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$X$ is c-closed iff every countably compact subset of $X$ is closed
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I need help to understand definition of c-closed. In here https://www.sciencedirect.com/science/article/pii/0166864180900279, author said that $X$ is c-closed iff every countably compact subset of $X$ is closed. Equivalently, every non-closed subset $A$ of $X$ contains a sequence which has no a cluster point in $A$.
I can not understand the equivalence. Is every non-closed containing a sequence which has no a cluster point in $X$?
Then every space is c-closed?
I know it's wrong. For example, the ordinal space $omega_1+1$ with the order topology is not c-space since $omega_1$ is countably compact but isn't closed. How to understand this space isn't c-closed using second definition?
general-topology compactness
asked Jul 14 at 22:27
flourence
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up vote
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favorite
I need help to understand definition of c-closed. In here https://www.sciencedirect.com/science/article/pii/0166864180900279, author said that $X$ is c-closed iff every countably compact subset of $X$ is closed. Equivalently, every non-closed subset $A$ of $X$ contains a sequence which has no a cluster point in $A$.
I can not understand the equivalence. Is every non-closed containing a sequence which has no a cluster point in $X$?
Then every space is c-closed?
I know it's wrong. For example, the ordinal space $omega_1+1$ with the order topology is not c-space since $omega_1$ is countably compact but isn't closed. How to understand this space isn't c-closed using second definition?
general-topology compactness
asked Jul 14 at 22:27
flourence
1548
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I need help to understand definition of c-closed. In here https://www.sciencedirect.com/science/article/pii/0166864180900279, author said that $X$ is c-closed iff every countably compact subset of $X$ is closed. Equivalently, every non-closed subset $A$ of $X$ contains a sequence which has no a cluster point in $A$.
I can not understand the equivalence. Is every non-closed containing a sequence which has no a cluster point in $X$?
Then every space is c-closed?
I know it's wrong. For example, the ordinal space $omega_1+1$ with the order topology is not c-space since $omega_1$ is countably compact but isn't closed. How to understand this space isn't c-closed using second definition?
general-topology compactness
asked Jul 14 at 22:27
flourence
1548
I need help to understand definition of c-closed. In here https://www.sciencedirect.com/science/article/pii/0166864180900279, author said that $X$ is c-closed iff every countably compact subset of $X$ is closed. Equivalently, every non-closed subset $A$ of $X$ contains a sequence which has no a cluster point in $A$.
I can not understand the equivalence. Is every non-closed containing a sequence which has no a cluster point in $X$?
Then every space is c-closed?
I know it's wrong. For example, the ordinal space $omega_1+1$ with the order topology is not c-space since $omega_1$ is countably compact but isn't closed. How to understand this space isn't c-closed using second definition?
general-topology compactness
asked Jul 14 at 22:27
flourence
1548
asked Jul 14 at 22:27
flourence
1548
asked Jul 14 at 22:27
flourence
1548
asked Jul 14 at 22:27
flourence
1548
1548
add a comment |Â
add a comment |Â
3 Answers
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The cluster point of the non-closed set $A$ is considered to be in $A$, as you first wrote it.
The second definition is just the contrapositive of the first:
If every sequence of $A$ has a cluster point in $A$, then it is countably compact, hence it must be closed if $X$ is c-closed.
For the specific example, we would need to find a sequence in $omega_1$ that tends to $omega_1$ (the only missing point), which doesn't exist, as $omega_1$ is regular.
answered Jul 15 at 1:45
Berci
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A set $A$ is countably compact iff every countable subset (or sequence of distinct points, if you prefer) of $A$ has a cluster point in $A$.
The alternative formulation is
$(ast$) For all $A subseteq X$: if $A$ is not closed there exist $a_n in A$ such that $A$ contains no cluster point for $(a_n)$.
The proposed equivalence is simply a matter of contrapositive reasoning:
Suppose $X$ is $C$-closed. Let $A$ be non-closed. This means that $A$ is not countably compact. So there is a sequence $x_n$ from $A$ without a cluster point in $A$. Hence $X$ obeys ($ast$). For the reverse, suppose $X$ obeys the property $(ast)$. Then $X$ is $C$-closed: let $A$ be countably compact. Then $A$ is closed. For suppose it were not closed, then by the assumption $(ast)$ on $X$ we'd have a sequence $a_n$ from $A$ without a cluster point in $A$. This contradicts $A$ being countably compact. So $A$ is closed and $X$ is $C$-closed.
