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Show that $A$ or $B$ must be bounded in $omega_1$
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This is a question which took me longer to figure out than I would have liked. I could not find it on this website so I thought I would share my solution here.
Let $A$ and $B$ be disjoint closed subsets of $omega_1$, the first uncountable ordinal. Show that either $A$ or $B$ must be bounded in $omega_1$.
I have left my solution below.
general-topology ordinals
asked Jul 20 at 0:09
Julian Benali
25712
add a comment |Â
up vote
3
down vote
favorite
This is a question which took me longer to figure out than I would have liked. I could not find it on this website so I thought I would share my solution here.
Let $A$ and $B$ be disjoint closed subsets of $omega_1$, the first uncountable ordinal. Show that either $A$ or $B$ must be bounded in $omega_1$.
I have left my solution below.
general-topology ordinals
asked Jul 20 at 0:09
Julian Benali
25712
1
+1: Interesting! So, put another way, you've proved that $omega_1$ has no pairwise disjoint clubs (CLosed UnBounded Subsets).
– Cameron Buie
Jul 20 at 0:29
1
@CameronBuie it's not called the "club-filter" for nothing...
– Henno Brandsma
Jul 20 at 4:53
1
In fact if $F$ is a non-empty countable family of clubs of $omega_1$ then $cap F$ is club.
– DanielWainfleet
Jul 20 at 9:01
1
@HennoBrandsma: Once I saw this result, I started looking up more information on clubs of $omega_1.$ The club-filter was the first (new(-to-me)) concept I found! I'm a bit disappointed in myself that I didn't immediately recognize that this result meant that the clubs of $omega_1$ are a filter....
– Cameron Buie
Jul 20 at 23:18
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This is a question which took me longer to figure out than I would have liked. I could not find it on this website so I thought I would share my solution here.
Let $A$ and $B$ be disjoint closed subsets of $omega_1$, the first uncountable ordinal. Show that either $A$ or $B$ must be bounded in $omega_1$.
I have left my solution below.
general-topology ordinals
asked Jul 20 at 0:09
Julian Benali
25712
This is a question which took me longer to figure out than I would have liked. I could not find it on this website so I thought I would share my solution here.
Let $A$ and $B$ be disjoint closed subsets of $omega_1$, the first uncountable ordinal. Show that either $A$ or $B$ must be bounded in $omega_1$.
I have left my solution below.
general-topology ordinals
asked Jul 20 at 0:09
Julian Benali
25712
asked Jul 20 at 0:09
Julian Benali
25712
asked Jul 20 at 0:09
Julian Benali
25712
asked Jul 20 at 0:09
Julian Benali
25712
25712
1
+1: Interesting! So, put another way, you've proved that $omega_1$ has no pairwise disjoint clubs (CLosed UnBounded Subsets).
– Cameron Buie
Jul 20 at 0:29
1
@CameronBuie it's not called the "club-filter" for nothing...
– Henno Brandsma
Jul 20 at 4:53
1
In fact if $F$ is a non-empty countable family of clubs of $omega_1$ then $cap F$ is club.
– DanielWainfleet
Jul 20 at 9:01
1
@HennoBrandsma: Once I saw this result, I started looking up more information on clubs of $omega_1.$ The club-filter was the first (new(-to-me)) concept I found! I'm a bit disappointed in myself that I didn't immediately recognize that this result meant that the clubs of $omega_1$ are a filter....
– Cameron Buie
Jul 20 at 23:18
add a comment |Â
1
+1: Interesting! So, put another way, you've proved that $omega_1$ has no pairwise disjoint clubs (CLosed UnBounded Subsets).
– Cameron Buie
Jul 20 at 0:29
1
@CameronBuie it's not called the "club-filter" for nothing...
– Henno Brandsma
Jul 20 at 4:53
1
In fact if $F$ is a non-empty countable family of clubs of $omega_1$ then $cap F$ is club.
– DanielWainfleet
Jul 20 at 9:01
1
@HennoBrandsma: Once I saw this result, I started looking up more information on clubs of $omega_1.$ The club-filter was the first (new(-to-me)) concept I found! I'm a bit disappointed in myself that I didn't immediately recognize that this result meant that the clubs of $omega_1$ are a filter....
– Cameron Buie
Jul 20 at 23:18
1
1
+1: Interesting! So, put another way, you've proved that $omega_1$ has no pairwise disjoint clubs (CLosed UnBounded Subsets).
– Cameron Buie
Jul 20 at 0:29
+1: Interesting! So, put another way, you've proved that $omega_1$ has no pairwise disjoint clubs (CLosed UnBounded Subsets).
– Cameron Buie
Jul 20 at 0:29
1
1
@CameronBuie it's not called the "club-filter" for nothing...
– Henno Brandsma
Jul 20 at 4:53
@CameronBuie it's not called the "club-filter" for nothing...
– Henno Brandsma
Jul 20 at 4:53
1
1
In fact if $F$ is a non-empty countable family of clubs of $omega_1$ then $cap F$ is club.
