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A subset $Asubset mathbb R^n$ is compact iff every nested sequence of relatively closed, non-empty subsets of $A$ has non-empty intersection.
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This is the problem.
Let $Asubsetmathbb R^n$. I want to show that $A$ is compact if and only if every nested sequence $A_n_n=1^infty$ of relatively closed, non-empty subsets of $A$ has non-empty intersection.
I know how to prove "$Rightarrow$" direction, using the fact that $A$ has the Bolzano-Weierstrass property, or more simply, using a theorem relating the compactness and the finite intersection property.
But I am not sure how to prove "$Leftarrow$". I tried the following. I assume that $A$ is not compact, so $A$ does not have the Bolzano-Weierstrass proprty. So there is an infinite subset $B$ which does not have a limit point in $A$. So there exists a collection of open sets $O_x_xin B$ such that
$$O_xcap B=x$$
for all $xin B$. Then one can find a countable subset $x_i_i=1^inftysubset B$, and the corresponding collection of open sets $O_i_i=1^infty$ with $O_icap B=x_i$. Further more, one constructs a sequence of relatively closed, non-empty subsets
$$A_n=A-cup_i=1^nO_i$$.
$A_n_n=1^infty$ is a nested sequence. So $cap_n=1^infty A_nneemptyset$. But
$$cap_n=1^infty A_n=cap_n=1^infty(A-cup_i=1^nO_i)=A-cup_n=1^infty O_n$$
which is surely non-empty, in general, since $x_i_i=1^infty$ is generally a proper subset of $B$.
I cannot find any contradiction. Please help me out!
After searching online, I found out that it may be useful to use the property of being sequentially compact, etc.. But I am reading Fred H. Croom's Principles of Topology, which does not discuss it. Please use some different methods.
general-topology metric-spaces proof-writing compactness
edited Jul 19 at 14:14
Shervin Sorouri
334111
asked Jul 19 at 14:03
Drake Marquis
488413
add a comment |Â
up vote
1
down vote
favorite
This is the problem.
Let $Asubsetmathbb R^n$. I want to show that $A$ is compact if and only if every nested sequence $A_n_n=1^infty$ of relatively closed, non-empty subsets of $A$ has non-empty intersection.
I know how to prove "$Rightarrow$" direction, using the fact that $A$ has the Bolzano-Weierstrass property, or more simply, using a theorem relating the compactness and the finite intersection property.
But I am not sure how to prove "$Leftarrow$". I tried the following. I assume that $A$ is not compact, so $A$ does not have the Bolzano-Weierstrass proprty. So there is an infinite subset $B$ which does not have a limit point in $A$. So there exists a collection of open sets $O_x_xin B$ such that
$$O_xcap B=x$$
for all $xin B$. Then one can find a countable subset $x_i_i=1^inftysubset B$, and the corresponding collection of open sets $O_i_i=1^infty$ with $O_icap B=x_i$. Further more, one constructs a sequence of relatively closed, non-empty subsets
$$A_n=A-cup_i=1^nO_i$$.
$A_n_n=1^infty$ is a nested sequence. So $cap_n=1^infty A_nneemptyset$. But
$$cap_n=1^infty A_n=cap_n=1^infty(A-cup_i=1^nO_i)=A-cup_n=1^infty O_n$$
which is surely non-empty, in general, since $x_i_i=1^infty$ is generally a proper subset of $B$.
I cannot find any contradiction. Please help me out!
After searching online, I found out that it may be useful to use the property of being sequentially compact, etc.. But I am reading Fred H. Croom's Principles of Topology, which does not discuss it. Please use some different methods.
general-topology metric-spaces proof-writing compactness
edited Jul 19 at 14:14
Shervin Sorouri
334111
asked Jul 19 at 14:03
Drake Marquis
488413
BTW: your proof of the existence of $x_i$ is the proof that relates compactness to sequential compactness. So in a sense, you already supplied the "missing piece" that you mused about in the final comment of your question.