The example $X= omega_1+1$ is no contradiction. It is indeed not $C$-closed. The negation of $C$-closed in the alternative formulation ($ast$) is:
There exists a subset $A$ of $X$ that is not closed, but such that all sequences from $A$ have a cluster point in $A$.
And indeed $A = omega_1$ is as required. So $X$ is also not $C$-compact in the other definition ($ast$).
I think it's mostly a matter of taking the negation of the alternative definition in the right way.
answered Jul 15 at 8:36
Henno Brandsma
91.7k342100
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Terminology: "Open" and "closed" mean "open in $X$" and "closed in $X$".
For one direction: Let $X$ be $c$-closed and let $Asubset X$ be non-closed. Then $A$ is not countably compact, so let $C=C_k: kin Bbb N$ be an open cover of $A$ with no finite sub-cover.
For $kin Bbb N$ let $D_k=cup_jleq kC_j$ and $E=D_k: neg [exists k'<k;(Acap D_k'=Acap D_k)];.$
We can enumerate $E$ as $E=E_n:nin Bbb N$ where $Acap E_nsubsetneqq Acap E_n+1$ for each $nin Bbb N.$ For each $nin Bbb N$ choose $a_nin Acap (E_n+1backslash E_n).$
Any $ain A$ belongs to $E_F(a)$ for some $F(a)in Bbb N$ but the set $a_n:n>F(a)$ is disjoint from $Acap E_F(a).$ So $F(a)$ is a nbhd of $a$ such that $n: a_nin E_F(a)$ is finite. So $a$ is not a cluster point of the sequence $(a_n)_nin Bbb N.$
Addendum: Re the space $omega_1+1$ with the $epsilon$-order topology: If $(a_n)_nin Bbb N$ is any sequence of members of $A=omega_1$ then $a_n:nin Bbb N subset B=(cup a_n:nin Bbb N)+1in omega_1.$ The sub$space $ $ B $ is compact & Hausdorff so $(a_n)_nin Bbb N$ has a cluster point $ain Bsubset A.$
answered Jul 16 at 0:08
DanielWainfleet
31.7k31644
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The cluster point of the non-closed set $A$ is considered to be in $A$, as you first wrote it.
The second definition is just the contrapositive of the first:
If every sequence of $A$ has a cluster point in $A$, then it is countably compact, hence it must be closed if $X$ is c-closed.
For the specific example, we would need to find a sequence in $omega_1$ that tends to $omega_1$ (the only missing point), which doesn't exist, as $omega_1$ is regular.
answered Jul 15 at 1:45
Berci
56.4k23570
add a comment |Â
up vote
0
down vote
The cluster point of the non-closed set $A$ is considered to be in $A$, as you first wrote it.
The second definition is just the contrapositive of the first:
If every sequence of $A$ has a cluster point in $A$, then it is countably compact, hence it must be closed if $X$ is c-closed.
For the specific example, we would need to find a sequence in $omega_1$ that tends to $omega_1$ (the only missing point), which doesn't exist, as $omega_1$ is regular.
answered Jul 15 at 1:45
Berci
56.4k23570
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The cluster point of the non-closed set $A$ is considered to be in $A$, as you first wrote it.
The second definition is just the contrapositive of the first:
If every sequence of $A$ has a cluster point in $A$, then it is countably compact, hence it must be closed if $X$ is c-closed.
For the specific example, we would need to find a sequence in $omega_1$ that tends to $omega_1$ (the only missing point), which doesn't exist, as $omega_1$ is regular.
answered Jul 15 at 1:45
Berci
56.4k23570
The cluster point of the non-closed set $A$ is considered to be in $A$, as you first wrote it.
The second definition is just the contrapositive of the first:
If every sequence of $A$ has a cluster point in $A$, then it is countably compact, hence it must be closed if $X$ is c-closed.
For the specific example, we would need to find a sequence in $omega_1$ that tends to $omega_1$ (the only missing point), which doesn't exist, as $omega_1$ is regular.
answered Jul 15 at 1:45
Berci
56.4k23570
edited Jul 15 at 1:59
answered Jul 15 at 1:45
Berci
56.4k23570
answered Jul 15 at 1:45
Berci
56.4k23570
answered Jul 15 at 1:45
Berci
56.4k23570
56.4k23570
add a comment |Â
add a comment |Â
up vote
0
down vote
A set $A$ is countably compact iff every countable subset (or sequence of distinct points, if you prefer) of $A$ has a cluster point in $A$.
