Let A be a closed discrete set inside a compact set K. Show that A is finite.
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1
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Clearly, A must be compact also, so I'm assuming this proof requires the use of all 3 factors - closed, discrete & compact - but I'm not quite sure how to actually use these in a proper proof.
Any help/hints/explanation would be greatly appreciated!
complex-analysis discrete-mathematics metric-spaces compactness
asked Dec 3 '16 at 14:27
Le Coq
104
add a comment |
up vote
1
down vote
favorite
Clearly, A must be compact also, so I'm assuming this proof requires the use of all 3 factors - closed, discrete & compact - but I'm not quite sure how to actually use these in a proper proof.
Any help/hints/explanation would be greatly appreciated!
complex-analysis discrete-mathematics metric-spaces compactness
asked Dec 3 '16 at 14:27
Le Coq
104
Try proving "A discrete compact space is finite".
– Daniel Fischer♦
Dec 3 '16 at 14:30
@DanielFischer Do you need closed? See my answer.
– Jack Bauer
Nov 15 at 13:21
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Clearly, A must be compact also, so I'm assuming this proof requires the use of all 3 factors - closed, discrete & compact - but I'm not quite sure how to actually use these in a proper proof.
Any help/hints/explanation would be greatly appreciated!
complex-analysis discrete-mathematics metric-spaces compactness
asked Dec 3 '16 at 14:27
Le Coq
104
Clearly, A must be compact also, so I'm assuming this proof requires the use of all 3 factors - closed, discrete & compact - but I'm not quite sure how to actually use these in a proper proof.
Any help/hints/explanation would be greatly appreciated!
complex-analysis discrete-mathematics metric-spaces compactness
complex-analysis discrete-mathematics metric-spaces compactness
asked Dec 3 '16 at 14:27
Le Coq
104
asked Dec 3 '16 at 14:27
Le Coq
104
asked Dec 3 '16 at 14:27
Le Coq
104
asked Dec 3 '16 at 14:27
Le Coq
104
asked Dec 3 '16 at 14:27
Le Coq
104
104
Try proving "A discrete compact space is finite".
– Daniel Fischer♦
Dec 3 '16 at 14:30
@DanielFischer Do you need closed? See my answer.
– Jack Bauer
Nov 15 at 13:21
add a comment |
Try proving "A discrete compact space is finite".
– Daniel Fischer♦
Dec 3 '16 at 14:30
@DanielFischer Do you need closed? See my answer.
– Jack Bauer
Nov 15 at 13:21
Try proving "A discrete compact space is finite".
– Daniel Fischer♦
Dec 3 '16 at 14:30
Try proving "A discrete compact space is finite".
– Daniel Fischer♦
Dec 3 '16 at 14:30
@DanielFischer Do you need closed? See my answer.
– Jack Bauer
Nov 15 at 13:21
@DanielFischer Do you need closed? See my answer.
– Jack Bauer
Nov 15 at 13:21
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
We know (or can prove) that a closed subset of a compact set is compact. Therefore, $A$ is compact. We are also given that $A$ is discrete, which means that, for each $x in A$, we can find a neighborhood $N_x$ of $x$ such that $N_x cap A = {x}$.
Notice that ${N_x }_{x in X}$ is an open cover of $A$...
answered Dec 3 '16 at 14:56
Kaj Hansen
27.2k43779
ahhhh yea I see! thank you
– Le Coq
Dec 3 '16 at 15:02
KajHansen Do you need closed? See my answer.
– Jack Bauer
Nov 15 at 13:22
add a comment |
up vote
0
down vote
If $A$ is infinite, then $A$ has a limit point because $K$ is limit point compact. I did not make use of $A$ being closed.
answered Nov 15 at 11:29
Jack Bauer
1,2461531
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
We know (or can prove) that a closed subset of a compact set is compact. Therefore, $A$ is compact. We are also given that $A$ is discrete, which means that, for each $x in A$, we can find a neighborhood $N_x$ of $x$ such that $N_x cap A = {x}$.
