Showing that $[0, omega_1]$ is compact. [duplicate]
up vote
0
down vote
favorite
This question already has an answer here:
First uncountable ordinal
1 answer
I want to show that $[0, omega_1]$ is compact, where $omega_1$ is the least uncountable ordinal, and I have just been introduced to the concepts of ordinals.
The tips I have seen to showing this is that $[0, omega_1]$:
- contains a maximal element
- there are no infinite strictly decreasing sequences of ordinals.
Take an open cover of the desired set. My thoughts are that some open set must contain $omega_1$, which in turn contains all ordinals smaller than it by the definition of an ordinal, and so this open set is a cover of the entire space. Any further insights appreciated.
general-topology compactness ordinals well-orders
marked as duplicate by Asaf Karagila♦
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 23 at 8:50
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 |
up vote
0
down vote
favorite
This question already has an answer here:
First uncountable ordinal
1 answer
I want to show that $[0, omega_1]$ is compact, where $omega_1$ is the least uncountable ordinal, and I have just been introduced to the concepts of ordinals.
The tips I have seen to showing this is that $[0, omega_1]$:
- contains a maximal element
- there are no infinite strictly decreasing sequences of ordinals.
Take an open cover of the desired set. My thoughts are that some open set must contain $omega_1$, which in turn contains all ordinals smaller than it by the definition of an ordinal, and so this open set is a cover of the entire space. Any further insights appreciated.
general-topology compactness ordinals well-orders
marked as duplicate by Asaf Karagila♦
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 23 at 8:50
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.
An alternative way is to use that for a countable ordinal $beta$, the set $[0,beta]$ is compact in the order topology using some criterion for compactness in ordered spaces, ifyou know such theorems.
– Henno Brandsma
Nov 23 at 4:56
@HennoBrandsma I have a theorem in my notes (with a proof) that states : Let $X$ be a simply ordered set with the least upper bound property with respect to the order topology. Then each closed interval is compact. Does this apply? I guess really I am asking if the least upper bound property is present here?
– IntegrateThis
Nov 23 at 5:00
1
Yes, you could use that theorem to show that for all ordinals $[0,beta]$ is compact. A subset of ordinals wtih an upper bound has a sup, as this we can take the minimum of the set of upperbounds, which exists as all non-empty sets of ordinals have a minimum (equivalent to your property it hss no infinite strictly decreasing subsets).
– Henno Brandsma
Nov 23 at 5:11
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
First uncountable ordinal
1 answer
I want to show that $[0, omega_1]$ is compact, where $omega_1$ is the least uncountable ordinal, and I have just been introduced to the concepts of ordinals.
The tips I have seen to showing this is that $[0, omega_1]$:
- contains a maximal element
- there are no infinite strictly decreasing sequences of ordinals.
Take an open cover of the desired set. My thoughts are that some open set must contain $omega_1$, which in turn contains all ordinals smaller than it by the definition of an ordinal, and so this open set is a cover of the entire space. Any further insights appreciated.
general-topology compactness ordinals well-orders
This question already has an answer here:
First uncountable ordinal
1 answer
I want to show that $[0, omega_1]$ is compact, where $omega_1$ is the least uncountable ordinal, and I have just been introduced to the concepts of ordinals.
The tips I have seen to showing this is that $[0, omega_1]$:
- contains a maximal element
- there are no infinite strictly decreasing sequences of ordinals.
Take an open cover of the desired set. My thoughts are that some open set must contain $omega_1$, which in turn contains all ordinals smaller than it by the definition of an ordinal, and so this open set is a cover of the entire space. Any further insights appreciated.
This question already has an answer here:
First uncountable ordinal
1 answer
general-topology compactness ordinals well-orders
general-topology compactness ordinals well-orders
edited Nov 23 at 8:48
Asaf Karagila♦
301k32422753
301k32422753
asked Nov 23 at 3:40
IntegrateThis
1,7081717
1,7081717
marked as duplicate by Asaf Karagila♦
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 23 at 8:50
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 Asaf Karagila♦
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 23 at 8:50
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.
