Mixing
Divergence on definitions of limit point compactness
$begingroup$
A topological space $Omega$ is said to be weakly sequentially compact if every sequence in $Omega$ has a limit point.
A topological space $Omega$ is said to be limit point compact if every infinite subset of $Omega$ has a limit point.
Are these two definitions equivalent? I'm doing a summer course on general topology and my professor just started talking about compactness, and the first definition is the one he uses, the second is from Munkres' textbook. My professor said they were the same thing but I'm not completely convinced yet and it's much more pleasant to work with the second one. It's clear the first definition implies the second one, but I'm not sure about the converse.
I think there's a problem in saying the definitions are equivalent because the second one deals with infinite subsets and the first only cares about sequences, so I could have a sequence that's eventually constant or only has a finite number of terms and it would have a limit point, which I think isn't necessarily true if all we're given is the second definition.
Am I right or are they really equivalent?
general-topology
asked Feb 1 at 20:25
Matheus AndradeMatheus Andrade
1,390418
$endgroup$
|
show 4 more comments
$begingroup$
A topological space $Omega$ is said to be weakly sequentially compact if every sequence in $Omega$ has a limit point.
A topological space $Omega$ is said to be limit point compact if every infinite subset of $Omega$ has a limit point.
Are these two definitions equivalent? I'm doing a summer course on general topology and my professor just started talking about compactness, and the first definition is the one he uses, the second is from Munkres' textbook. My professor said they were the same thing but I'm not completely convinced yet and it's much more pleasant to work with the second one. It's clear the first definition implies the second one, but I'm not sure about the converse.
I think there's a problem in saying the definitions are equivalent because the second one deals with infinite subsets and the first only cares about sequences, so I could have a sequence that's eventually constant or only has a finite number of terms and it would have a limit point, which I think isn't necessarily true if all we're given is the second definition.
Am I right or are they really equivalent?
general-topology
asked Feb 1 at 20:25
Matheus AndradeMatheus Andrade
1,390418
$endgroup$
$begingroup$
By “every sequence has a limit point” do you mean a convergent subsequence (with limit in the space)? If so then yes—they are equivalent. Suppose limit point compactness is taken first. A sequence either has a trivial convergent subsequence if its range is finite or, if the range is infinite, then you can use limit point compactness to get a limit point of the range which a subsequence then converges to.
$endgroup$
– LoveTooNap29
Feb 1 at 20:38
1
$begingroup$
ok well they are still equivalent. You can always find a limit point of the range of the sequence. If the range is finite, it’s trivial, if the range is infinite, appeal to limit point compactness. Where’s the confusion?
$endgroup$
– LoveTooNap29
Feb 1 at 20:45
1
$begingroup$
Also don’t you mean the limit point belongs in $Omega$ not $S$? Take $Omega=[0,1]$ and $x_n=1/n$. Then $0$ is a limit point of $S={x_n: ninmathbb{N}}$ but not a member of $S$ while $0in Omega$.
$endgroup$
– LoveTooNap29
Feb 1 at 20:50
1
$begingroup$
@LoveTooNap29 A limit point of a sequence $(x_n)$ in $X$ is a point $p in X$ such that for every neighbourhood $O$ of $p$ and every $n_0 in mathbb{N}$ there is some $n_1 > n_0$ such that $x_{n_1} in O$. It does not imply a convergent subsequence (it does in a metric or first countable space, but not in general). A limit point of that sequence need not be a limit point of the set ${x_n: n in mathbb{N}}$, confusingly.
$endgroup$
– Henno Brandsma
Feb 1 at 22:34
1
$begingroup$
@LoveTooNap29 I added a counterexample.
$endgroup$
– Henno Brandsma
Feb 1 at 22:54
|
show 4 more comments
$begingroup$
A topological space $Omega$ is said to be weakly sequentially compact if every sequence in $Omega$ has a limit point.
A topological space $Omega$ is said to be limit point compact if every infinite subset of $Omega$ has a limit point.
