B}={{q}:qinmathbb{Q}cap[0,1]}cup{]a,b[:0leqslant a<bleqslant 1}}$
Prove that $mathscr{B}$ is a base for the topology $tau$ in $[0,1]$ and show $([0,1],tau)$ is compact.
Proving it is basis of topology I wrote it down on my notebook but it took me a while to deal with all the cases of intersection. I proved the union of all the open sets of the basis equals $[0,1]$ and I intersect two arbitray open sets of the following kind ${q_i}cup]a_i,b_i[$ and $q_jcup ]a_j,b_j[$ which gave me 8 cases.
When I tried to prove $[0,1]$ to be a compact I thought the basis at [0,1] ,since it would contain the interval and the rationals are dense in $mathbb{R}$, I thoughtit would be the same basis as the Euclidian basis. That would allow me to use Heine-Borel Theorem and finally prove $[0,1]$ is compact.
Question:
I am aware my answer is imprecise however it is imprecise as my ideas regarding the problem. I guess it is wrong.
How should I then solve the exercise?
general-topology proof-verification proof-writing
asked Jan 30 at 22:54
Pedro GomesPedro Gomes
1,9942721
$endgroup$
add a comment |
$begingroup$
Consider $mathscr{B}={{q}:qinmathbb{Q}cap[0,1]}cup{]a,b[:0leqslant a<bleqslant 1}}$
Prove that $mathscr{B}$ is a base for the topology $tau$ in $[0,1]$ and show $([0,1],tau)$ is compact.
Proving it is basis of topology I wrote it down on my notebook but it took me a while to deal with all the cases of intersection. I proved the union of all the open sets of the basis equals $[0,1]$ and I intersect two arbitray open sets of the following kind ${q_i}cup]a_i,b_i[$ and $q_jcup ]a_j,b_j[$ which gave me 8 cases.
When I tried to prove $[0,1]$ to be a compact I thought the basis at [0,1] ,since it would contain the interval and the rationals are dense in $mathbb{R}$, I thoughtit would be the same basis as the Euclidian basis. That would allow me to use Heine-Borel Theorem and finally prove $[0,1]$ is compact.
Question:
I am aware my answer is imprecise however it is imprecise as my ideas regarding the problem. I guess it is wrong.
How should I then solve the exercise?
general-topology proof-verification proof-writing
asked Jan 30 at 22:54
Pedro GomesPedro Gomes
1,9942721
$endgroup$
1
$begingroup$
Two deeper results: (1).If $(X,tau) , (X',tau')$ are compact Hausdorff spaces and if $f:Xto X'$ is a continuous bijection then $f$ is a homeomorphism...(2). Corollary: If $(X,tau')$ is compact Hausdorff and if $tau$ is a strictly stronger topology on $X$ then $(X,tau)$ is not compact . (Proof: $id_X$ is a continuous bijection from $(X,tau)$ to $(X,tau')$ because $tau' subset tau,$ so if $ tau$ is a compact topology then by (1) with $X'=X$ we would have $tau'=tau)$....Let $X=[0,1],$ let $tau'$ be the usual topology on $[0,1]. $
$endgroup$
– DanielWainfleet
Jan 31 at 2:22
add a comment |
$begingroup$
Consider $mathscr{B}={{q}:qinmathbb{Q}cap[0,1]}cup{]a,b[:0leqslant a<bleqslant 1}}$
Prove that $mathscr{B}$ is a base for the topology $tau$ in $[0,1]$ and show $([0,1],tau)$ is compact.
Proving it is basis of topology I wrote it down on my notebook but it took me a while to deal with all the cases of intersection. I proved the union of all the open sets of the basis equals $[0,1]$ and I intersect two arbitray open sets of the following kind ${q_i}cup]a_i,b_i[$ and $q_jcup ]a_j,b_j[$ which gave me 8 cases.
When I tried to prove $[0,1]$ to be a compact I thought the basis at [0,1] ,since it would contain the interval and the rationals are dense in $mathbb{R}$, I thoughtit would be the same basis as the Euclidian basis. That would allow me to use Heine-Borel Theorem and finally prove $[0,1]$ is compact.
Question:
I am aware my answer is imprecise however it is imprecise as my ideas regarding the problem. I guess it is wrong.
How should I then solve the exercise?
general-topology proof-verification proof-writing
asked Jan 30 at 22:54
Pedro GomesPedro Gomes
1,9942721
$endgroup$
Consider $mathscr{B}={{q}:qinmathbb{Q}cap[0,1]}cup{]a,b[:0leqslant a<bleqslant 1}}$
Prove that $mathscr{B}$ is a base for the topology $tau$ in $[0,1]$ and show $([0,1],tau)$ is compact.