An alternative way is to use that for a countable ordinal $beta$, the set $[0,beta]$ is compact in the order topology using some criterion for compactness in ordered spaces, ifyou know such theorems.
– Henno Brandsma
Nov 23 at 4:56
@HennoBrandsma I have a theorem in my notes (with a proof) that states : Let $X$ be a simply ordered set with the least upper bound property with respect to the order topology. Then each closed interval is compact. Does this apply? I guess really I am asking if the least upper bound property is present here?
– IntegrateThis
Nov 23 at 5:00
1
Yes, you could use that theorem to show that for all ordinals $[0,beta]$ is compact. A subset of ordinals wtih an upper bound has a sup, as this we can take the minimum of the set of upperbounds, which exists as all non-empty sets of ordinals have a minimum (equivalent to your property it hss no infinite strictly decreasing subsets).
– Henno Brandsma
Nov 23 at 5:11
add a comment |
An alternative way is to use that for a countable ordinal $beta$, the set $[0,beta]$ is compact in the order topology using some criterion for compactness in ordered spaces, ifyou know such theorems.
– Henno Brandsma
Nov 23 at 4:56
@HennoBrandsma I have a theorem in my notes (with a proof) that states : Let $X$ be a simply ordered set with the least upper bound property with respect to the order topology. Then each closed interval is compact. Does this apply? I guess really I am asking if the least upper bound property is present here?
– IntegrateThis
Nov 23 at 5:00
1
Yes, you could use that theorem to show that for all ordinals $[0,beta]$ is compact. A subset of ordinals wtih an upper bound has a sup, as this we can take the minimum of the set of upperbounds, which exists as all non-empty sets of ordinals have a minimum (equivalent to your property it hss no infinite strictly decreasing subsets).
– Henno Brandsma
Nov 23 at 5:11
An alternative way is to use that for a countable ordinal $beta$, the set $[0,beta]$ is compact in the order topology using some criterion for compactness in ordered spaces, ifyou know such theorems.
– Henno Brandsma
Nov 23 at 4:56
An alternative way is to use that for a countable ordinal $beta$, the set $[0,beta]$ is compact in the order topology using some criterion for compactness in ordered spaces, ifyou know such theorems.
– Henno Brandsma
Nov 23 at 4:56
@HennoBrandsma I have a theorem in my notes (with a proof) that states : Let $X$ be a simply ordered set with the least upper bound property with respect to the order topology. Then each closed interval is compact. Does this apply? I guess really I am asking if the least upper bound property is present here?
– IntegrateThis
Nov 23 at 5:00
@HennoBrandsma I have a theorem in my notes (with a proof) that states : Let $X$ be a simply ordered set with the least upper bound property with respect to the order topology. Then each closed interval is compact. Does this apply? I guess really I am asking if the least upper bound property is present here?
– IntegrateThis
Nov 23 at 5:00
1
1
Yes, you could use that theorem to show that for all ordinals $[0,beta]$ is compact. A subset of ordinals wtih an upper bound has a sup, as this we can take the minimum of the set of upperbounds, which exists as all non-empty sets of ordinals have a minimum (equivalent to your property it hss no infinite strictly decreasing subsets).
– Henno Brandsma
Nov 23 at 5:11
Yes, you could use that theorem to show that for all ordinals $[0,beta]$ is compact. A subset of ordinals wtih an upper bound has a sup, as this we can take the minimum of the set of upperbounds, which exists as all non-empty sets of ordinals have a minimum (equivalent to your property it hss no infinite strictly decreasing subsets).
– Henno Brandsma
Nov 23 at 5:11
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
We can show for an arbitrary ordinal $alpha$ that $[0,alpha]$ is compact with the order topology.