The alternative formulation is
$(ast$) For all $A subseteq X$: if $A$ is not closed there exist $a_n in A$ such that $A$ contains no cluster point for $(a_n)$.
The proposed equivalence is simply a matter of contrapositive reasoning:
Suppose $X$ is $C$-closed. Let $A$ be non-closed. This means that $A$ is not countably compact. So there is a sequence $x_n$ from $A$ without a cluster point in $A$. Hence $X$ obeys ($ast$). For the reverse, suppose $X$ obeys the property $(ast)$. Then $X$ is $C$-closed: let $A$ be countably compact. Then $A$ is closed. For suppose it were not closed, then by the assumption $(ast)$ on $X$ we'd have a sequence $a_n$ from $A$ without a cluster point in $A$. This contradicts $A$ being countably compact. So $A$ is closed and $X$ is $C$-closed.
The example $X= omega_1+1$ is no contradiction. It is indeed not $C$-closed. The negation of $C$-closed in the alternative formulation ($ast$) is:
There exists a subset $A$ of $X$ that is not closed, but such that all sequences from $A$ have a cluster point in $A$.
And indeed $A = omega_1$ is as required. So $X$ is also not $C$-compact in the other definition ($ast$).
I think it's mostly a matter of taking the negation of the alternative definition in the right way.
answered Jul 15 at 8:36
Henno Brandsma
91.7k342100
add a comment |Â
up vote
0
down vote
A set $A$ is countably compact iff every countable subset (or sequence of distinct points, if you prefer) of $A$ has a cluster point in $A$.
The alternative formulation is
$(ast$) For all $A subseteq X$: if $A$ is not closed there exist $a_n in A$ such that $A$ contains no cluster point for $(a_n)$.
The proposed equivalence is simply a matter of contrapositive reasoning:
Suppose $X$ is $C$-closed. Let $A$ be non-closed. This means that $A$ is not countably compact. So there is a sequence $x_n$ from $A$ without a cluster point in $A$. Hence $X$ obeys ($ast$). For the reverse, suppose $X$ obeys the property $(ast)$. Then $X$ is $C$-closed: let $A$ be countably compact. Then $A$ is closed. For suppose it were not closed, then by the assumption $(ast)$ on $X$ we'd have a sequence $a_n$ from $A$ without a cluster point in $A$. This contradicts $A$ being countably compact. So $A$ is closed and $X$ is $C$-closed.
The example $X= omega_1+1$ is no contradiction. It is indeed not $C$-closed. The negation of $C$-closed in the alternative formulation ($ast$) is:
There exists a subset $A$ of $X$ that is not closed, but such that all sequences from $A$ have a cluster point in $A$.
And indeed $A = omega_1$ is as required. So $X$ is also not $C$-compact in the other definition ($ast$).
I think it's mostly a matter of taking the negation of the alternative definition in the right way.
answered Jul 15 at 8:36
Henno Brandsma
91.7k342100
add a comment |Â
up vote
0
down vote
up vote
0
down vote
A set $A$ is countably compact iff every countable subset (or sequence of distinct points, if you prefer) of $A$ has a cluster point in $A$.
The alternative formulation is
$(ast$) For all $A subseteq X$: if $A$ is not closed there exist $a_n in A$ such that $A$ contains no cluster point for $(a_n)$.
The proposed equivalence is simply a matter of contrapositive reasoning:
Suppose $X$ is $C$-closed. Let $A$ be non-closed. This means that $A$ is not countably compact. So there is a sequence $x_n$ from $A$ without a cluster point in $A$. Hence $X$ obeys ($ast$). For the reverse, suppose $X$ obeys the property $(ast)$. Then $X$ is $C$-closed: let $A$ be countably compact. Then $A$ is closed. For suppose it were not closed, then by the assumption $(ast)$ on $X$ we'd have a sequence $a_n$ from $A$ without a cluster point in $A$. This contradicts $A$ being countably compact. So $A$ is closed and $X$ is $C$-closed.
The example $X= omega_1+1$ is no contradiction. It is indeed not $C$-closed. The negation of $C$-closed in the alternative formulation ($ast$) is:
There exists a subset $A$ of $X$ that is not closed, but such that all sequences from $A$ have a cluster point in $A$.