Proving it is basis of topology I wrote it down on my notebook but it took me a while to deal with all the cases of intersection. I proved the union of all the open sets of the basis equals $[0,1]$ and I intersect two arbitray open sets of the following kind ${q_i}cup]a_i,b_i[$ and $q_jcup ]a_j,b_j[$ which gave me 8 cases.
When I tried to prove $[0,1]$ to be a compact I thought the basis at [0,1] ,since it would contain the interval and the rationals are dense in $mathbb{R}$, I thoughtit would be the same basis as the Euclidian basis. That would allow me to use Heine-Borel Theorem and finally prove $[0,1]$ is compact.
Question:
I am aware my answer is imprecise however it is imprecise as my ideas regarding the problem. I guess it is wrong.
How should I then solve the exercise?
general-topology proof-verification proof-writing
general-topology proof-verification proof-writing
asked Jan 30 at 22:54
Pedro GomesPedro Gomes
1,9942721
asked Jan 30 at 22:54
Pedro GomesPedro Gomes
1,9942721
asked Jan 30 at 22:54
Pedro GomesPedro Gomes
1,9942721
asked Jan 30 at 22:54
Pedro GomesPedro Gomes
1,9942721
asked Jan 30 at 22:54
Pedro GomesPedro Gomes
1,9942721
1,9942721
1
$begingroup$
Two deeper results: (1).If $(X,tau) , (X',tau')$ are compact Hausdorff spaces and if $f:Xto X'$ is a continuous bijection then $f$ is a homeomorphism...(2). Corollary: If $(X,tau')$ is compact Hausdorff and if $tau$ is a strictly stronger topology on $X$ then $(X,tau)$ is not compact . (Proof: $id_X$ is a continuous bijection from $(X,tau)$ to $(X,tau')$ because $tau' subset tau,$ so if $ tau$ is a compact topology then by (1) with $X'=X$ we would have $tau'=tau)$....Let $X=[0,1],$ let $tau'$ be the usual topology on $[0,1]. $
$endgroup$
– DanielWainfleet
Jan 31 at 2:22
add a comment |
1
$begingroup$
Two deeper results: (1).If $(X,tau) , (X',tau')$ are compact Hausdorff spaces and if $f:Xto X'$ is a continuous bijection then $f$ is a homeomorphism...(2). Corollary: If $(X,tau')$ is compact Hausdorff and if $tau$ is a strictly stronger topology on $X$ then $(X,tau)$ is not compact . (Proof: $id_X$ is a continuous bijection from $(X,tau)$ to $(X,tau')$ because $tau' subset tau,$ so if $ tau$ is a compact topology then by (1) with $X'=X$ we would have $tau'=tau)$....Let $X=[0,1],$ let $tau'$ be the usual topology on $[0,1]. $
$endgroup$
– DanielWainfleet
Jan 31 at 2:22
1
1
$begingroup$
Two deeper results: (1).If $(X,tau) , (X',tau')$ are compact Hausdorff spaces and if $f:Xto X'$ is a continuous bijection then $f$ is a homeomorphism...(2). Corollary: If $(X,tau')$ is compact Hausdorff and if $tau$ is a strictly stronger topology on $X$ then $(X,tau)$ is not compact . (Proof: $id_X$ is a continuous bijection from $(X,tau)$ to $(X,tau')$ because $tau' subset tau,$ so if $ tau$ is a compact topology then by (1) with $X'=X$ we would have $tau'=tau)$....Let $X=[0,1],$ let $tau'$ be the usual topology on $[0,1]. $
$endgroup$
– DanielWainfleet
Jan 31 at 2:22
$begingroup$
Two deeper results: (1).If $(X,tau) , (X',tau')$ are compact Hausdorff spaces and if $f:Xto X'$ is a continuous bijection then $f$ is a homeomorphism...(2). Corollary: If $(X,tau')$ is compact Hausdorff and if $tau$ is a strictly stronger topology on $X$ then $(X,tau)$ is not compact . (Proof: $id_X$ is a continuous bijection from $(X,tau)$ to $(X,tau')$ because $tau' subset tau,$ so if $ tau$ is a compact topology then by (1) with $X'=X$ we would have $tau'=tau)$....Let $X=[0,1],$ let $tau'$ be the usual topology on $[0,1]. $
$endgroup$
– DanielWainfleet
Jan 31 at 2:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The showing a base part is not very hard: they clearly cover $[0,1]$ and any two basic subsets that intersect have their intersection in the base again (open intervals or singleton rationals). That makes the check pretty easy.
