If $(X,d)$ is a metric space, prove that intersection of any collection of compact sets in $(X,d)$ is...
$begingroup$
This question already has an answer here:
Proving that the Intersection of any Collection of Compact Sets is Compact
2 answers
If $(X,d)$ be a compact metric space then, arbitrary intersection of compact subsets is compact.
Is it true if $(X,d)$ is a metric space, but not compact?
There is similar type of questions already have been asked before, but those are from real-analysis where any closed and bounded set is compact. But in metric space closed and bounded set may not be compact. Hence this question is different from previously asked questions.
metric-spaces
asked Nov 25 '18 at 6:14
Tamas Kanti GaraiTamas Kanti Garai
32
$endgroup$
marked as duplicate by Nosrati, Saad, José Carlos Santos, Shailesh, Cesareo Nov 26 '18 at 1:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Proving that the Intersection of any Collection of Compact Sets is Compact
2 answers
If $(X,d)$ be a compact metric space then, arbitrary intersection of compact subsets is compact.
Is it true if $(X,d)$ is a metric space, but not compact?
There is similar type of questions already have been asked before, but those are from real-analysis where any closed and bounded set is compact. But in metric space closed and bounded set may not be compact. Hence this question is different from previously asked questions.
metric-spaces
asked Nov 25 '18 at 6:14
Tamas Kanti GaraiTamas Kanti Garai
32
$endgroup$
marked as duplicate by Nosrati, Saad, José Carlos Santos, Shailesh, Cesareo Nov 26 '18 at 1:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
You may find answer here: math.stackexchange.com/questions/1447506/…
$endgroup$
– Aniruddha Deshmukh
Nov 25 '18 at 6:35
$begingroup$
If you are taking arbitrary intersections of compact sets, that will be a closed set (because each of the compact sets are closed sets and arbitrary intersection of closed sets is closed). As the intersection is a subset of a compact set, it's a compact set.
$endgroup$
– John_Wick
Nov 25 '18 at 6:37
add a comment |
$begingroup$
This question already has an answer here:
Proving that the Intersection of any Collection of Compact Sets is Compact
2 answers
If $(X,d)$ be a compact metric space then, arbitrary intersection of compact subsets is compact.
Is it true if $(X,d)$ is a metric space, but not compact?
There is similar type of questions already have been asked before, but those are from real-analysis where any closed and bounded set is compact. But in metric space closed and bounded set may not be compact. Hence this question is different from previously asked questions.
metric-spaces
asked Nov 25 '18 at 6:14
Tamas Kanti GaraiTamas Kanti Garai
32
$endgroup$
This question already has an answer here:
Proving that the Intersection of any Collection of Compact Sets is Compact
2 answers
If $(X,d)$ be a compact metric space then, arbitrary intersection of compact subsets is compact.
Is it true if $(X,d)$ is a metric space, but not compact?
There is similar type of questions already have been asked before, but those are from real-analysis where any closed and bounded set is compact. But in metric space closed and bounded set may not be compact. Hence this question is different from previously asked questions.
This question already has an answer here:
Proving that the Intersection of any Collection of Compact Sets is Compact
2 answers
metric-spaces
metric-spaces
asked Nov 25 '18 at 6:14
Tamas Kanti GaraiTamas Kanti Garai
32
asked Nov 25 '18 at 6:14
Tamas Kanti GaraiTamas Kanti Garai
32
edited Nov 26 '18 at 3:44
Tamas Kanti Garai
asked Nov 25 '18 at 6:14
Tamas Kanti GaraiTamas Kanti Garai
32
asked Nov 25 '18 at 6:14
Tamas Kanti GaraiTamas Kanti Garai
32
asked Nov 25 '18 at 6:14
Tamas Kanti GaraiTamas Kanti Garai
32
32
marked as duplicate by Nosrati, Saad, José Carlos Santos, Shailesh, Cesareo Nov 26 '18 at 1:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Nosrati, Saad, José Carlos Santos, Shailesh, Cesareo Nov 26 '18 at 1:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
You may find answer here: math.stackexchange.com/questions/1447506/…
$endgroup$
– Aniruddha Deshmukh
Nov 25 '18 at 6:35
$begingroup$
If you are taking arbitrary intersections of compact sets, that will be a closed set (because each of the compact sets are closed sets and arbitrary intersection of closed sets is closed). As the intersection is a subset of a compact set, it's a compact set.
$endgroup$
– John_Wick
Nov 25 '18 at 6:37
add a comment |
$begingroup$
You may find answer here: math.stackexchange.com/questions/1447506/…
$endgroup$
– Aniruddha Deshmukh
Nov 25 '18 at 6:35
$begingroup$
If you are taking arbitrary intersections of compact sets, that will be a closed set (because each of the compact sets are closed sets and arbitrary intersection of closed sets is closed). As the intersection is a subset of a compact set, it's a compact set.
$endgroup$
– John_Wick
Nov 25 '18 at 6:37
$begingroup$
You may find answer here: math.stackexchange.com/questions/1447506/…
$endgroup$
– Aniruddha Deshmukh
Nov 25 '18 at 6:35
$begingroup$
You may find answer here: math.stackexchange.com/questions/1447506/…
$endgroup$
– Aniruddha Deshmukh
Nov 25 '18 at 6:35
$begingroup$
If you are taking arbitrary intersections of compact sets, that will be a closed set (because each of the compact sets are closed sets and arbitrary intersection of closed sets is closed). As the intersection is a subset of a compact set, it's a compact set.