Let $mathcal U$ be an open cover of $[0,alpha]$, and let $beta$ be the smallest ordinal such that $[beta,alpha]$ is covered by a finite subcover of $mathcal U$. (Certainly there are some $beta$s with this property, such as $alpha$ itself, so there must be a smallest one).
Now if $beta$ is a successor ordinal $beta=delta+1$, then there must be an element of $mathcal U$ that contains $delta$, and adding that to our finite cover of $[beta,alpha]$ produces a finite cover of $[delta,alpha]$, a contradiction.
On the other hand, if $beta$ is a limit ordinal, then every open set that contains $beta$ must also contain some $delta<beta$, and therefore our finite subcover of $[beta,alpha]$ is already a cover of $[delta,alpha]$. Again a contradiction.
The only possibility remaining is that $beta=0$.
add a comment |
up vote
1
down vote
Suppose that $mathcal{U}$ is an open cover of $[0,omega_1]$ by basic open sets of the order topology (that's the topology we are putting on this ordered set), and suppose for a contradiction that it has no finite subcover. (It suffices to consider bcovers by base elements, as is easy to see in all topological spaces, if you know Alexander's subbase theorem, the proof will be easier even).
Some element of the cover has to contain the maximale element $omega_1$, and so some $U_0 in mathcal{U}$ must be of the form $(beta_0, omega_1]$ (such are the basic open subsets in the order topology that contain the maximal element).
Also, $0$ must be covered so there is some $U'_0 in mathcal{U}$ of the form $[0,alpha_0)$ that contains $0$.
Now we know that $U_0, U'_0$ cannot be a finite subcover (which would happen when $alpha_0 > beta_0$), and so that some $U_1 in mathcal{U}$ must cover $beta_0$, say $beta_0 in U_1 = (alpha_2, beta_1)$, which implies that $alpha_2 < beta_0 < beta_1$.
Now consider what happens if the sets $U_0, U'_0, U_1$ do not form a subcover yet, find a new element to cover etc., and define an infinite decreasing sequence of ordinals that way.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
We can show for an arbitrary ordinal $alpha$ that $[0,alpha]$ is compact with the order topology.
Let $mathcal U$ be an open cover of $[0,alpha]$, and let $beta$ be the smallest ordinal such that $[beta,alpha]$ is covered by a finite subcover of $mathcal U$. (Certainly there are some $beta$s with this property, such as $alpha$ itself, so there must be a smallest one).
Now if $beta$ is a successor ordinal $beta=delta+1$, then there must be an element of $mathcal U$ that contains $delta$, and adding that to our finite cover of $[beta,alpha]$ produces a finite cover of $[delta,alpha]$, a contradiction.
On the other hand, if $beta$ is a limit ordinal, then every open set that contains $beta$ must also contain some $delta<beta$, and therefore our finite subcover of $[beta,alpha]$ is already a cover of $[delta,alpha]$. Again a contradiction.
The only possibility remaining is that $beta=0$.
add a comment |
up vote
3
down vote
accepted
We can show for an arbitrary ordinal $alpha$ that $[0,alpha]$ is compact with the order topology.
Let $mathcal U$ be an open cover of $[0,alpha]$, and let $beta$ be the smallest ordinal such that $[beta,alpha]$ is covered by a finite subcover of $mathcal U$. (Certainly there are some $beta$s with this property, such as $alpha$ itself, so there must be a smallest one).
Now if $beta$ is a successor ordinal $beta=delta+1$, then there must be an element of $mathcal U$ that contains $delta$, and adding that to our finite cover of $[beta,alpha]$ produces a finite cover of $[delta,alpha]$, a contradiction.
On the other hand, if $beta$ is a limit ordinal, then every open set that contains $beta$ must also contain some $delta<beta$, and therefore our finite subcover of $[beta,alpha]$ is already a cover of $[delta,alpha]$. Again a contradiction.
The only possibility remaining is that $beta=0$.