– Willie Wong
Jul 19 at 14:22
@WillieWong Unfortuntely, I did not prove the existence of $x_i$ but I found it natural to have such $x_i$ because one can always label some points in $B$, at least in $mathbb R^n$. So from your comment, it is possible that there does not exist such kind countable subset for an infinite set in a generic metric space or even a generic topological space. Right?
– Drake Marquis
Jul 19 at 14:29
I see, I misread your argument. In a metric space pretty much all notions of compactness are equivalent. So if $A$ were not compact, you will be able to find a sequence of points in $A$ which does not converge in $A$.
– Willie Wong
Jul 19 at 14:40
And yes, in general topological spaces the equivalence can fail. A set $A$ which is non-compact but for which every nested sequence $A_n$ ... holds is the long line. You can rescue things by changing the index set from the naturals to arbitrary directed sets (essentially doing the "net" construction).
– Willie Wong
Jul 19 at 14:53
@WillieWong Thanks for your comments. I think I need more time to learn and digest your comments. But I just want to confirm that there always exists a countable subset of an infinite set. The existence should not depend on the topology of the space. Right?
– Drake Marquis
Jul 20 at 2:10
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is the problem.
Let $Asubsetmathbb R^n$. I want to show that $A$ is compact if and only if every nested sequence $A_n_n=1^infty$ of relatively closed, non-empty subsets of $A$ has non-empty intersection.
I know how to prove "$Rightarrow$" direction, using the fact that $A$ has the Bolzano-Weierstrass property, or more simply, using a theorem relating the compactness and the finite intersection property.
But I am not sure how to prove "$Leftarrow$". I tried the following. I assume that $A$ is not compact, so $A$ does not have the Bolzano-Weierstrass proprty. So there is an infinite subset $B$ which does not have a limit point in $A$. So there exists a collection of open sets $O_x_xin B$ such that
$$O_xcap B=x$$
for all $xin B$. Then one can find a countable subset $x_i_i=1^inftysubset B$, and the corresponding collection of open sets $O_i_i=1^infty$ with $O_icap B=x_i$. Further more, one constructs a sequence of relatively closed, non-empty subsets
$$A_n=A-cup_i=1^nO_i$$.
$A_n_n=1^infty$ is a nested sequence. So $cap_n=1^infty A_nneemptyset$. But
$$cap_n=1^infty A_n=cap_n=1^infty(A-cup_i=1^nO_i)=A-cup_n=1^infty O_n$$
which is surely non-empty, in general, since $x_i_i=1^infty$ is generally a proper subset of $B$.
I cannot find any contradiction. Please help me out!
After searching online, I found out that it may be useful to use the property of being sequentially compact, etc.. But I am reading Fred H. Croom's Principles of Topology, which does not discuss it. Please use some different methods.
general-topology metric-spaces proof-writing compactness
edited Jul 19 at 14:14
Shervin Sorouri
334111
asked Jul 19 at 14:03
Drake Marquis
488413
This is the problem.
Let $Asubsetmathbb R^n$. I want to show that $A$ is compact if and only if every nested sequence $A_n_n=1^infty$ of relatively closed, non-empty subsets of $A$ has non-empty intersection.
I know how to prove "$Rightarrow$" direction, using the fact that $A$ has the Bolzano-Weierstrass property, or more simply, using a theorem relating the compactness and the finite intersection property.
But I am not sure how to prove "$Leftarrow$". I tried the following. I assume that $A$ is not compact, so $A$ does not have the Bolzano-Weierstrass proprty. So there is an infinite subset $B$ which does not have a limit point in $A$. So there exists a collection of open sets $O_x_xin B$ such that
$$O_xcap B=x$$
for all $xin B$. Then one can find a countable subset $x_i_i=1^inftysubset B$, and the corresponding collection of open sets $O_i_i=1^infty$ with $O_icap B=x_i$. Further more, one constructs a sequence of relatively closed, non-empty subsets
$$A_n=A-cup_i=1^nO_i$$.