Are these two definitions equivalent? I'm doing a summer course on general topology and my professor just started talking about compactness, and the first definition is the one he uses, the second is from Munkres' textbook. My professor said they were the same thing but I'm not completely convinced yet and it's much more pleasant to work with the second one. It's clear the first definition implies the second one, but I'm not sure about the converse.
I think there's a problem in saying the definitions are equivalent because the second one deals with infinite subsets and the first only cares about sequences, so I could have a sequence that's eventually constant or only has a finite number of terms and it would have a limit point, which I think isn't necessarily true if all we're given is the second definition.
Am I right or are they really equivalent?
general-topology
asked Feb 1 at 20:25
Matheus AndradeMatheus Andrade
1,390418
$endgroup$
A topological space $Omega$ is said to be weakly sequentially compact if every sequence in $Omega$ has a limit point.
A topological space $Omega$ is said to be limit point compact if every infinite subset of $Omega$ has a limit point.
Are these two definitions equivalent? I'm doing a summer course on general topology and my professor just started talking about compactness, and the first definition is the one he uses, the second is from Munkres' textbook. My professor said they were the same thing but I'm not completely convinced yet and it's much more pleasant to work with the second one. It's clear the first definition implies the second one, but I'm not sure about the converse.
I think there's a problem in saying the definitions are equivalent because the second one deals with infinite subsets and the first only cares about sequences, so I could have a sequence that's eventually constant or only has a finite number of terms and it would have a limit point, which I think isn't necessarily true if all we're given is the second definition.
Am I right or are they really equivalent?
general-topology
general-topology
asked Feb 1 at 20:25
Matheus AndradeMatheus Andrade
1,390418
asked Feb 1 at 20:25
Matheus AndradeMatheus Andrade
1,390418
edited Feb 1 at 20:31
Matheus Andrade
asked Feb 1 at 20:25
Matheus AndradeMatheus Andrade
1,390418
asked Feb 1 at 20:25
Matheus AndradeMatheus Andrade
1,390418
asked Feb 1 at 20:25
Matheus AndradeMatheus Andrade
1,390418
1,390418
$begingroup$
By “every sequence has a limit point” do you mean a convergent subsequence (with limit in the space)? If so then yes—they are equivalent. Suppose limit point compactness is taken first. A sequence either has a trivial convergent subsequence if its range is finite or, if the range is infinite, then you can use limit point compactness to get a limit point of the range which a subsequence then converges to.
$endgroup$
– LoveTooNap29
Feb 1 at 20:38
1
$begingroup$
ok well they are still equivalent. You can always find a limit point of the range of the sequence. If the range is finite, it’s trivial, if the range is infinite, appeal to limit point compactness. Where’s the confusion?
$endgroup$
– LoveTooNap29
Feb 1 at 20:45
1
$begingroup$
Also don’t you mean the limit point belongs in $Omega$ not $S$? Take $Omega=[0,1]$ and $x_n=1/n$. Then $0$ is a limit point of $S={x_n: ninmathbb{N}}$ but not a member of $S$ while $0in Omega$.
$endgroup$
– LoveTooNap29
Feb 1 at 20:50
1
$begingroup$
@LoveTooNap29 A limit point of a sequence $(x_n)$ in $X$ is a point $p in X$ such that for every neighbourhood $O$ of $p$ and every $n_0 in mathbb{N}$ there is some $n_1 > n_0$ such that $x_{n_1} in O$. It does not imply a convergent subsequence (it does in a metric or first countable space, but not in general). A limit point of that sequence need not be a limit point of the set ${x_n: n in mathbb{N}}$, confusingly.
$endgroup$
– Henno Brandsma
Feb 1 at 22:34
1
$begingroup$
@LoveTooNap29 I added a counterexample.
$endgroup$
– Henno Brandsma
Feb 1 at 22:54
|
show 4 more comments
$begingroup$
By “every sequence has a limit point” do you mean a convergent subsequence (with limit in the space)? If so then yes—they are equivalent. Suppose limit point compactness is taken first. A sequence either has a trivial convergent subsequence if its range is finite or, if the range is infinite, then you can use limit point compactness to get a limit point of the range which a subsequence then converges to.
