How is this definition of closedness compatible with the order topology?
$begingroup$
Let $kappa$ be a limit ordinal.
Taken from the definition of a closed unbounded set,
we say a subset $Csubseteqkappa$ is closed in $kappa$ if and only if
$sup(Ccapalpha)=alpha<kappaimpliesalphain C$
Is this definition equivalent to being closed with respect to the order topology on $kappa$?
Wikipedia claims it should be, but I cannot prove it.
set-theory order-topology
edited Jan 6 at 13:35
Andrés E. Caicedo
65.8k8160251
asked Jan 6 at 12:20
LilalasLilalas
54
$endgroup$
add a comment |
$begingroup$
Let $kappa$ be a limit ordinal.
Taken from the definition of a closed unbounded set,
we say a subset $Csubseteqkappa$ is closed in $kappa$ if and only if
$sup(Ccapalpha)=alpha<kappaimpliesalphain C$
Is this definition equivalent to being closed with respect to the order topology on $kappa$?
Wikipedia claims it should be, but I cannot prove it.
set-theory order-topology
edited Jan 6 at 13:35
Andrés E. Caicedo
65.8k8160251
asked Jan 6 at 12:20
LilalasLilalas
54
$endgroup$
add a comment |
$begingroup$
Let $kappa$ be a limit ordinal.
Taken from the definition of a closed unbounded set,
we say a subset $Csubseteqkappa$ is closed in $kappa$ if and only if
$sup(Ccapalpha)=alpha<kappaimpliesalphain C$
Is this definition equivalent to being closed with respect to the order topology on $kappa$?
Wikipedia claims it should be, but I cannot prove it.
set-theory order-topology
edited Jan 6 at 13:35
Andrés E. Caicedo
65.8k8160251
asked Jan 6 at 12:20
LilalasLilalas
54
$endgroup$
Let $kappa$ be a limit ordinal.
Taken from the definition of a closed unbounded set,
we say a subset $Csubseteqkappa$ is closed in $kappa$ if and only if
$sup(Ccapalpha)=alpha<kappaimpliesalphain C$
Is this definition equivalent to being closed with respect to the order topology on $kappa$?
Wikipedia claims it should be, but I cannot prove it.
set-theory order-topology
set-theory order-topology
edited Jan 6 at 13:35
Andrés E. Caicedo
65.8k8160251
asked Jan 6 at 12:20
LilalasLilalas
54
edited Jan 6 at 13:35
Andrés E. Caicedo
65.8k8160251
asked Jan 6 at 12:20
LilalasLilalas
54
edited Jan 6 at 13:35
Andrés E. Caicedo
65.8k8160251
edited Jan 6 at 13:35
Andrés E. Caicedo
65.8k8160251
edited Jan 6 at 13:35
Andrés E. Caicedo
65.8k8160251
65.8k8160251
asked Jan 6 at 12:20
LilalasLilalas
54
asked Jan 6 at 12:20
LilalasLilalas
54
asked Jan 6 at 12:20
LilalasLilalas
54
54
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The order topology has as its sub-base all upper and lower sets, $U(alpha) = {beta: beta > alpha}$ and $L(alpha) = {beta: beta < alpha}$.
This implies that in an ordinal $kappa$, all successor ordinals $alpha+1in kappa$ are isolated points, because ${alpha+1}= (alpha, alpha+2)=U(alpha} cap L(alpha+2)$.
At limit ordinals $betain kappa$ we have that a local base is all
sets of the form $(alpha, beta]$ (this is open as $U(alpha)cap L(beta+1)$), so at limit ordinals the topology "looks to the left".
Now suppose that $C$ is order-closed and that $sup(Ccap alpha) = alpha < kappa$ and $alpha$ is a limit. Then if $(gamma, beta]$ is any basic open neighbourhood of $alpha$, with some $gamma < alpha$, we note that $gamma$ is no upperbound for $Ccap alpha$ and so some $gamma'in Ccap alpha$ exists with $gamma' > gamma$. This means that $gamma' in (gamma, alpha]$ so that we know that every basic neighbourhood of $alpha$ intersects $C$, so $alpha in overline{C}=C$, as required.