$A_n_n=1^infty$ is a nested sequence. So $cap_n=1^infty A_nneemptyset$. But
$$cap_n=1^infty A_n=cap_n=1^infty(A-cup_i=1^nO_i)=A-cup_n=1^infty O_n$$
which is surely non-empty, in general, since $x_i_i=1^infty$ is generally a proper subset of $B$.
I cannot find any contradiction. Please help me out!
After searching online, I found out that it may be useful to use the property of being sequentially compact, etc.. But I am reading Fred H. Croom's Principles of Topology, which does not discuss it. Please use some different methods.
general-topology metric-spaces proof-writing compactness
edited Jul 19 at 14:14
Shervin Sorouri
334111
asked Jul 19 at 14:03
Drake Marquis
488413
edited Jul 19 at 14:14
Shervin Sorouri
334111
edited Jul 19 at 14:14
Shervin Sorouri
334111
edited Jul 19 at 14:14
Shervin Sorouri
334111
334111
asked Jul 19 at 14:03
Drake Marquis
488413
asked Jul 19 at 14:03
Drake Marquis
488413
asked Jul 19 at 14:03
Drake Marquis
488413
488413
BTW: your proof of the existence of $x_i$ is the proof that relates compactness to sequential compactness. So in a sense, you already supplied the "missing piece" that you mused about in the final comment of your question.
– Willie Wong
Jul 19 at 14:22
@WillieWong Unfortuntely, I did not prove the existence of $x_i$ but I found it natural to have such $x_i$ because one can always label some points in $B$, at least in $mathbb R^n$. So from your comment, it is possible that there does not exist such kind countable subset for an infinite set in a generic metric space or even a generic topological space. Right?
– Drake Marquis
Jul 19 at 14:29
I see, I misread your argument. In a metric space pretty much all notions of compactness are equivalent. So if $A$ were not compact, you will be able to find a sequence of points in $A$ which does not converge in $A$.
– Willie Wong
Jul 19 at 14:40
And yes, in general topological spaces the equivalence can fail. A set $A$ which is non-compact but for which every nested sequence $A_n$ ... holds is the long line. You can rescue things by changing the index set from the naturals to arbitrary directed sets (essentially doing the "net" construction).
– Willie Wong
Jul 19 at 14:53
@WillieWong Thanks for your comments. I think I need more time to learn and digest your comments. But I just want to confirm that there always exists a countable subset of an infinite set. The existence should not depend on the topology of the space. Right?
– Drake Marquis
Jul 20 at 2:10
add a comment |Â
BTW: your proof of the existence of $x_i$ is the proof that relates compactness to sequential compactness. So in a sense, you already supplied the "missing piece" that you mused about in the final comment of your question.
– Willie Wong
Jul 19 at 14:22
@WillieWong Unfortuntely, I did not prove the existence of $x_i$ but I found it natural to have such $x_i$ because one can always label some points in $B$, at least in $mathbb R^n$. So from your comment, it is possible that there does not exist such kind countable subset for an infinite set in a generic metric space or even a generic topological space. Right?
– Drake Marquis
Jul 19 at 14:29
I see, I misread your argument. In a metric space pretty much all notions of compactness are equivalent. So if $A$ were not compact, you will be able to find a sequence of points in $A$ which does not converge in $A$.
– Willie Wong
Jul 19 at 14:40
And yes, in general topological spaces the equivalence can fail. A set $A$ which is non-compact but for which every nested sequence $A_n$ ... holds is the long line. You can rescue things by changing the index set from the naturals to arbitrary directed sets (essentially doing the "net" construction).
– Willie Wong
Jul 19 at 14:53
@WillieWong Thanks for your comments. I think I need more time to learn and digest your comments. But I just want to confirm that there always exists a countable subset of an infinite set. The existence should not depend on the topology of the space. Right?