$endgroup$
– LoveTooNap29
Feb 1 at 20:38
1
$begingroup$
ok well they are still equivalent. You can always find a limit point of the range of the sequence. If the range is finite, it’s trivial, if the range is infinite, appeal to limit point compactness. Where’s the confusion?
$endgroup$
– LoveTooNap29
Feb 1 at 20:45
1
$begingroup$
Also don’t you mean the limit point belongs in $Omega$ not $S$? Take $Omega=[0,1]$ and $x_n=1/n$. Then $0$ is a limit point of $S={x_n: ninmathbb{N}}$ but not a member of $S$ while $0in Omega$.
$endgroup$
– LoveTooNap29
Feb 1 at 20:50
1
$begingroup$
@LoveTooNap29 A limit point of a sequence $(x_n)$ in $X$ is a point $p in X$ such that for every neighbourhood $O$ of $p$ and every $n_0 in mathbb{N}$ there is some $n_1 > n_0$ such that $x_{n_1} in O$. It does not imply a convergent subsequence (it does in a metric or first countable space, but not in general). A limit point of that sequence need not be a limit point of the set ${x_n: n in mathbb{N}}$, confusingly.
$endgroup$
– Henno Brandsma
Feb 1 at 22:34
1
$begingroup$
@LoveTooNap29 I added a counterexample.
$endgroup$
– Henno Brandsma
Feb 1 at 22:54
$begingroup$
By “every sequence has a limit point” do you mean a convergent subsequence (with limit in the space)? If so then yes—they are equivalent. Suppose limit point compactness is taken first. A sequence either has a trivial convergent subsequence if its range is finite or, if the range is infinite, then you can use limit point compactness to get a limit point of the range which a subsequence then converges to.
$endgroup$
– LoveTooNap29
Feb 1 at 20:38
$begingroup$
By “every sequence has a limit point” do you mean a convergent subsequence (with limit in the space)? If so then yes—they are equivalent. Suppose limit point compactness is taken first. A sequence either has a trivial convergent subsequence if its range is finite or, if the range is infinite, then you can use limit point compactness to get a limit point of the range which a subsequence then converges to.
$endgroup$
– LoveTooNap29
Feb 1 at 20:38
1
1
$begingroup$
ok well they are still equivalent. You can always find a limit point of the range of the sequence. If the range is finite, it’s trivial, if the range is infinite, appeal to limit point compactness. Where’s the confusion?
$endgroup$
– LoveTooNap29
Feb 1 at 20:45
$begingroup$
ok well they are still equivalent. You can always find a limit point of the range of the sequence. If the range is finite, it’s trivial, if the range is infinite, appeal to limit point compactness. Where’s the confusion?
$endgroup$
– LoveTooNap29
Feb 1 at 20:45
1
1
$begingroup$
Also don’t you mean the limit point belongs in $Omega$ not $S$? Take $Omega=[0,1]$ and $x_n=1/n$. Then $0$ is a limit point of $S={x_n: ninmathbb{N}}$ but not a member of $S$ while $0in Omega$.
$endgroup$
– LoveTooNap29
Feb 1 at 20:50
$begingroup$
Also don’t you mean the limit point belongs in $Omega$ not $S$? Take $Omega=[0,1]$ and $x_n=1/n$. Then $0$ is a limit point of $S={x_n: ninmathbb{N}}$ but not a member of $S$ while $0in Omega$.
$endgroup$
– LoveTooNap29
Feb 1 at 20:50
1
1
$begingroup$
@LoveTooNap29 A limit point of a sequence $(x_n)$ in $X$ is a point $p in X$ such that for every neighbourhood $O$ of $p$ and every $n_0 in mathbb{N}$ there is some $n_1 > n_0$ such that $x_{n_1} in O$. It does not imply a convergent subsequence (it does in a metric or first countable space, but not in general). A limit point of that sequence need not be a limit point of the set ${x_n: n in mathbb{N}}$, confusingly.