It seems to me that a set like $C={frac{1}{n} mid n ge 1}$ is actually closed and discrete in this topology, so that would make the topology non-compact.
It certainly is not the Euclidean topology on $[0,1]$, as it has plenty of isolated points; a dense set of them. It's also not connected for that reason.
answered Jan 30 at 23:25
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
1
$begingroup$
We can also observe that ${{0,1}}cup {(1/n,1): nin Bbb N}$ is an open cover with no finite sub-cover.
$endgroup$
– DanielWainfleet
Jan 31 at 2:00
$begingroup$
Thanks for your answer. $C$ is obviously closed in $([0,1],tau)$. I know there is a theorem that states that every closed set in a compact space is compact. If $C$ was not compact that would be a contradiction hence $([0,1],tau)$ would be not compact. But I do not really know why $C$ is not compact once ${0}in[0,1]$ so the sequence and subsequence should converge. What is the reasoning behind the answer? Thanks in advance!
$endgroup$
– Pedro Gomes
Jan 31 at 11:18
$begingroup$
@PedroGomes it’s a discrete subspace ( all points are rational) and an infinite discrete space is never compact.
$endgroup$
– Henno Brandsma
Jan 31 at 16:13
$begingroup$
Sorry for disturbing. But I have been thinking about the answer and it has been a little bit hard to accept $C={frac{1}{n} mid n ge 1}$ to be closed once each rational number is defined as open on the basis. Is this true?
$endgroup$
– Pedro Gomes
Feb 1 at 17:46
$begingroup$
@PedroGomes $C$ is also open for that reason. So $[0,1$ is also not connected.
$endgroup$
– Henno Brandsma
Feb 1 at 17:51
|
show 2 more comments
Your Answer
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1 Answer
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1 Answer
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votes
$begingroup$
The showing a base part is not very hard: they clearly cover $[0,1]$ and any two basic subsets that intersect have their intersection in the base again (open intervals or singleton rationals). That makes the check pretty easy.
It seems to me that a set like $C={frac{1}{n} mid n ge 1}$ is actually closed and discrete in this topology, so that would make the topology non-compact.
It certainly is not the Euclidean topology on $[0,1]$, as it has plenty of isolated points; a dense set of them. It's also not connected for that reason.
answered Jan 30 at 23:25
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
1
$begingroup$
We can also observe that ${{0,1}}cup {(1/n,1): nin Bbb N}$ is an open cover with no finite sub-cover.
$endgroup$
– DanielWainfleet
Jan 31 at 2:00
$begingroup$
Thanks for your answer. $C$ is obviously closed in $([0,1],tau)$. I know there is a theorem that states that every closed set in a compact space is compact. If $C$ was not compact that would be a contradiction hence $([0,1],tau)$ would be not compact. But I do not really know why $C$ is not compact once ${0}in[0,1]$ so the sequence and subsequence should converge. What is the reasoning behind the answer? Thanks in advance!
$endgroup$
– Pedro Gomes
Jan 31 at 11:18
$begingroup$
@PedroGomes it’s a discrete subspace ( all points are rational) and an infinite discrete space is never compact.
$endgroup$
– Henno Brandsma
Jan 31 at 16:13
$begingroup$
Sorry for disturbing. But I have been thinking about the answer and it has been a little bit hard to accept $C={frac{1}{n} mid n ge 1}$ to be closed once each rational number is defined as open on the basis. Is this true?
$endgroup$
– Pedro Gomes
Feb 1 at 17:46
$begingroup$
@PedroGomes $C$ is also open for that reason. So $[0,1$ is also not connected.
$endgroup$
– Henno Brandsma
Feb 1 at 17:51
|
show 2 more comments
$begingroup$
The showing a base part is not very hard: they clearly cover $[0,1]$ and any two basic subsets that intersect have their intersection in the base again (open intervals or singleton rationals). That makes the check pretty easy.
It seems to me that a set like $C={frac{1}{n} mid n ge 1}$ is actually closed and discrete in this topology, so that would make the topology non-compact.
It certainly is not the Euclidean topology on $[0,1]$, as it has plenty of isolated points; a dense set of them. It's also not connected for that reason.
answered Jan 30 at 23:25
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
1
$begingroup$
We can also observe that ${{0,1}}cup {(1/n,1): nin Bbb N}$ is an open cover with no finite sub-cover.