– Drake Marquis
Jul 20 at 2:10
BTW: your proof of the existence of $x_i$ is the proof that relates compactness to sequential compactness. So in a sense, you already supplied the "missing piece" that you mused about in the final comment of your question.
– Willie Wong
Jul 19 at 14:22
BTW: your proof of the existence of $x_i$ is the proof that relates compactness to sequential compactness. So in a sense, you already supplied the "missing piece" that you mused about in the final comment of your question.
– Willie Wong
Jul 19 at 14:22
@WillieWong Unfortuntely, I did not prove the existence of $x_i$ but I found it natural to have such $x_i$ because one can always label some points in $B$, at least in $mathbb R^n$. So from your comment, it is possible that there does not exist such kind countable subset for an infinite set in a generic metric space or even a generic topological space. Right?
– Drake Marquis
Jul 19 at 14:29
@WillieWong Unfortuntely, I did not prove the existence of $x_i$ but I found it natural to have such $x_i$ because one can always label some points in $B$, at least in $mathbb R^n$. So from your comment, it is possible that there does not exist such kind countable subset for an infinite set in a generic metric space or even a generic topological space. Right?
– Drake Marquis
Jul 19 at 14:29
I see, I misread your argument. In a metric space pretty much all notions of compactness are equivalent. So if $A$ were not compact, you will be able to find a sequence of points in $A$ which does not converge in $A$.
– Willie Wong
Jul 19 at 14:40
I see, I misread your argument. In a metric space pretty much all notions of compactness are equivalent. So if $A$ were not compact, you will be able to find a sequence of points in $A$ which does not converge in $A$.
– Willie Wong
Jul 19 at 14:40
And yes, in general topological spaces the equivalence can fail. A set $A$ which is non-compact but for which every nested sequence $A_n$ ... holds is the long line. You can rescue things by changing the index set from the naturals to arbitrary directed sets (essentially doing the "net" construction).
– Willie Wong
Jul 19 at 14:53
And yes, in general topological spaces the equivalence can fail. A set $A$ which is non-compact but for which every nested sequence $A_n$ ... holds is the long line. You can rescue things by changing the index set from the naturals to arbitrary directed sets (essentially doing the "net" construction).
– Willie Wong
Jul 19 at 14:53
@WillieWong Thanks for your comments. I think I need more time to learn and digest your comments. But I just want to confirm that there always exists a countable subset of an infinite set. The existence should not depend on the topology of the space. Right?
– Drake Marquis
Jul 20 at 2:10
@WillieWong Thanks for your comments. I think I need more time to learn and digest your comments. But I just want to confirm that there always exists a countable subset of an infinite set. The existence should not depend on the topology of the space. Right?
– Drake Marquis
Jul 20 at 2:10
add a comment |Â
2 Answers
2
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up vote
1
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You've successfully found a countable collection of points $x_isubset B subset A$ such that $x_i$ has no limit points in $A$.
Now let $C_j = x_i_i = j^infty$.
- Why are the $C_j$ nested?
- Why are the $C_j$ relatively closed?
- What's the intersection $cap C_j$?
answered Jul 19 at 14:20
Willie Wong
54.2k8103206
Wow... it is so simple! Thanks!
– Drake Marquis
Jul 19 at 14:27
$cap C_j$ should be empty, inituitively. But I want to prove it. One way to do so is to assume that $cap C_jneemptyset$, then there is $xincap C_j$, and $x$ should be one of $x_j$'s. Say $x=x_m$ for some $minmathbb N$. Then $xnotin C_m+1$, so $xnotincap C_j$. So $cap C_j=emptyset$. Is this OK?
– Drake Marquis
Jul 20 at 2:13
@DrakeMarquis: yes. I would however forgo the double negative. "If $yin C_1$, then $y = x_i$ for some $i geq 1$. But then $y notin C_i+1$, hence $ynotin cap C_j$." No need to "assume for contradiction."