$endgroup$
– Henno Brandsma
Feb 1 at 22:34
$begingroup$
@LoveTooNap29 A limit point of a sequence $(x_n)$ in $X$ is a point $p in X$ such that for every neighbourhood $O$ of $p$ and every $n_0 in mathbb{N}$ there is some $n_1 > n_0$ such that $x_{n_1} in O$. It does not imply a convergent subsequence (it does in a metric or first countable space, but not in general). A limit point of that sequence need not be a limit point of the set ${x_n: n in mathbb{N}}$, confusingly.
$endgroup$
– Henno Brandsma
Feb 1 at 22:34
1
1
$begingroup$
@LoveTooNap29 I added a counterexample.
$endgroup$
– Henno Brandsma
Feb 1 at 22:54
$begingroup$
@LoveTooNap29 I added a counterexample.
$endgroup$
– Henno Brandsma
Feb 1 at 22:54
|
show 4 more comments
2 Answers
2
active
oldest
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$begingroup$
They are not in general equivalent: let $X=mathbb{N} times {0,1}$ in the product topology were the natural numbers carry the discrete topology and ${0,1}$ the trivial/indiscrete one.
If $A subseteq X$ is infinite and $(n,i) in A$, then $(n,i')$ (where $i' neq i$) is a limit point of $A$, as every open set containing $(n,i')$ also contains $(n,i)$, so $O setminus {(n,i')}$ intersects $A$.
If $(x_n)$ is a bijective enumeration of $X$ (say $x_1 = (1,0), x_2=(1,1), x_3=(2,0), x_4 = (2,1), ldots$ it's clear that every point $(n,i)$ in $X$ has a neighbourhood ${n} times {0,1}$ that only contains two points of the sequence, so the sequence has no limit point in $X$.
In any space $X$ that is weakly sequentially compact we can see that every infinite subset $A$ of $X$ has a point $p$ such that every neighbourhood of $p$ intersects $A$ in an infinite set. ($p$ is then called an $omega$-limit point of $A$), so quite a bit stronger than just limit point compact.
If $X$ is $T_1$ (so finite sets are closed) then a limit point compact $X$ has the property that every infinite subset has an $omega$-limit point in $X$. (thi is sometimes called "strongly limit point compact" or $omega$-limit compact or something similar)
And so for $T_1$ spaces the notions are equivalent, but not quite in general.
For all spaces strong limit point compactness is equivalent to "every countable open cover of $X$ has a finite subcover" (something that my example space indeed fails), which is usually called "countably compact" (i.e. compactness for countable covers).
answered Feb 1 at 22:54
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
The two are equivalent (edit: in metric spaces or first countable spaces). Here's one direction, briefly: if $Omega$ is limit-point compact, then consider a sequence $(x_n)$. The range $S={x_n: nin mathbb{N}}$, if it is finite, trivially, has a limit point. If it is infinite, then since $Omega$ is limit-point compact, we can find a limit point $x$ of $S$ that is in $Omega$.
To emphasize that limit points are either elements of the space or the underlying ambient space: if $Omega =[0,1]$ and $x_n=1/n$ then $0$ is a limit point of $S={x_n : nin mathbb{N}} subset Omega}$ but $0$ is not an element of $S$, while $0in Omega$. Compare to if $Omega=(0,1]$ is considered as a subspace of $mathbb{R}$ then $0$ is a limit point of $S$ still but now $0$ is in neither $S$ nor $Omega$, we just know $0in mathbb{R}$. Exactly which sets/spaces limit points are elements of is inextricably tied up with the notions of closedness, completeness. etc.
answered Feb 1 at 21:04
LoveTooNap29LoveTooNap29
1,2001614
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
They are not in general equivalent: let $X=mathbb{N} times {0,1}$ in the product topology were the natural numbers carry the discrete topology and ${0,1}$ the trivial/indiscrete one.
If $A subseteq X$ is infinite and $(n,i) in A$, then $(n,i')$ (where $i' neq i$) is a limit point of $A$, as every open set containing $(n,i')$ also contains $(n,i)$, so $O setminus {(n,i')}$ intersects $A$.
If $(x_n)$ is a bijective enumeration of $X$ (say $x_1 = (1,0), x_2=(1,1), x_3=(2,0), x_4 = (2,1), ldots$ it's clear that every point $(n,i)$ in $X$ has a neighbourhood ${n} times {0,1}$ that only contains two points of the sequence, so the sequence has no limit point in $X$.