– DanielWainfleet
Jul 20 at 9:01
In fact if $F$ is a non-empty countable family of clubs of $omega_1$ then $cap F$ is club.
– DanielWainfleet
Jul 20 at 9:01
1
1
@HennoBrandsma: Once I saw this result, I started looking up more information on clubs of $omega_1.$ The club-filter was the first (new(-to-me)) concept I found! I'm a bit disappointed in myself that I didn't immediately recognize that this result meant that the clubs of $omega_1$ are a filter....
– Cameron Buie
Jul 20 at 23:18
@HennoBrandsma: Once I saw this result, I started looking up more information on clubs of $omega_1.$ The club-filter was the first (new(-to-me)) concept I found! I'm a bit disappointed in myself that I didn't immediately recognize that this result meant that the clubs of $omega_1$ are a filter....
– Cameron Buie
Jul 20 at 23:18
add a comment |Â
1 Answer
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3
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Assume that both $A$ and $B$ are unbounded. Because $A$ and $B$ are unbounded, we may recursively define a strictly increasing sequence such that every other element is in $A$ or $B$ respectively. Choose some element $a_0in A$. If $iinmathbbZ^+$ is odd, let $a_iin B$ such that $a_i-1<a_i$ and if $i$ is even, let $a_iin A$ such that $a_i-1<a_i$. Now, let $a=cup_iinmathbbNa_i=supa_i:iinmathbbN$. $a$ is a countable union of countable ordinals and is therefore a countable ordinal (and thus an element of $omega_1$).
Let $(x,y)$ be an open interval in $omega_1$ containing $a$ so that $x<a<y$. Then $xin a$ which means there must exist $iinmathbbN$ such that $xin a_i$ so that $x<a_i$. Clearly $a_i+1<y$, for if it were not, we would have $a_i+1<a<yleq a_i+1$. Thus, $x<a_i<a_i+1<y$, so $a_i$ and $a_i+1$ are both in $(x,y)$, which means every open interval containing $a$ must contain an element of $A$ and an element of $B$.
Since $a$ is clearly not $0$, every open set containing $a$ contains an open interval and therefore intersects both $A$ and $B$. Hence, $a$ is a limit point of both $A$ and $B$. Since $A$ and $B$ are closed, this implies $a$ is an element of both $A$ and $B$ which contradicts the fact that $A$ and $B$ are disjoint. So, by way of contradiction, either $A$ or $B$ must be bounded.
answered Jul 20 at 0:09
Julian Benali
25712
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Assume that both $A$ and $B$ are unbounded. Because $A$ and $B$ are unbounded, we may recursively define a strictly increasing sequence such that every other element is in $A$ or $B$ respectively. Choose some element $a_0in A$. If $iinmathbbZ^+$ is odd, let $a_iin B$ such that $a_i-1<a_i$ and if $i$ is even, let $a_iin A$ such that $a_i-1<a_i$. Now, let $a=cup_iinmathbbNa_i=supa_i:iinmathbbN$. $a$ is a countable union of countable ordinals and is therefore a countable ordinal (and thus an element of $omega_1$).
Let $(x,y)$ be an open interval in $omega_1$ containing $a$ so that $x<a<y$. Then $xin a$ which means there must exist $iinmathbbN$ such that $xin a_i$ so that $x<a_i$. Clearly $a_i+1<y$, for if it were not, we would have $a_i+1<a<yleq a_i+1$. Thus, $x<a_i<a_i+1<y$, so $a_i$ and $a_i+1$ are both in $(x,y)$, which means every open interval containing $a$ must contain an element of $A$ and an element of $B$.
Since $a$ is clearly not $0$, every open set containing $a$ contains an open interval and therefore intersects both $A$ and $B$. Hence, $a$ is a limit point of both $A$ and $B$. Since $A$ and $B$ are closed, this implies $a$ is an element of both $A$ and $B$ which contradicts the fact that $A$ and $B$ are disjoint. So, by way of contradiction, either $A$ or $B$ must be bounded.
answered Jul 20 at 0:09
Julian Benali
25712
add a comment |Â
up vote
3
down vote
accepted
Assume that both $A$ and $B$ are unbounded. Because $A$ and $B$ are unbounded, we may recursively define a strictly increasing sequence such that every other element is in $A$ or $B$ respectively. Choose some element $a_0in A$. If $iinmathbbZ^+$ is odd, let $a_iin B$ such that $a_i-1<a_i$ and if $i$ is even, let $a_iin A$ such that $a_i-1<a_i$. Now, let $a=cup_iinmathbbNa_i=supa_i:iinmathbbN$. $a$ is a countable union of countable ordinals and is therefore a countable ordinal (and thus an element of $omega_1$).