– Willie Wong
Jul 20 at 13:28
This is very nice! Thank you!
– Drake Marquis
Jul 24 at 0:30
add a comment |Â
up vote
0
down vote
Let $F$ be a family of subsets of $X.$ We say that $F$ has the Finite Intersection Property (F.I.P.) iff (i) $Fne emptyset,$ and (ii) $;cap Gne emptyset$ whenever $G$ is a finite non-empty subset of $F.$
Theorem: A space $X$ is compact iff $cap Fne emptyset$ for every family $F$ of closed sets such that $F$ has the F.I.P.
Proof: (1). If $X$ is not compact,let $C$ be an open cover of $X$ with no finite sub-cover. Let $F=Xbackslash c: cin C.$ Then $F$ is a family of closed sets and $F$ has the F.I.P. but $cap F=emptyset.$
(2). If $F$ is a family of closed sets and $F$ has the F.I.P. but $cap F$ is empty then $Xbackslash f: fin F$ is an open cover of $X$ with no finite sub-cover, so $X$ is not compact.
answered Jul 20 at 5:34
DanielWainfleet
31.7k31643
But your theorem only applies to the question the OP asked in one direction, as far as I can tell. (And the OP stated that he used this result in his question.) The condition that every nested sequence of nonempty closed subsets have non-trivial intersection does not imply F.I.P.
– Willie Wong
Jul 20 at 13:32
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You've successfully found a countable collection of points $x_isubset B subset A$ such that $x_i$ has no limit points in $A$.
Now let $C_j = x_i_i = j^infty$.
- Why are the $C_j$ nested?
- Why are the $C_j$ relatively closed?
- What's the intersection $cap C_j$?
answered Jul 19 at 14:20
Willie Wong
54.2k8103206
Wow... it is so simple! Thanks!
– Drake Marquis
Jul 19 at 14:27
$cap C_j$ should be empty, inituitively. But I want to prove it. One way to do so is to assume that $cap C_jneemptyset$, then there is $xincap C_j$, and $x$ should be one of $x_j$'s. Say $x=x_m$ for some $minmathbb N$. Then $xnotin C_m+1$, so $xnotincap C_j$. So $cap C_j=emptyset$. Is this OK?
– Drake Marquis
Jul 20 at 2:13
@DrakeMarquis: yes. I would however forgo the double negative. "If $yin C_1$, then $y = x_i$ for some $i geq 1$. But then $y notin C_i+1$, hence $ynotin cap C_j$." No need to "assume for contradiction."
– Willie Wong
Jul 20 at 13:28
This is very nice! Thank you!
– Drake Marquis
Jul 24 at 0:30
add a comment |Â
up vote
1
down vote
You've successfully found a countable collection of points $x_isubset B subset A$ such that $x_i$ has no limit points in $A$.
Now let $C_j = x_i_i = j^infty$.
- Why are the $C_j$ nested?
- Why are the $C_j$ relatively closed?
- What's the intersection $cap C_j$?
answered Jul 19 at 14:20
Willie Wong
54.2k8103206
Wow... it is so simple! Thanks!
– Drake Marquis
Jul 19 at 14:27
$cap C_j$ should be empty, inituitively. But I want to prove it. One way to do so is to assume that $cap C_jneemptyset$, then there is $xincap C_j$, and $x$ should be one of $x_j$'s. Say $x=x_m$ for some $minmathbb N$. Then $xnotin C_m+1$, so $xnotincap C_j$. So $cap C_j=emptyset$. Is this OK?
– Drake Marquis
Jul 20 at 2:13
@DrakeMarquis: yes. I would however forgo the double negative. "If $yin C_1$, then $y = x_i$ for some $i geq 1$. But then $y notin C_i+1$, hence $ynotin cap C_j$." No need to "assume for contradiction."
– Willie Wong
Jul 20 at 13:28
This is very nice! Thank you!