In any space $X$ that is weakly sequentially compact we can see that every infinite subset $A$ of $X$ has a point $p$ such that every neighbourhood of $p$ intersects $A$ in an infinite set. ($p$ is then called an $omega$-limit point of $A$), so quite a bit stronger than just limit point compact.
If $X$ is $T_1$ (so finite sets are closed) then a limit point compact $X$ has the property that every infinite subset has an $omega$-limit point in $X$. (thi is sometimes called "strongly limit point compact" or $omega$-limit compact or something similar)
And so for $T_1$ spaces the notions are equivalent, but not quite in general.
For all spaces strong limit point compactness is equivalent to "every countable open cover of $X$ has a finite subcover" (something that my example space indeed fails), which is usually called "countably compact" (i.e. compactness for countable covers).
answered Feb 1 at 22:54
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
They are not in general equivalent: let $X=mathbb{N} times {0,1}$ in the product topology were the natural numbers carry the discrete topology and ${0,1}$ the trivial/indiscrete one.
If $A subseteq X$ is infinite and $(n,i) in A$, then $(n,i')$ (where $i' neq i$) is a limit point of $A$, as every open set containing $(n,i')$ also contains $(n,i)$, so $O setminus {(n,i')}$ intersects $A$.
If $(x_n)$ is a bijective enumeration of $X$ (say $x_1 = (1,0), x_2=(1,1), x_3=(2,0), x_4 = (2,1), ldots$ it's clear that every point $(n,i)$ in $X$ has a neighbourhood ${n} times {0,1}$ that only contains two points of the sequence, so the sequence has no limit point in $X$.
In any space $X$ that is weakly sequentially compact we can see that every infinite subset $A$ of $X$ has a point $p$ such that every neighbourhood of $p$ intersects $A$ in an infinite set. ($p$ is then called an $omega$-limit point of $A$), so quite a bit stronger than just limit point compact.
If $X$ is $T_1$ (so finite sets are closed) then a limit point compact $X$ has the property that every infinite subset has an $omega$-limit point in $X$. (thi is sometimes called "strongly limit point compact" or $omega$-limit compact or something similar)
And so for $T_1$ spaces the notions are equivalent, but not quite in general.
For all spaces strong limit point compactness is equivalent to "every countable open cover of $X$ has a finite subcover" (something that my example space indeed fails), which is usually called "countably compact" (i.e. compactness for countable covers).
answered Feb 1 at 22:54
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
They are not in general equivalent: let $X=mathbb{N} times {0,1}$ in the product topology were the natural numbers carry the discrete topology and ${0,1}$ the trivial/indiscrete one.
If $A subseteq X$ is infinite and $(n,i) in A$, then $(n,i')$ (where $i' neq i$) is a limit point of $A$, as every open set containing $(n,i')$ also contains $(n,i)$, so $O setminus {(n,i')}$ intersects $A$.
If $(x_n)$ is a bijective enumeration of $X$ (say $x_1 = (1,0), x_2=(1,1), x_3=(2,0), x_4 = (2,1), ldots$ it's clear that every point $(n,i)$ in $X$ has a neighbourhood ${n} times {0,1}$ that only contains two points of the sequence, so the sequence has no limit point in $X$.
In any space $X$ that is weakly sequentially compact we can see that every infinite subset $A$ of $X$ has a point $p$ such that every neighbourhood of $p$ intersects $A$ in an infinite set. ($p$ is then called an $omega$-limit point of $A$), so quite a bit stronger than just limit point compact.
If $X$ is $T_1$ (so finite sets are closed) then a limit point compact $X$ has the property that every infinite subset has an $omega$-limit point in $X$. (thi is sometimes called "strongly limit point compact" or $omega$-limit compact or something similar)
And so for $T_1$ spaces the notions are equivalent, but not quite in general.