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
We can show for an arbitrary ordinal $alpha$ that $[0,alpha]$ is compact with the order topology.
Let $mathcal U$ be an open cover of $[0,alpha]$, and let $beta$ be the smallest ordinal such that $[beta,alpha]$ is covered by a finite subcover of $mathcal U$. (Certainly there are some $beta$s with this property, such as $alpha$ itself, so there must be a smallest one).
Now if $beta$ is a successor ordinal $beta=delta+1$, then there must be an element of $mathcal U$ that contains $delta$, and adding that to our finite cover of $[beta,alpha]$ produces a finite cover of $[delta,alpha]$, a contradiction.
On the other hand, if $beta$ is a limit ordinal, then every open set that contains $beta$ must also contain some $delta<beta$, and therefore our finite subcover of $[beta,alpha]$ is already a cover of $[delta,alpha]$. Again a contradiction.
The only possibility remaining is that $beta=0$.
We can show for an arbitrary ordinal $alpha$ that $[0,alpha]$ is compact with the order topology.
Let $mathcal U$ be an open cover of $[0,alpha]$, and let $beta$ be the smallest ordinal such that $[beta,alpha]$ is covered by a finite subcover of $mathcal U$. (Certainly there are some $beta$s with this property, such as $alpha$ itself, so there must be a smallest one).
Now if $beta$ is a successor ordinal $beta=delta+1$, then there must be an element of $mathcal U$ that contains $delta$, and adding that to our finite cover of $[beta,alpha]$ produces a finite cover of $[delta,alpha]$, a contradiction.
On the other hand, if $beta$ is a limit ordinal, then every open set that contains $beta$ must also contain some $delta<beta$, and therefore our finite subcover of $[beta,alpha]$ is already a cover of $[delta,alpha]$. Again a contradiction.
The only possibility remaining is that $beta=0$.
answered Nov 23 at 5:08
Henning Makholm
236k16300534
236k16300534
add a comment |
add a comment |
up vote
1
down vote
Suppose that $mathcal{U}$ is an open cover of $[0,omega_1]$ by basic open sets of the order topology (that's the topology we are putting on this ordered set), and suppose for a contradiction that it has no finite subcover. (It suffices to consider bcovers by base elements, as is easy to see in all topological spaces, if you know Alexander's subbase theorem, the proof will be easier even).
Some element of the cover has to contain the maximale element $omega_1$, and so some $U_0 in mathcal{U}$ must be of the form $(beta_0, omega_1]$ (such are the basic open subsets in the order topology that contain the maximal element).
Also, $0$ must be covered so there is some $U'_0 in mathcal{U}$ of the form $[0,alpha_0)$ that contains $0$.
Now we know that $U_0, U'_0$ cannot be a finite subcover (which would happen when $alpha_0 > beta_0$), and so that some $U_1 in mathcal{U}$ must cover $beta_0$, say $beta_0 in U_1 = (alpha_2, beta_1)$, which implies that $alpha_2 < beta_0 < beta_1$.
Now consider what happens if the sets $U_0, U'_0, U_1$ do not form a subcover yet, find a new element to cover etc., and define an infinite decreasing sequence of ordinals that way.
add a comment |
up vote
1
down vote
Suppose that $mathcal{U}$ is an open cover of $[0,omega_1]$ by basic open sets of the order topology (that's the topology we are putting on this ordered set), and suppose for a contradiction that it has no finite subcover. (It suffices to consider bcovers by base elements, as is easy to see in all topological spaces, if you know Alexander's subbase theorem, the proof will be easier even).
Some element of the cover has to contain the maximale element $omega_1$, and so some $U_0 in mathcal{U}$ must be of the form $(beta_0, omega_1]$ (such are the basic open subsets in the order topology that contain the maximal element).
Also, $0$ must be covered so there is some $U'_0 in mathcal{U}$ of the form $[0,alpha_0)$ that contains $0$.