A similar argument shows that if $C$ obeys the "$sup$-property", it is order closed: let $alpha in overline{C}$. If $alpha$ is isolated it has a neighbourhood ${alpha}$ that must intersect $C$, so $alpha in C$. So now assume $alpha$ is a limit. $alpha in overline{C}$ implies that for every $gamma < alpha$ the basic neighbourhood $(gamma, alpha]$ intersects $C$. This readily implies that $alpha = sup(C cap alpha)$ (as $alpha$ is surely an upperbound for $C cap alpha$ and the previous remarks shows no smaller one can exist. The property then implies $alpha in C$ and so $overline{C} subseteq C$ and $C$ is order-closed.
edited Jan 8 at 12:32
Namaste
1
answered Jan 6 at 23:33
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
$begingroup$
Unfortunately, my edit got rejected, so I am writing this in a comment because it took me some while to realise it. You are using these two statements from general topology: 1. $C$ is closed if and only if $overline{C}subseteq C$, and 2. $alphainoverline{C}$ if and only if every open neighbourhood of $alpha$ intersects $C$
$endgroup$
– Lilalas
Jan 11 at 10:21
$begingroup$
@Lilalas Indeed, that’s what I use. It’s pretty elementary I’d say, topology 101.
$endgroup$
– Henno Brandsma
Jan 11 at 11:46
add a comment |
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$begingroup$
The order topology has as its sub-base all upper and lower sets, $U(alpha) = {beta: beta > alpha}$ and $L(alpha) = {beta: beta < alpha}$.
This implies that in an ordinal $kappa$, all successor ordinals $alpha+1in kappa$ are isolated points, because ${alpha+1}= (alpha, alpha+2)=U(alpha} cap L(alpha+2)$.
At limit ordinals $betain kappa$ we have that a local base is all
sets of the form $(alpha, beta]$ (this is open as $U(alpha)cap L(beta+1)$), so at limit ordinals the topology "looks to the left".
Now suppose that $C$ is order-closed and that $sup(Ccap alpha) = alpha < kappa$ and $alpha$ is a limit. Then if $(gamma, beta]$ is any basic open neighbourhood of $alpha$, with some $gamma < alpha$, we note that $gamma$ is no upperbound for $Ccap alpha$ and so some $gamma'in Ccap alpha$ exists with $gamma' > gamma$. This means that $gamma' in (gamma, alpha]$ so that we know that every basic neighbourhood of $alpha$ intersects $C$, so $alpha in overline{C}=C$, as required.
A similar argument shows that if $C$ obeys the "$sup$-property", it is order closed: let $alpha in overline{C}$. If $alpha$ is isolated it has a neighbourhood ${alpha}$ that must intersect $C$, so $alpha in C$. So now assume $alpha$ is a limit. $alpha in overline{C}$ implies that for every $gamma < alpha$ the basic neighbourhood $(gamma, alpha]$ intersects $C$. This readily implies that $alpha = sup(C cap alpha)$ (as $alpha$ is surely an upperbound for $C cap alpha$ and the previous remarks shows no smaller one can exist. The property then implies $alpha in C$ and so $overline{C} subseteq C$ and $C$ is order-closed.
edited Jan 8 at 12:32
Namaste
1
answered Jan 6 at 23:33
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
$begingroup$
Unfortunately, my edit got rejected, so I am writing this in a comment because it took me some while to realise it. You are using these two statements from general topology: 1. $C$ is closed if and only if $overline{C}subseteq C$, and 2. $alphainoverline{C}$ if and only if every open neighbourhood of $alpha$ intersects $C$
$endgroup$
– Lilalas
Jan 11 at 10:21
$begingroup$
@Lilalas Indeed, that’s what I use. It’s pretty elementary I’d say, topology 101.
$endgroup$
– Henno Brandsma
Jan 11 at 11:46
add a comment |
$begingroup$
The order topology has as its sub-base all upper and lower sets, $U(alpha) = {beta: beta > alpha}$ and $L(alpha) = {beta: beta < alpha}$.
This implies that in an ordinal $kappa$, all successor ordinals $alpha+1in kappa$ are isolated points, because ${alpha+1}= (alpha, alpha+2)=U(alpha} cap L(alpha+2)$.