$endgroup$
– DanielWainfleet
Jan 31 at 2:00
$begingroup$
Thanks for your answer. $C$ is obviously closed in $([0,1],tau)$. I know there is a theorem that states that every closed set in a compact space is compact. If $C$ was not compact that would be a contradiction hence $([0,1],tau)$ would be not compact. But I do not really know why $C$ is not compact once ${0}in[0,1]$ so the sequence and subsequence should converge. What is the reasoning behind the answer? Thanks in advance!
$endgroup$
– Pedro Gomes
Jan 31 at 11:18
$begingroup$
@PedroGomes it’s a discrete subspace ( all points are rational) and an infinite discrete space is never compact.
$endgroup$
– Henno Brandsma
Jan 31 at 16:13
$begingroup$
Sorry for disturbing. But I have been thinking about the answer and it has been a little bit hard to accept $C={frac{1}{n} mid n ge 1}$ to be closed once each rational number is defined as open on the basis. Is this true?
$endgroup$
– Pedro Gomes
Feb 1 at 17:46
$begingroup$
@PedroGomes $C$ is also open for that reason. So $[0,1$ is also not connected.
$endgroup$
– Henno Brandsma
Feb 1 at 17:51
|
show 2 more comments
$begingroup$
The showing a base part is not very hard: they clearly cover $[0,1]$ and any two basic subsets that intersect have their intersection in the base again (open intervals or singleton rationals). That makes the check pretty easy.
It seems to me that a set like $C={frac{1}{n} mid n ge 1}$ is actually closed and discrete in this topology, so that would make the topology non-compact.
It certainly is not the Euclidean topology on $[0,1]$, as it has plenty of isolated points; a dense set of them. It's also not connected for that reason.
answered Jan 30 at 23:25
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
The showing a base part is not very hard: they clearly cover $[0,1]$ and any two basic subsets that intersect have their intersection in the base again (open intervals or singleton rationals). That makes the check pretty easy.
It seems to me that a set like $C={frac{1}{n} mid n ge 1}$ is actually closed and discrete in this topology, so that would make the topology non-compact.
It certainly is not the Euclidean topology on $[0,1]$, as it has plenty of isolated points; a dense set of them. It's also not connected for that reason.
answered Jan 30 at 23:25
Henno BrandsmaHenno Brandsma
115k349125
answered Jan 30 at 23:25
Henno BrandsmaHenno Brandsma
115k349125
answered Jan 30 at 23:25
Henno BrandsmaHenno Brandsma
115k349125
answered Jan 30 at 23:25
Henno BrandsmaHenno Brandsma
115k349125
115k349125
1
$begingroup$
We can also observe that ${{0,1}}cup {(1/n,1): nin Bbb N}$ is an open cover with no finite sub-cover.
$endgroup$
– DanielWainfleet
Jan 31 at 2:00
$begingroup$
Thanks for your answer. $C$ is obviously closed in $([0,1],tau)$. I know there is a theorem that states that every closed set in a compact space is compact. If $C$ was not compact that would be a contradiction hence $([0,1],tau)$ would be not compact. But I do not really know why $C$ is not compact once ${0}in[0,1]$ so the sequence and subsequence should converge. What is the reasoning behind the answer? Thanks in advance!
$endgroup$
– Pedro Gomes
Jan 31 at 11:18
$begingroup$
@PedroGomes it’s a discrete subspace ( all points are rational) and an infinite discrete space is never compact.
$endgroup$
– Henno Brandsma
Jan 31 at 16:13
$begingroup$
Sorry for disturbing. But I have been thinking about the answer and it has been a little bit hard to accept $C={frac{1}{n} mid n ge 1}$ to be closed once each rational number is defined as open on the basis. Is this true?
$endgroup$
– Pedro Gomes
Feb 1 at 17:46
$begingroup$
@PedroGomes $C$ is also open for that reason. So $[0,1$ is also not connected.
$endgroup$
– Henno Brandsma
Feb 1 at 17:51
|
show 2 more comments
1
$begingroup$
We can also observe that ${{0,1}}cup {(1/n,1): nin Bbb N}$ is an open cover with no finite sub-cover.