And indeed $A = omega_1$ is as required. So $X$ is also not $C$-compact in the other definition ($ast$).
I think it's mostly a matter of taking the negation of the alternative definition in the right way.
answered Jul 15 at 8:36
Henno Brandsma
91.7k342100
A set $A$ is countably compact iff every countable subset (or sequence of distinct points, if you prefer) of $A$ has a cluster point in $A$.
The alternative formulation is
$(ast$) For all $A subseteq X$: if $A$ is not closed there exist $a_n in A$ such that $A$ contains no cluster point for $(a_n)$.
The proposed equivalence is simply a matter of contrapositive reasoning:
Suppose $X$ is $C$-closed. Let $A$ be non-closed. This means that $A$ is not countably compact. So there is a sequence $x_n$ from $A$ without a cluster point in $A$. Hence $X$ obeys ($ast$). For the reverse, suppose $X$ obeys the property $(ast)$. Then $X$ is $C$-closed: let $A$ be countably compact. Then $A$ is closed. For suppose it were not closed, then by the assumption $(ast)$ on $X$ we'd have a sequence $a_n$ from $A$ without a cluster point in $A$. This contradicts $A$ being countably compact. So $A$ is closed and $X$ is $C$-closed.
The example $X= omega_1+1$ is no contradiction. It is indeed not $C$-closed. The negation of $C$-closed in the alternative formulation ($ast$) is:
There exists a subset $A$ of $X$ that is not closed, but such that all sequences from $A$ have a cluster point in $A$.
And indeed $A = omega_1$ is as required. So $X$ is also not $C$-compact in the other definition ($ast$).
I think it's mostly a matter of taking the negation of the alternative definition in the right way.
answered Jul 15 at 8:36
Henno Brandsma
91.7k342100
edited Jul 15 at 15:50
answered Jul 15 at 8:36
Henno Brandsma
91.7k342100
answered Jul 15 at 8:36
Henno Brandsma
91.7k342100
answered Jul 15 at 8:36
Henno Brandsma
91.7k342100
91.7k342100
add a comment |Â
add a comment |Â
up vote
0
down vote
Terminology: "Open" and "closed" mean "open in $X$" and "closed in $X$".
For one direction: Let $X$ be $c$-closed and let $Asubset X$ be non-closed. Then $A$ is not countably compact, so let $C=C_k: kin Bbb N$ be an open cover of $A$ with no finite sub-cover.
For $kin Bbb N$ let $D_k=cup_jleq kC_j$ and $E=D_k: neg [exists k'<k;(Acap D_k'=Acap D_k)];.$
We can enumerate $E$ as $E=E_n:nin Bbb N$ where $Acap E_nsubsetneqq Acap E_n+1$ for each $nin Bbb N.$ For each $nin Bbb N$ choose $a_nin Acap (E_n+1backslash E_n).$
Any $ain A$ belongs to $E_F(a)$ for some $F(a)in Bbb N$ but the set $a_n:n>F(a)$ is disjoint from $Acap E_F(a).$ So $F(a)$ is a nbhd of $a$ such that $n: a_nin E_F(a)$ is finite. So $a$ is not a cluster point of the sequence $(a_n)_nin Bbb N.$
Addendum: Re the space $omega_1+1$ with the $epsilon$-order topology: If $(a_n)_nin Bbb N$ is any sequence of members of $A=omega_1$ then $a_n:nin Bbb N subset B=(cup a_n:nin Bbb N)+1in omega_1.$ The sub$space $ $ B $ is compact & Hausdorff so $(a_n)_nin Bbb N$ has a cluster point $ain Bsubset A.$
answered Jul 16 at 0:08
DanielWainfleet
31.7k31644
add a comment |Â
up vote
0
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Terminology: "Open" and "closed" mean "open in $X$" and "closed in $X$".
For one direction: Let $X$ be $c$-closed and let $Asubset X$ be non-closed. Then $A$ is not countably compact, so let $C=C_k: kin Bbb N$ be an open cover of $A$ with no finite sub-cover.