Let $(x,y)$ be an open interval in $omega_1$ containing $a$ so that $x<a<y$. Then $xin a$ which means there must exist $iinmathbbN$ such that $xin a_i$ so that $x<a_i$. Clearly $a_i+1<y$, for if it were not, we would have $a_i+1<a<yleq a_i+1$. Thus, $x<a_i<a_i+1<y$, so $a_i$ and $a_i+1$ are both in $(x,y)$, which means every open interval containing $a$ must contain an element of $A$ and an element of $B$.
Since $a$ is clearly not $0$, every open set containing $a$ contains an open interval and therefore intersects both $A$ and $B$. Hence, $a$ is a limit point of both $A$ and $B$. Since $A$ and $B$ are closed, this implies $a$ is an element of both $A$ and $B$ which contradicts the fact that $A$ and $B$ are disjoint. So, by way of contradiction, either $A$ or $B$ must be bounded.
answered Jul 20 at 0:09
Julian Benali
25712
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Assume that both $A$ and $B$ are unbounded. Because $A$ and $B$ are unbounded, we may recursively define a strictly increasing sequence such that every other element is in $A$ or $B$ respectively. Choose some element $a_0in A$. If $iinmathbbZ^+$ is odd, let $a_iin B$ such that $a_i-1<a_i$ and if $i$ is even, let $a_iin A$ such that $a_i-1<a_i$. Now, let $a=cup_iinmathbbNa_i=supa_i:iinmathbbN$. $a$ is a countable union of countable ordinals and is therefore a countable ordinal (and thus an element of $omega_1$).
Let $(x,y)$ be an open interval in $omega_1$ containing $a$ so that $x<a<y$. Then $xin a$ which means there must exist $iinmathbbN$ such that $xin a_i$ so that $x<a_i$. Clearly $a_i+1<y$, for if it were not, we would have $a_i+1<a<yleq a_i+1$. Thus, $x<a_i<a_i+1<y$, so $a_i$ and $a_i+1$ are both in $(x,y)$, which means every open interval containing $a$ must contain an element of $A$ and an element of $B$.
Since $a$ is clearly not $0$, every open set containing $a$ contains an open interval and therefore intersects both $A$ and $B$. Hence, $a$ is a limit point of both $A$ and $B$. Since $A$ and $B$ are closed, this implies $a$ is an element of both $A$ and $B$ which contradicts the fact that $A$ and $B$ are disjoint. So, by way of contradiction, either $A$ or $B$ must be bounded.
answered Jul 20 at 0:09
Julian Benali
25712
Assume that both $A$ and $B$ are unbounded. Because $A$ and $B$ are unbounded, we may recursively define a strictly increasing sequence such that every other element is in $A$ or $B$ respectively. Choose some element $a_0in A$. If $iinmathbbZ^+$ is odd, let $a_iin B$ such that $a_i-1<a_i$ and if $i$ is even, let $a_iin A$ such that $a_i-1<a_i$. Now, let $a=cup_iinmathbbNa_i=supa_i:iinmathbbN$. $a$ is a countable union of countable ordinals and is therefore a countable ordinal (and thus an element of $omega_1$).
Let $(x,y)$ be an open interval in $omega_1$ containing $a$ so that $x<a<y$. Then $xin a$ which means there must exist $iinmathbbN$ such that $xin a_i$ so that $x<a_i$. Clearly $a_i+1<y$, for if it were not, we would have $a_i+1<a<yleq a_i+1$. Thus, $x<a_i<a_i+1<y$, so $a_i$ and $a_i+1$ are both in $(x,y)$, which means every open interval containing $a$ must contain an element of $A$ and an element of $B$.
Since $a$ is clearly not $0$, every open set containing $a$ contains an open interval and therefore intersects both $A$ and $B$. Hence, $a$ is a limit point of both $A$ and $B$. Since $A$ and $B$ are closed, this implies $a$ is an element of both $A$ and $B$ which contradicts the fact that $A$ and $B$ are disjoint. So, by way of contradiction, either $A$ or $B$ must be bounded.
answered Jul 20 at 0:09
Julian Benali
25712
answered Jul 20 at 0:09
Julian Benali
25712
answered Jul 20 at 0:09
Julian Benali
25712
answered Jul 20 at 0:09
Julian Benali
25712
25712
add a comment |Â
add a comment |Â
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1
+1: Interesting! So, put another way, you've proved that $omega_1$ has no pairwise disjoint clubs (CLosed UnBounded Subsets).
– Cameron Buie
Jul 20 at 0:29
1
@CameronBuie it's not called the "club-filter" for nothing...
– Henno Brandsma
Jul 20 at 4:53
1
In fact if $F$ is a non-empty countable family of clubs of $omega_1$ then $cap F$ is club.
– DanielWainfleet
Jul 20 at 9:01
1
@HennoBrandsma: Once I saw this result, I started looking up more information on clubs of $omega_1.$ The club-filter was the first (new(-to-me)) concept I found! I'm a bit disappointed in myself that I didn't immediately recognize that this result meant that the clubs of $omega_1$ are a filter....
– Cameron Buie
Jul 20 at 23:18