At limit ordinals $betain kappa$ we have that a local base is all
sets of the form $(alpha, beta]$ (this is open as $U(alpha)cap L(beta+1)$), so at limit ordinals the topology "looks to the left".
Now suppose that $C$ is order-closed and that $sup(Ccap alpha) = alpha < kappa$ and $alpha$ is a limit. Then if $(gamma, beta]$ is any basic open neighbourhood of $alpha$, with some $gamma < alpha$, we note that $gamma$ is no upperbound for $Ccap alpha$ and so some $gamma'in Ccap alpha$ exists with $gamma' > gamma$. This means that $gamma' in (gamma, alpha]$ so that we know that every basic neighbourhood of $alpha$ intersects $C$, so $alpha in overline{C}=C$, as required.
A similar argument shows that if $C$ obeys the "$sup$-property", it is order closed: let $alpha in overline{C}$. If $alpha$ is isolated it has a neighbourhood ${alpha}$ that must intersect $C$, so $alpha in C$. So now assume $alpha$ is a limit. $alpha in overline{C}$ implies that for every $gamma < alpha$ the basic neighbourhood $(gamma, alpha]$ intersects $C$. This readily implies that $alpha = sup(C cap alpha)$ (as $alpha$ is surely an upperbound for $C cap alpha$ and the previous remarks shows no smaller one can exist. The property then implies $alpha in C$ and so $overline{C} subseteq C$ and $C$ is order-closed.
edited Jan 8 at 12:32
Namaste
1
answered Jan 6 at 23:33
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
$begingroup$
Unfortunately, my edit got rejected, so I am writing this in a comment because it took me some while to realise it. You are using these two statements from general topology: 1. $C$ is closed if and only if $overline{C}subseteq C$, and 2. $alphainoverline{C}$ if and only if every open neighbourhood of $alpha$ intersects $C$
$endgroup$
– Lilalas
Jan 11 at 10:21
$begingroup$
@Lilalas Indeed, that’s what I use. It’s pretty elementary I’d say, topology 101.
$endgroup$
– Henno Brandsma
Jan 11 at 11:46
add a comment |
$begingroup$
The order topology has as its sub-base all upper and lower sets, $U(alpha) = {beta: beta > alpha}$ and $L(alpha) = {beta: beta < alpha}$.
This implies that in an ordinal $kappa$, all successor ordinals $alpha+1in kappa$ are isolated points, because ${alpha+1}= (alpha, alpha+2)=U(alpha} cap L(alpha+2)$.
At limit ordinals $betain kappa$ we have that a local base is all
sets of the form $(alpha, beta]$ (this is open as $U(alpha)cap L(beta+1)$), so at limit ordinals the topology "looks to the left".
Now suppose that $C$ is order-closed and that $sup(Ccap alpha) = alpha < kappa$ and $alpha$ is a limit. Then if $(gamma, beta]$ is any basic open neighbourhood of $alpha$, with some $gamma < alpha$, we note that $gamma$ is no upperbound for $Ccap alpha$ and so some $gamma'in Ccap alpha$ exists with $gamma' > gamma$. This means that $gamma' in (gamma, alpha]$ so that we know that every basic neighbourhood of $alpha$ intersects $C$, so $alpha in overline{C}=C$, as required.
A similar argument shows that if $C$ obeys the "$sup$-property", it is order closed: let $alpha in overline{C}$. If $alpha$ is isolated it has a neighbourhood ${alpha}$ that must intersect $C$, so $alpha in C$. So now assume $alpha$ is a limit. $alpha in overline{C}$ implies that for every $gamma < alpha$ the basic neighbourhood $(gamma, alpha]$ intersects $C$. This readily implies that $alpha = sup(C cap alpha)$ (as $alpha$ is surely an upperbound for $C cap alpha$ and the previous remarks shows no smaller one can exist. The property then implies $alpha in C$ and so $overline{C} subseteq C$ and $C$ is order-closed.
edited Jan 8 at 12:32
Namaste
1
answered Jan 6 at 23:33
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
The order topology has as its sub-base all upper and lower sets, $U(alpha) = {beta: beta > alpha}$ and $L(alpha) = {beta: beta < alpha}$.
This implies that in an ordinal $kappa$, all successor ordinals $alpha+1in kappa$ are isolated points, because ${alpha+1}= (alpha, alpha+2)=U(alpha} cap L(alpha+2)$.