For $kin Bbb N$ let $D_k=cup_jleq kC_j$ and $E=D_k: neg [exists k'<k;(Acap D_k'=Acap D_k)];.$
We can enumerate $E$ as $E=E_n:nin Bbb N$ where $Acap E_nsubsetneqq Acap E_n+1$ for each $nin Bbb N.$ For each $nin Bbb N$ choose $a_nin Acap (E_n+1backslash E_n).$
Any $ain A$ belongs to $E_F(a)$ for some $F(a)in Bbb N$ but the set $a_n:n>F(a)$ is disjoint from $Acap E_F(a).$ So $F(a)$ is a nbhd of $a$ such that $n: a_nin E_F(a)$ is finite. So $a$ is not a cluster point of the sequence $(a_n)_nin Bbb N.$
Addendum: Re the space $omega_1+1$ with the $epsilon$-order topology: If $(a_n)_nin Bbb N$ is any sequence of members of $A=omega_1$ then $a_n:nin Bbb N subset B=(cup a_n:nin Bbb N)+1in omega_1.$ The sub$space $ $ B $ is compact & Hausdorff so $(a_n)_nin Bbb N$ has a cluster point $ain Bsubset A.$
answered Jul 16 at 0:08
DanielWainfleet
31.7k31644
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Terminology: "Open" and "closed" mean "open in $X$" and "closed in $X$".
For one direction: Let $X$ be $c$-closed and let $Asubset X$ be non-closed. Then $A$ is not countably compact, so let $C=C_k: kin Bbb N$ be an open cover of $A$ with no finite sub-cover.
For $kin Bbb N$ let $D_k=cup_jleq kC_j$ and $E=D_k: neg [exists k'<k;(Acap D_k'=Acap D_k)];.$
We can enumerate $E$ as $E=E_n:nin Bbb N$ where $Acap E_nsubsetneqq Acap E_n+1$ for each $nin Bbb N.$ For each $nin Bbb N$ choose $a_nin Acap (E_n+1backslash E_n).$
Any $ain A$ belongs to $E_F(a)$ for some $F(a)in Bbb N$ but the set $a_n:n>F(a)$ is disjoint from $Acap E_F(a).$ So $F(a)$ is a nbhd of $a$ such that $n: a_nin E_F(a)$ is finite. So $a$ is not a cluster point of the sequence $(a_n)_nin Bbb N.$
Addendum: Re the space $omega_1+1$ with the $epsilon$-order topology: If $(a_n)_nin Bbb N$ is any sequence of members of $A=omega_1$ then $a_n:nin Bbb N subset B=(cup a_n:nin Bbb N)+1in omega_1.$ The sub$space $ $ B $ is compact & Hausdorff so $(a_n)_nin Bbb N$ has a cluster point $ain Bsubset A.$
answered Jul 16 at 0:08
DanielWainfleet
31.7k31644
Terminology: "Open" and "closed" mean "open in $X$" and "closed in $X$".
For one direction: Let $X$ be $c$-closed and let $Asubset X$ be non-closed. Then $A$ is not countably compact, so let $C=C_k: kin Bbb N$ be an open cover of $A$ with no finite sub-cover.
For $kin Bbb N$ let $D_k=cup_jleq kC_j$ and $E=D_k: neg [exists k'<k;(Acap D_k'=Acap D_k)];.$
We can enumerate $E$ as $E=E_n:nin Bbb N$ where $Acap E_nsubsetneqq Acap E_n+1$ for each $nin Bbb N.$ For each $nin Bbb N$ choose $a_nin Acap (E_n+1backslash E_n).$
Any $ain A$ belongs to $E_F(a)$ for some $F(a)in Bbb N$ but the set $a_n:n>F(a)$ is disjoint from $Acap E_F(a).$ So $F(a)$ is a nbhd of $a$ such that $n: a_nin E_F(a)$ is finite. So $a$ is not a cluster point of the sequence $(a_n)_nin Bbb N.$
Addendum: Re the space $omega_1+1$ with the $epsilon$-order topology: If $(a_n)_nin Bbb N$ is any sequence of members of $A=omega_1$ then $a_n:nin Bbb N subset B=(cup a_n:nin Bbb N)+1in omega_1.$ The sub$space $ $ B $ is compact & Hausdorff so $(a_n)_nin Bbb N$ has a cluster point $ain Bsubset A.$
answered Jul 16 at 0:08
DanielWainfleet
31.7k31644
edited Jul 16 at 0:27
answered Jul 16 at 0:08
DanielWainfleet
31.7k31644
answered Jul 16 at 0:08
DanielWainfleet
31.7k31644
answered Jul 16 at 0:08
DanielWainfleet
31.7k31644
31.7k31644
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