$endgroup$
– DanielWainfleet
Jan 31 at 2:00
$begingroup$
Thanks for your answer. $C$ is obviously closed in $([0,1],tau)$. I know there is a theorem that states that every closed set in a compact space is compact. If $C$ was not compact that would be a contradiction hence $([0,1],tau)$ would be not compact. But I do not really know why $C$ is not compact once ${0}in[0,1]$ so the sequence and subsequence should converge. What is the reasoning behind the answer? Thanks in advance!
$endgroup$
– Pedro Gomes
Jan 31 at 11:18
$begingroup$
@PedroGomes it’s a discrete subspace ( all points are rational) and an infinite discrete space is never compact.
$endgroup$
– Henno Brandsma
Jan 31 at 16:13
$begingroup$
Sorry for disturbing. But I have been thinking about the answer and it has been a little bit hard to accept $C={frac{1}{n} mid n ge 1}$ to be closed once each rational number is defined as open on the basis. Is this true?
$endgroup$
– Pedro Gomes
Feb 1 at 17:46
$begingroup$
@PedroGomes $C$ is also open for that reason. So $[0,1$ is also not connected.
$endgroup$
– Henno Brandsma
Feb 1 at 17:51
1
1
$begingroup$
We can also observe that ${{0,1}}cup {(1/n,1): nin Bbb N}$ is an open cover with no finite sub-cover.
$endgroup$
– DanielWainfleet
Jan 31 at 2:00
$begingroup$
We can also observe that ${{0,1}}cup {(1/n,1): nin Bbb N}$ is an open cover with no finite sub-cover.
$endgroup$
– DanielWainfleet
Jan 31 at 2:00
$begingroup$
Thanks for your answer. $C$ is obviously closed in $([0,1],tau)$. I know there is a theorem that states that every closed set in a compact space is compact. If $C$ was not compact that would be a contradiction hence $([0,1],tau)$ would be not compact. But I do not really know why $C$ is not compact once ${0}in[0,1]$ so the sequence and subsequence should converge. What is the reasoning behind the answer? Thanks in advance!
$endgroup$
– Pedro Gomes
Jan 31 at 11:18
$begingroup$
Thanks for your answer. $C$ is obviously closed in $([0,1],tau)$. I know there is a theorem that states that every closed set in a compact space is compact. If $C$ was not compact that would be a contradiction hence $([0,1],tau)$ would be not compact. But I do not really know why $C$ is not compact once ${0}in[0,1]$ so the sequence and subsequence should converge. What is the reasoning behind the answer? Thanks in advance!
$endgroup$
– Pedro Gomes
Jan 31 at 11:18
$begingroup$
@PedroGomes it’s a discrete subspace ( all points are rational) and an infinite discrete space is never compact.
$endgroup$
– Henno Brandsma
Jan 31 at 16:13
$begingroup$
@PedroGomes it’s a discrete subspace ( all points are rational) and an infinite discrete space is never compact.
$endgroup$
– Henno Brandsma
Jan 31 at 16:13
$begingroup$
Sorry for disturbing. But I have been thinking about the answer and it has been a little bit hard to accept $C={frac{1}{n} mid n ge 1}$ to be closed once each rational number is defined as open on the basis. Is this true?
$endgroup$
– Pedro Gomes
Feb 1 at 17:46
$begingroup$
Sorry for disturbing. But I have been thinking about the answer and it has been a little bit hard to accept $C={frac{1}{n} mid n ge 1}$ to be closed once each rational number is defined as open on the basis. Is this true?
$endgroup$
– Pedro Gomes
Feb 1 at 17:46
$begingroup$
@PedroGomes $C$ is also open for that reason. So $[0,1$ is also not connected.
$endgroup$
– Henno Brandsma
Feb 1 at 17:51
$begingroup$
@PedroGomes $C$ is also open for that reason. So $[0,1$ is also not connected.
$endgroup$
– Henno Brandsma
Feb 1 at 17:51
|
show 2 more comments
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Two deeper results: (1).If $(X,tau) , (X',tau')$ are compact Hausdorff spaces and if $f:Xto X'$ is a continuous bijection then $f$ is a homeomorphism...(2). Corollary: If $(X,tau')$ is compact Hausdorff and if $tau$ is a strictly stronger topology on $X$ then $(X,tau)$ is not compact . (Proof: $id_X$ is a continuous bijection from $(X,tau)$ to $(X,tau')$ because $tau' subset tau,$ so if $ tau$ is a compact topology then by (1) with $X'=X$ we would have $tau'=tau)$....Let $X=[0,1],$ let $tau'$ be the usual topology on $[0,1]. $
$endgroup$
– DanielWainfleet
Jan 31 at 2:22