For all spaces strong limit point compactness is equivalent to "every countable open cover of $X$ has a finite subcover" (something that my example space indeed fails), which is usually called "countably compact" (i.e. compactness for countable covers).
answered Feb 1 at 22:54
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
They are not in general equivalent: let $X=mathbb{N} times {0,1}$ in the product topology were the natural numbers carry the discrete topology and ${0,1}$ the trivial/indiscrete one.
If $A subseteq X$ is infinite and $(n,i) in A$, then $(n,i')$ (where $i' neq i$) is a limit point of $A$, as every open set containing $(n,i')$ also contains $(n,i)$, so $O setminus {(n,i')}$ intersects $A$.
If $(x_n)$ is a bijective enumeration of $X$ (say $x_1 = (1,0), x_2=(1,1), x_3=(2,0), x_4 = (2,1), ldots$ it's clear that every point $(n,i)$ in $X$ has a neighbourhood ${n} times {0,1}$ that only contains two points of the sequence, so the sequence has no limit point in $X$.
In any space $X$ that is weakly sequentially compact we can see that every infinite subset $A$ of $X$ has a point $p$ such that every neighbourhood of $p$ intersects $A$ in an infinite set. ($p$ is then called an $omega$-limit point of $A$), so quite a bit stronger than just limit point compact.
If $X$ is $T_1$ (so finite sets are closed) then a limit point compact $X$ has the property that every infinite subset has an $omega$-limit point in $X$. (thi is sometimes called "strongly limit point compact" or $omega$-limit compact or something similar)
And so for $T_1$ spaces the notions are equivalent, but not quite in general.
For all spaces strong limit point compactness is equivalent to "every countable open cover of $X$ has a finite subcover" (something that my example space indeed fails), which is usually called "countably compact" (i.e. compactness for countable covers).
answered Feb 1 at 22:54
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 1 at 22:54
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 1 at 22:54
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 1 at 22:54
Henno BrandsmaHenno Brandsma
116k349127
116k349127
add a comment |
add a comment |
$begingroup$
The two are equivalent (edit: in metric spaces or first countable spaces). Here's one direction, briefly: if $Omega$ is limit-point compact, then consider a sequence $(x_n)$. The range $S={x_n: nin mathbb{N}}$, if it is finite, trivially, has a limit point. If it is infinite, then since $Omega$ is limit-point compact, we can find a limit point $x$ of $S$ that is in $Omega$.
To emphasize that limit points are either elements of the space or the underlying ambient space: if $Omega =[0,1]$ and $x_n=1/n$ then $0$ is a limit point of $S={x_n : nin mathbb{N}} subset Omega}$ but $0$ is not an element of $S$, while $0in Omega$. Compare to if $Omega=(0,1]$ is considered as a subspace of $mathbb{R}$ then $0$ is a limit point of $S$ still but now $0$ is in neither $S$ nor $Omega$, we just know $0in mathbb{R}$. Exactly which sets/spaces limit points are elements of is inextricably tied up with the notions of closedness, completeness. etc.
answered Feb 1 at 21:04
LoveTooNap29LoveTooNap29
1,2001614
$endgroup$
add a comment |
$begingroup$
The two are equivalent (edit: in metric spaces or first countable spaces). Here's one direction, briefly: if $Omega$ is limit-point compact, then consider a sequence $(x_n)$. The range $S={x_n: nin mathbb{N}}$, if it is finite, trivially, has a limit point. If it is infinite, then since $Omega$ is limit-point compact, we can find a limit point $x$ of $S$ that is in $Omega$.
To emphasize that limit points are either elements of the space or the underlying ambient space: if $Omega =[0,1]$ and $x_n=1/n$ then $0$ is a limit point of $S={x_n : nin mathbb{N}} subset Omega}$ but $0$ is not an element of $S$, while $0in Omega$. Compare to if $Omega=(0,1]$ is considered as a subspace of $mathbb{R}$ then $0$ is a limit point of $S$ still but now $0$ is in neither $S$ nor $Omega$, we just know $0in mathbb{R}$. Exactly which sets/spaces limit points are elements of is inextricably tied up with the notions of closedness, completeness. etc.
answered Feb 1 at 21:04
LoveTooNap29LoveTooNap29
1,2001614
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add a comment |
$begingroup$
The two are equivalent (edit: in metric spaces or first countable spaces). Here's one direction, briefly: if $Omega$ is limit-point compact, then consider a sequence $(x_n)$. The range $S={x_n: nin mathbb{N}}$, if it is finite, trivially, has a limit point. If it is infinite, then since $Omega$ is limit-point compact, we can find a limit point $x$ of $S$ that is in $Omega$.