$endgroup$
– John_Wick
Nov 25 '18 at 6:37
$begingroup$
If you are taking arbitrary intersections of compact sets, that will be a closed set (because each of the compact sets are closed sets and arbitrary intersection of closed sets is closed). As the intersection is a subset of a compact set, it's a compact set.
$endgroup$
– John_Wick
Nov 25 '18 at 6:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $(X,d)$ be a metric space and ${C_i}_{iin I}$ be some collection of compact subsets of $(X,d)$ . Let $alphain I$ be fixed and consider the collection ${C_icap C_{alpha}}_{iin I}$ of subsets of $C_{alpha}$ . Note that for each $iin I$ we have $C_icap C_{alpha}$ is a closed subset of the compact set $C_{alpha}$.Therefore $bigcap_{iin I} C_i=bigcap_{iin I} (C_icap C_{alpha})$ is also a closed subset of $C_{alpha}$ being arbitrary intersection of closed subsets. But closed subset of compact set is compact , hence $bigcap_{iin I} C_i$ is compact subset of $(X,d)$ . Note that compactness is not relative property.
answered Nov 25 '18 at 6:31
UserSUserS
1,5391112
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $(X,d)$ be a metric space and ${C_i}_{iin I}$ be some collection of compact subsets of $(X,d)$ . Let $alphain I$ be fixed and consider the collection ${C_icap C_{alpha}}_{iin I}$ of subsets of $C_{alpha}$ . Note that for each $iin I$ we have $C_icap C_{alpha}$ is a closed subset of the compact set $C_{alpha}$.Therefore $bigcap_{iin I} C_i=bigcap_{iin I} (C_icap C_{alpha})$ is also a closed subset of $C_{alpha}$ being arbitrary intersection of closed subsets. But closed subset of compact set is compact , hence $bigcap_{iin I} C_i$ is compact subset of $(X,d)$ . Note that compactness is not relative property.
answered Nov 25 '18 at 6:31
UserSUserS
1,5391112
$endgroup$
add a comment |
$begingroup$
Let $(X,d)$ be a metric space and ${C_i}_{iin I}$ be some collection of compact subsets of $(X,d)$ . Let $alphain I$ be fixed and consider the collection ${C_icap C_{alpha}}_{iin I}$ of subsets of $C_{alpha}$ . Note that for each $iin I$ we have $C_icap C_{alpha}$ is a closed subset of the compact set $C_{alpha}$.Therefore $bigcap_{iin I} C_i=bigcap_{iin I} (C_icap C_{alpha})$ is also a closed subset of $C_{alpha}$ being arbitrary intersection of closed subsets. But closed subset of compact set is compact , hence $bigcap_{iin I} C_i$ is compact subset of $(X,d)$ . Note that compactness is not relative property.
answered Nov 25 '18 at 6:31
UserSUserS
1,5391112
$endgroup$
add a comment |
$begingroup$
Let $(X,d)$ be a metric space and ${C_i}_{iin I}$ be some collection of compact subsets of $(X,d)$ . Let $alphain I$ be fixed and consider the collection ${C_icap C_{alpha}}_{iin I}$ of subsets of $C_{alpha}$ . Note that for each $iin I$ we have $C_icap C_{alpha}$ is a closed subset of the compact set $C_{alpha}$.Therefore $bigcap_{iin I} C_i=bigcap_{iin I} (C_icap C_{alpha})$ is also a closed subset of $C_{alpha}$ being arbitrary intersection of closed subsets. But closed subset of compact set is compact , hence $bigcap_{iin I} C_i$ is compact subset of $(X,d)$ . Note that compactness is not relative property.
answered Nov 25 '18 at 6:31
UserSUserS
1,5391112
$endgroup$
Let $(X,d)$ be a metric space and ${C_i}_{iin I}$ be some collection of compact subsets of $(X,d)$ . Let $alphain I$ be fixed and consider the collection ${C_icap C_{alpha}}_{iin I}$ of subsets of $C_{alpha}$ . Note that for each $iin I$ we have $C_icap C_{alpha}$ is a closed subset of the compact set $C_{alpha}$.Therefore $bigcap_{iin I} C_i=bigcap_{iin I} (C_icap C_{alpha})$ is also a closed subset of $C_{alpha}$ being arbitrary intersection of closed subsets. But closed subset of compact set is compact , hence $bigcap_{iin I} C_i$ is compact subset of $(X,d)$ . Note that compactness is not relative property.
answered Nov 25 '18 at 6:31
UserSUserS
1,5391112
edited Nov 25 '18 at 6:46
answered Nov 25 '18 at 6:31
UserSUserS
1,5391112
answered Nov 25 '18 at 6:31
UserSUserS
1,5391112
answered Nov 25 '18 at 6:31
UserSUserS
1,5391112
1,5391112
add a comment |
add a comment |
$begingroup$
You may find answer here: math.stackexchange.com/questions/1447506/…
$endgroup$
– Aniruddha Deshmukh
Nov 25 '18 at 6:35
$begingroup$
If you are taking arbitrary intersections of compact sets, that will be a closed set (because each of the compact sets are closed sets and arbitrary intersection of closed sets is closed). As the intersection is a subset of a compact set, it's a compact set.
$endgroup$
– John_Wick
Nov 25 '18 at 6:37