Now we know that $U_0, U'_0$ cannot be a finite subcover (which would happen when $alpha_0 > beta_0$), and so that some $U_1 in mathcal{U}$ must cover $beta_0$, say $beta_0 in U_1 = (alpha_2, beta_1)$, which implies that $alpha_2 < beta_0 < beta_1$.
Now consider what happens if the sets $U_0, U'_0, U_1$ do not form a subcover yet, find a new element to cover etc., and define an infinite decreasing sequence of ordinals that way.
add a comment |
up vote
1
down vote
up vote
1
down vote
Suppose that $mathcal{U}$ is an open cover of $[0,omega_1]$ by basic open sets of the order topology (that's the topology we are putting on this ordered set), and suppose for a contradiction that it has no finite subcover. (It suffices to consider bcovers by base elements, as is easy to see in all topological spaces, if you know Alexander's subbase theorem, the proof will be easier even).
Some element of the cover has to contain the maximale element $omega_1$, and so some $U_0 in mathcal{U}$ must be of the form $(beta_0, omega_1]$ (such are the basic open subsets in the order topology that contain the maximal element).
Also, $0$ must be covered so there is some $U'_0 in mathcal{U}$ of the form $[0,alpha_0)$ that contains $0$.
Now we know that $U_0, U'_0$ cannot be a finite subcover (which would happen when $alpha_0 > beta_0$), and so that some $U_1 in mathcal{U}$ must cover $beta_0$, say $beta_0 in U_1 = (alpha_2, beta_1)$, which implies that $alpha_2 < beta_0 < beta_1$.
Now consider what happens if the sets $U_0, U'_0, U_1$ do not form a subcover yet, find a new element to cover etc., and define an infinite decreasing sequence of ordinals that way.
Suppose that $mathcal{U}$ is an open cover of $[0,omega_1]$ by basic open sets of the order topology (that's the topology we are putting on this ordered set), and suppose for a contradiction that it has no finite subcover. (It suffices to consider bcovers by base elements, as is easy to see in all topological spaces, if you know Alexander's subbase theorem, the proof will be easier even).
Some element of the cover has to contain the maximale element $omega_1$, and so some $U_0 in mathcal{U}$ must be of the form $(beta_0, omega_1]$ (such are the basic open subsets in the order topology that contain the maximal element).
Also, $0$ must be covered so there is some $U'_0 in mathcal{U}$ of the form $[0,alpha_0)$ that contains $0$.
Now we know that $U_0, U'_0$ cannot be a finite subcover (which would happen when $alpha_0 > beta_0$), and so that some $U_1 in mathcal{U}$ must cover $beta_0$, say $beta_0 in U_1 = (alpha_2, beta_1)$, which implies that $alpha_2 < beta_0 < beta_1$.
Now consider what happens if the sets $U_0, U'_0, U_1$ do not form a subcover yet, find a new element to cover etc., and define an infinite decreasing sequence of ordinals that way.
edited Nov 23 at 5:08
answered Nov 23 at 4:54
Henno Brandsma
103k345113
103k345113
add a comment |
add a comment |
An alternative way is to use that for a countable ordinal $beta$, the set $[0,beta]$ is compact in the order topology using some criterion for compactness in ordered spaces, ifyou know such theorems.
– Henno Brandsma
Nov 23 at 4:56
@HennoBrandsma I have a theorem in my notes (with a proof) that states : Let $X$ be a simply ordered set with the least upper bound property with respect to the order topology. Then each closed interval is compact. Does this apply? I guess really I am asking if the least upper bound property is present here?
– IntegrateThis
Nov 23 at 5:00
1
Yes, you could use that theorem to show that for all ordinals $[0,beta]$ is compact. A subset of ordinals wtih an upper bound has a sup, as this we can take the minimum of the set of upperbounds, which exists as all non-empty sets of ordinals have a minimum (equivalent to your property it hss no infinite strictly decreasing subsets).
– Henno Brandsma
Nov 23 at 5:11