– Drake Marquis
Jul 24 at 0:30
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You've successfully found a countable collection of points $x_isubset B subset A$ such that $x_i$ has no limit points in $A$.
Now let $C_j = x_i_i = j^infty$.
- Why are the $C_j$ nested?
- Why are the $C_j$ relatively closed?
- What's the intersection $cap C_j$?
answered Jul 19 at 14:20
Willie Wong
54.2k8103206
You've successfully found a countable collection of points $x_isubset B subset A$ such that $x_i$ has no limit points in $A$.
Now let $C_j = x_i_i = j^infty$.
- Why are the $C_j$ nested?
- Why are the $C_j$ relatively closed?
- What's the intersection $cap C_j$?
answered Jul 19 at 14:20
Willie Wong
54.2k8103206
answered Jul 19 at 14:20
Willie Wong
54.2k8103206
answered Jul 19 at 14:20
Willie Wong
54.2k8103206
answered Jul 19 at 14:20
Willie Wong
54.2k8103206
54.2k8103206
Wow... it is so simple! Thanks!
– Drake Marquis
Jul 19 at 14:27
$cap C_j$ should be empty, inituitively. But I want to prove it. One way to do so is to assume that $cap C_jneemptyset$, then there is $xincap C_j$, and $x$ should be one of $x_j$'s. Say $x=x_m$ for some $minmathbb N$. Then $xnotin C_m+1$, so $xnotincap C_j$. So $cap C_j=emptyset$. Is this OK?
– Drake Marquis
Jul 20 at 2:13
@DrakeMarquis: yes. I would however forgo the double negative. "If $yin C_1$, then $y = x_i$ for some $i geq 1$. But then $y notin C_i+1$, hence $ynotin cap C_j$." No need to "assume for contradiction."
– Willie Wong
Jul 20 at 13:28
This is very nice! Thank you!
– Drake Marquis
Jul 24 at 0:30
add a comment |Â
Wow... it is so simple! Thanks!
– Drake Marquis
Jul 19 at 14:27
$cap C_j$ should be empty, inituitively. But I want to prove it. One way to do so is to assume that $cap C_jneemptyset$, then there is $xincap C_j$, and $x$ should be one of $x_j$'s. Say $x=x_m$ for some $minmathbb N$. Then $xnotin C_m+1$, so $xnotincap C_j$. So $cap C_j=emptyset$. Is this OK?
– Drake Marquis
Jul 20 at 2:13
@DrakeMarquis: yes. I would however forgo the double negative. "If $yin C_1$, then $y = x_i$ for some $i geq 1$. But then $y notin C_i+1$, hence $ynotin cap C_j$." No need to "assume for contradiction."
– Willie Wong
Jul 20 at 13:28
This is very nice! Thank you!
– Drake Marquis
Jul 24 at 0:30
Wow... it is so simple! Thanks!
– Drake Marquis
Jul 19 at 14:27
Wow... it is so simple! Thanks!
– Drake Marquis
Jul 19 at 14:27
$cap C_j$ should be empty, inituitively. But I want to prove it. One way to do so is to assume that $cap C_jneemptyset$, then there is $xincap C_j$, and $x$ should be one of $x_j$'s. Say $x=x_m$ for some $minmathbb N$. Then $xnotin C_m+1$, so $xnotincap C_j$. So $cap C_j=emptyset$. Is this OK?
– Drake Marquis
Jul 20 at 2:13
$cap C_j$ should be empty, inituitively. But I want to prove it. One way to do so is to assume that $cap C_jneemptyset$, then there is $xincap C_j$, and $x$ should be one of $x_j$'s. Say $x=x_m$ for some $minmathbb N$. Then $xnotin C_m+1$, so $xnotincap C_j$. So $cap C_j=emptyset$. Is this OK?