At limit ordinals $betain kappa$ we have that a local base is all
sets of the form $(alpha, beta]$ (this is open as $U(alpha)cap L(beta+1)$), so at limit ordinals the topology "looks to the left".
Now suppose that $C$ is order-closed and that $sup(Ccap alpha) = alpha < kappa$ and $alpha$ is a limit. Then if $(gamma, beta]$ is any basic open neighbourhood of $alpha$, with some $gamma < alpha$, we note that $gamma$ is no upperbound for $Ccap alpha$ and so some $gamma'in Ccap alpha$ exists with $gamma' > gamma$. This means that $gamma' in (gamma, alpha]$ so that we know that every basic neighbourhood of $alpha$ intersects $C$, so $alpha in overline{C}=C$, as required.
A similar argument shows that if $C$ obeys the "$sup$-property", it is order closed: let $alpha in overline{C}$. If $alpha$ is isolated it has a neighbourhood ${alpha}$ that must intersect $C$, so $alpha in C$. So now assume $alpha$ is a limit. $alpha in overline{C}$ implies that for every $gamma < alpha$ the basic neighbourhood $(gamma, alpha]$ intersects $C$. This readily implies that $alpha = sup(C cap alpha)$ (as $alpha$ is surely an upperbound for $C cap alpha$ and the previous remarks shows no smaller one can exist. The property then implies $alpha in C$ and so $overline{C} subseteq C$ and $C$ is order-closed.
edited Jan 8 at 12:32
Namaste
1
answered Jan 6 at 23:33
Henno BrandsmaHenno Brandsma
113k348123
edited Jan 8 at 12:32
Namaste
1
edited Jan 8 at 12:32
Namaste
1
edited Jan 8 at 12:32
Namaste
1
1
answered Jan 6 at 23:33
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 6 at 23:33
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 6 at 23:33
Henno BrandsmaHenno Brandsma
113k348123
113k348123
$begingroup$
Unfortunately, my edit got rejected, so I am writing this in a comment because it took me some while to realise it. You are using these two statements from general topology: 1. $C$ is closed if and only if $overline{C}subseteq C$, and 2. $alphainoverline{C}$ if and only if every open neighbourhood of $alpha$ intersects $C$
$endgroup$
– Lilalas
Jan 11 at 10:21
$begingroup$
@Lilalas Indeed, that’s what I use. It’s pretty elementary I’d say, topology 101.
$endgroup$
– Henno Brandsma
Jan 11 at 11:46
add a comment |
$begingroup$
Unfortunately, my edit got rejected, so I am writing this in a comment because it took me some while to realise it. You are using these two statements from general topology: 1. $C$ is closed if and only if $overline{C}subseteq C$, and 2. $alphainoverline{C}$ if and only if every open neighbourhood of $alpha$ intersects $C$
$endgroup$
– Lilalas
Jan 11 at 10:21
$begingroup$
@Lilalas Indeed, that’s what I use. It’s pretty elementary I’d say, topology 101.
$endgroup$
– Henno Brandsma
Jan 11 at 11:46
$begingroup$
Unfortunately, my edit got rejected, so I am writing this in a comment because it took me some while to realise it. You are using these two statements from general topology: 1. $C$ is closed if and only if $overline{C}subseteq C$, and 2. $alphainoverline{C}$ if and only if every open neighbourhood of $alpha$ intersects $C$
$endgroup$
– Lilalas
Jan 11 at 10:21
$begingroup$
Unfortunately, my edit got rejected, so I am writing this in a comment because it took me some while to realise it. You are using these two statements from general topology: 1. $C$ is closed if and only if $overline{C}subseteq C$, and 2. $alphainoverline{C}$ if and only if every open neighbourhood of $alpha$ intersects $C$
$endgroup$
– Lilalas
Jan 11 at 10:21
$begingroup$
@Lilalas Indeed, that’s what I use. It’s pretty elementary I’d say, topology 101.
$endgroup$
– Henno Brandsma
Jan 11 at 11:46
$begingroup$
@Lilalas Indeed, that’s what I use. It’s pretty elementary I’d say, topology 101.
$endgroup$
– Henno Brandsma
Jan 11 at 11:46
add a comment |
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