Notice that ${N_x }_{x in X}$ is an open cover of $A$...
answered Dec 3 '16 at 14:56
Kaj Hansen
27.2k43779
ahhhh yea I see! thank you
– Le Coq
Dec 3 '16 at 15:02
KajHansen Do you need closed? See my answer.
– Jack Bauer
Nov 15 at 13:22
add a comment |
up vote
2
down vote
accepted
We know (or can prove) that a closed subset of a compact set is compact. Therefore, $A$ is compact. We are also given that $A$ is discrete, which means that, for each $x in A$, we can find a neighborhood $N_x$ of $x$ such that $N_x cap A = {x}$.
Notice that ${N_x }_{x in X}$ is an open cover of $A$...
answered Dec 3 '16 at 14:56
Kaj Hansen
27.2k43779
ahhhh yea I see! thank you
– Le Coq
Dec 3 '16 at 15:02
KajHansen Do you need closed? See my answer.
– Jack Bauer
Nov 15 at 13:22
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
We know (or can prove) that a closed subset of a compact set is compact. Therefore, $A$ is compact. We are also given that $A$ is discrete, which means that, for each $x in A$, we can find a neighborhood $N_x$ of $x$ such that $N_x cap A = {x}$.
Notice that ${N_x }_{x in X}$ is an open cover of $A$...
answered Dec 3 '16 at 14:56
Kaj Hansen
27.2k43779
We know (or can prove) that a closed subset of a compact set is compact. Therefore, $A$ is compact. We are also given that $A$ is discrete, which means that, for each $x in A$, we can find a neighborhood $N_x$ of $x$ such that $N_x cap A = {x}$.
Notice that ${N_x }_{x in X}$ is an open cover of $A$...
answered Dec 3 '16 at 14:56
Kaj Hansen
27.2k43779
edited Dec 4 '16 at 1:30
answered Dec 3 '16 at 14:56
Kaj Hansen
27.2k43779
answered Dec 3 '16 at 14:56
Kaj Hansen
27.2k43779
answered Dec 3 '16 at 14:56
Kaj Hansen
27.2k43779
27.2k43779
ahhhh yea I see! thank you
– Le Coq
Dec 3 '16 at 15:02
KajHansen Do you need closed? See my answer.
– Jack Bauer
Nov 15 at 13:22
add a comment |
ahhhh yea I see! thank you
– Le Coq
Dec 3 '16 at 15:02
KajHansen Do you need closed? See my answer.
– Jack Bauer
Nov 15 at 13:22
ahhhh yea I see! thank you
– Le Coq
Dec 3 '16 at 15:02
ahhhh yea I see! thank you
– Le Coq
Dec 3 '16 at 15:02
KajHansen Do you need closed? See my answer.
– Jack Bauer
Nov 15 at 13:22
KajHansen Do you need closed? See my answer.
– Jack Bauer
Nov 15 at 13:22
add a comment |
up vote
0
down vote
If $A$ is infinite, then $A$ has a limit point because $K$ is limit point compact. I did not make use of $A$ being closed.
answered Nov 15 at 11:29
Jack Bauer
1,2461531
add a comment |
up vote
0
down vote
If $A$ is infinite, then $A$ has a limit point because $K$ is limit point compact. I did not make use of $A$ being closed.
answered Nov 15 at 11:29
Jack Bauer
1,2461531
add a comment |
up vote
0
down vote
up vote
0
down vote
If $A$ is infinite, then $A$ has a limit point because $K$ is limit point compact. I did not make use of $A$ being closed.
answered Nov 15 at 11:29
Jack Bauer
1,2461531
If $A$ is infinite, then $A$ has a limit point because $K$ is limit point compact. I did not make use of $A$ being closed.
answered Nov 15 at 11:29
Jack Bauer
1,2461531
answered Nov 15 at 11:29
Jack Bauer
1,2461531
answered Nov 15 at 11:29
Jack Bauer
1,2461531
answered Nov 15 at 11:29
Jack Bauer
1,2461531
1,2461531
add a comment |
add a comment |
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Try proving "A discrete compact space is finite".
– Daniel Fischer♦
Dec 3 '16 at 14:30
@DanielFischer Do you need closed? See my answer.
– Jack Bauer
Nov 15 at 13:21