– Drake Marquis
Jul 20 at 2:13
@DrakeMarquis: yes. I would however forgo the double negative. "If $yin C_1$, then $y = x_i$ for some $i geq 1$. But then $y notin C_i+1$, hence $ynotin cap C_j$." No need to "assume for contradiction."
– Willie Wong
Jul 20 at 13:28
@DrakeMarquis: yes. I would however forgo the double negative. "If $yin C_1$, then $y = x_i$ for some $i geq 1$. But then $y notin C_i+1$, hence $ynotin cap C_j$." No need to "assume for contradiction."
– Willie Wong
Jul 20 at 13:28
This is very nice! Thank you!
– Drake Marquis
Jul 24 at 0:30
This is very nice! Thank you!
– Drake Marquis
Jul 24 at 0:30
add a comment |Â
up vote
0
down vote
Let $F$ be a family of subsets of $X.$ We say that $F$ has the Finite Intersection Property (F.I.P.) iff (i) $Fne emptyset,$ and (ii) $;cap Gne emptyset$ whenever $G$ is a finite non-empty subset of $F.$
Theorem: A space $X$ is compact iff $cap Fne emptyset$ for every family $F$ of closed sets such that $F$ has the F.I.P.
Proof: (1). If $X$ is not compact,let $C$ be an open cover of $X$ with no finite sub-cover. Let $F=Xbackslash c: cin C.$ Then $F$ is a family of closed sets and $F$ has the F.I.P. but $cap F=emptyset.$
(2). If $F$ is a family of closed sets and $F$ has the F.I.P. but $cap F$ is empty then $Xbackslash f: fin F$ is an open cover of $X$ with no finite sub-cover, so $X$ is not compact.
answered Jul 20 at 5:34
DanielWainfleet
31.7k31643
But your theorem only applies to the question the OP asked in one direction, as far as I can tell. (And the OP stated that he used this result in his question.) The condition that every nested sequence of nonempty closed subsets have non-trivial intersection does not imply F.I.P.
– Willie Wong
Jul 20 at 13:32
add a comment |Â
up vote
0
down vote
Let $F$ be a family of subsets of $X.$ We say that $F$ has the Finite Intersection Property (F.I.P.) iff (i) $Fne emptyset,$ and (ii) $;cap Gne emptyset$ whenever $G$ is a finite non-empty subset of $F.$
Theorem: A space $X$ is compact iff $cap Fne emptyset$ for every family $F$ of closed sets such that $F$ has the F.I.P.
Proof: (1). If $X$ is not compact,let $C$ be an open cover of $X$ with no finite sub-cover. Let $F=Xbackslash c: cin C.$ Then $F$ is a family of closed sets and $F$ has the F.I.P. but $cap F=emptyset.$
(2). If $F$ is a family of closed sets and $F$ has the F.I.P. but $cap F$ is empty then $Xbackslash f: fin F$ is an open cover of $X$ with no finite sub-cover, so $X$ is not compact.
answered Jul 20 at 5:34
DanielWainfleet
31.7k31643
But your theorem only applies to the question the OP asked in one direction, as far as I can tell. (And the OP stated that he used this result in his question.) The condition that every nested sequence of nonempty closed subsets have non-trivial intersection does not imply F.I.P.
– Willie Wong
Jul 20 at 13:32
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Let $F$ be a family of subsets of $X.$ We say that $F$ has the Finite Intersection Property (F.I.P.) iff (i) $Fne emptyset,$ and (ii) $;cap Gne emptyset$ whenever $G$ is a finite non-empty subset of $F.$
Theorem: A space $X$ is compact iff $cap Fne emptyset$ for every family $F$ of closed sets such that $F$ has the F.I.P.