To emphasize that limit points are either elements of the space or the underlying ambient space: if $Omega =[0,1]$ and $x_n=1/n$ then $0$ is a limit point of $S={x_n : nin mathbb{N}} subset Omega}$ but $0$ is not an element of $S$, while $0in Omega$. Compare to if $Omega=(0,1]$ is considered as a subspace of $mathbb{R}$ then $0$ is a limit point of $S$ still but now $0$ is in neither $S$ nor $Omega$, we just know $0in mathbb{R}$. Exactly which sets/spaces limit points are elements of is inextricably tied up with the notions of closedness, completeness. etc.
answered Feb 1 at 21:04
LoveTooNap29LoveTooNap29
1,2001614
$endgroup$
The two are equivalent (edit: in metric spaces or first countable spaces). Here's one direction, briefly: if $Omega$ is limit-point compact, then consider a sequence $(x_n)$. The range $S={x_n: nin mathbb{N}}$, if it is finite, trivially, has a limit point. If it is infinite, then since $Omega$ is limit-point compact, we can find a limit point $x$ of $S$ that is in $Omega$.
To emphasize that limit points are either elements of the space or the underlying ambient space: if $Omega =[0,1]$ and $x_n=1/n$ then $0$ is a limit point of $S={x_n : nin mathbb{N}} subset Omega}$ but $0$ is not an element of $S$, while $0in Omega$. Compare to if $Omega=(0,1]$ is considered as a subspace of $mathbb{R}$ then $0$ is a limit point of $S$ still but now $0$ is in neither $S$ nor $Omega$, we just know $0in mathbb{R}$. Exactly which sets/spaces limit points are elements of is inextricably tied up with the notions of closedness, completeness. etc.
answered Feb 1 at 21:04
LoveTooNap29LoveTooNap29
1,2001614
edited Feb 1 at 22:57
answered Feb 1 at 21:04
LoveTooNap29LoveTooNap29
1,2001614
answered Feb 1 at 21:04
LoveTooNap29LoveTooNap29
1,2001614
answered Feb 1 at 21:04
LoveTooNap29LoveTooNap29
1,2001614
1,2001614
add a comment |
add a comment |
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By “every sequence has a limit point” do you mean a convergent subsequence (with limit in the space)? If so then yes—they are equivalent. Suppose limit point compactness is taken first. A sequence either has a trivial convergent subsequence if its range is finite or, if the range is infinite, then you can use limit point compactness to get a limit point of the range which a subsequence then converges to.
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– LoveTooNap29
Feb 1 at 20:38
1
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ok well they are still equivalent. You can always find a limit point of the range of the sequence. If the range is finite, it’s trivial, if the range is infinite, appeal to limit point compactness. Where’s the confusion?
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– LoveTooNap29
Feb 1 at 20:45
1
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Also don’t you mean the limit point belongs in $Omega$ not $S$? Take $Omega=[0,1]$ and $x_n=1/n$. Then $0$ is a limit point of $S={x_n: ninmathbb{N}}$ but not a member of $S$ while $0in Omega$.
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– LoveTooNap29
Feb 1 at 20:50
1
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@LoveTooNap29 A limit point of a sequence $(x_n)$ in $X$ is a point $p in X$ such that for every neighbourhood $O$ of $p$ and every $n_0 in mathbb{N}$ there is some $n_1 > n_0$ such that $x_{n_1} in O$. It does not imply a convergent subsequence (it does in a metric or first countable space, but not in general). A limit point of that sequence need not be a limit point of the set ${x_n: n in mathbb{N}}$, confusingly.
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– Henno Brandsma
Feb 1 at 22:34
1
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@LoveTooNap29 I added a counterexample.
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– Henno Brandsma
Feb 1 at 22:54