Proof: (1). If $X$ is not compact,let $C$ be an open cover of $X$ with no finite sub-cover. Let $F=Xbackslash c: cin C.$ Then $F$ is a family of closed sets and $F$ has the F.I.P. but $cap F=emptyset.$
(2). If $F$ is a family of closed sets and $F$ has the F.I.P. but $cap F$ is empty then $Xbackslash f: fin F$ is an open cover of $X$ with no finite sub-cover, so $X$ is not compact.
answered Jul 20 at 5:34
DanielWainfleet
31.7k31643
Let $F$ be a family of subsets of $X.$ We say that $F$ has the Finite Intersection Property (F.I.P.) iff (i) $Fne emptyset,$ and (ii) $;cap Gne emptyset$ whenever $G$ is a finite non-empty subset of $F.$
Theorem: A space $X$ is compact iff $cap Fne emptyset$ for every family $F$ of closed sets such that $F$ has the F.I.P.
Proof: (1). If $X$ is not compact,let $C$ be an open cover of $X$ with no finite sub-cover. Let $F=Xbackslash c: cin C.$ Then $F$ is a family of closed sets and $F$ has the F.I.P. but $cap F=emptyset.$
(2). If $F$ is a family of closed sets and $F$ has the F.I.P. but $cap F$ is empty then $Xbackslash f: fin F$ is an open cover of $X$ with no finite sub-cover, so $X$ is not compact.
answered Jul 20 at 5:34
DanielWainfleet
31.7k31643
answered Jul 20 at 5:34
DanielWainfleet
31.7k31643
answered Jul 20 at 5:34
DanielWainfleet
31.7k31643
answered Jul 20 at 5:34
DanielWainfleet
31.7k31643
31.7k31643
But your theorem only applies to the question the OP asked in one direction, as far as I can tell. (And the OP stated that he used this result in his question.) The condition that every nested sequence of nonempty closed subsets have non-trivial intersection does not imply F.I.P.
– Willie Wong
Jul 20 at 13:32
add a comment |Â
But your theorem only applies to the question the OP asked in one direction, as far as I can tell. (And the OP stated that he used this result in his question.) The condition that every nested sequence of nonempty closed subsets have non-trivial intersection does not imply F.I.P.
– Willie Wong
Jul 20 at 13:32
But your theorem only applies to the question the OP asked in one direction, as far as I can tell. (And the OP stated that he used this result in his question.) The condition that every nested sequence of nonempty closed subsets have non-trivial intersection does not imply F.I.P.
– Willie Wong
Jul 20 at 13:32
But your theorem only applies to the question the OP asked in one direction, as far as I can tell. (And the OP stated that he used this result in his question.) The condition that every nested sequence of nonempty closed subsets have non-trivial intersection does not imply F.I.P.
– Willie Wong
Jul 20 at 13:32
add a comment |Â
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BTW: your proof of the existence of $x_i$ is the proof that relates compactness to sequential compactness. So in a sense, you already supplied the "missing piece" that you mused about in the final comment of your question.
– Willie Wong
Jul 19 at 14:22
@WillieWong Unfortuntely, I did not prove the existence of $x_i$ but I found it natural to have such $x_i$ because one can always label some points in $B$, at least in $mathbb R^n$. So from your comment, it is possible that there does not exist such kind countable subset for an infinite set in a generic metric space or even a generic topological space. Right?
– Drake Marquis
Jul 19 at 14:29
I see, I misread your argument. In a metric space pretty much all notions of compactness are equivalent. So if $A$ were not compact, you will be able to find a sequence of points in $A$ which does not converge in $A$.
– Willie Wong
Jul 19 at 14:40
And yes, in general topological spaces the equivalence can fail. A set $A$ which is non-compact but for which every nested sequence $A_n$ ... holds is the long line. You can rescue things by changing the index set from the naturals to arbitrary directed sets (essentially doing the "net" construction).
– Willie Wong
Jul 19 at 14:53
@WillieWong Thanks for your comments. I think I need more time to learn and digest your comments. But I just want to confirm that there always exists a countable subset of an infinite set. The existence should not depend on the topology of the space. Right?
– Drake Marquis
Jul 20 at 2:10