What is an infinite subset of a compact set?
$begingroup$
I am attempting to work on the following proof:
If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.
I know that that this proof has been answered here already, but I am more interested in understanding the statement itself. I am just struggling to comprehend what $E$ is in this question... an infinite subset of a compact set. For some reason, I just cannot visualize this. If anyone out there is able to help me see through the fog, or even provide a specific example for me to think about, I would greatly appreciate it.
general-topology
asked Dec 7 '18 at 5:05
automattikautomattik
406
$endgroup$
add a comment |
$begingroup$
I am attempting to work on the following proof:
If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.
I know that that this proof has been answered here already, but I am more interested in understanding the statement itself. I am just struggling to comprehend what $E$ is in this question... an infinite subset of a compact set. For some reason, I just cannot visualize this. If anyone out there is able to help me see through the fog, or even provide a specific example for me to think about, I would greatly appreciate it.
general-topology
asked Dec 7 '18 at 5:05
automattikautomattik
406
$endgroup$
add a comment |
$begingroup$
I am attempting to work on the following proof:
If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.
I know that that this proof has been answered here already, but I am more interested in understanding the statement itself. I am just struggling to comprehend what $E$ is in this question... an infinite subset of a compact set. For some reason, I just cannot visualize this. If anyone out there is able to help me see through the fog, or even provide a specific example for me to think about, I would greatly appreciate it.
general-topology
asked Dec 7 '18 at 5:05
automattikautomattik
406
$endgroup$
I am attempting to work on the following proof:
If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.
I know that that this proof has been answered here already, but I am more interested in understanding the statement itself. I am just struggling to comprehend what $E$ is in this question... an infinite subset of a compact set. For some reason, I just cannot visualize this. If anyone out there is able to help me see through the fog, or even provide a specific example for me to think about, I would greatly appreciate it.
general-topology
general-topology
asked Dec 7 '18 at 5:05
automattikautomattik
406
asked Dec 7 '18 at 5:05
automattikautomattik
406
asked Dec 7 '18 at 5:05
automattikautomattik
406
asked Dec 7 '18 at 5:05
automattikautomattik
406
asked Dec 7 '18 at 5:05
automattikautomattik
406
406
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
There are many possibilities. Take $(0,1)subseteq [0,1]$ or the circle embedded as the equator of the sphere: $S^1subseteq S^2$.
If you are familiar with it, take the middle thirds cantor set $Csubseteq [0,1]$.
answered Dec 7 '18 at 5:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
$endgroup$
$begingroup$
Thank you for the examples. Would it also be correct to say that $[0,1] subseteq [0,1]$ works too? I know this seems trivial, but I guess I am just wondering if the infinite subset can be closed as well.
$endgroup$
– automattik
Dec 8 '18 at 5:38
1
$begingroup$
Yes that's also correct.
$endgroup$
– Antonios-Alexandros Robotis
Dec 8 '18 at 5:39
add a comment |
$begingroup$
There are many examples as mentioned in the other answer. By Heine-Borel Theorem compactness is equivalent to the condition of being closed and bounded. Take the unit interval $I$ for example, you can pick the sequence $1,frac{1}{2},...,1/n...$ as an infinite subset, or any open(or closed) interval contained in $I$.
Although not mentioned in the description, a proof of the theorem can be found here:
Infinite subset of a compact set
answered Dec 7 '18 at 5:22
William SunWilliam Sun
471111
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
There are many possibilities. Take $(0,1)subseteq [0,1]$ or the circle embedded as the equator of the sphere: $S^1subseteq S^2$.
If you are familiar with it, take the middle thirds cantor set $Csubseteq [0,1]$.
answered Dec 7 '18 at 5:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
$endgroup$
$begingroup$
Thank you for the examples. Would it also be correct to say that $[0,1] subseteq [0,1]$ works too? I know this seems trivial, but I guess I am just wondering if the infinite subset can be closed as well.
$endgroup$
– automattik
Dec 8 '18 at 5:38
1
$begingroup$
Yes that's also correct.
$endgroup$
– Antonios-Alexandros Robotis
Dec 8 '18 at 5:39
add a comment |
$begingroup$
There are many possibilities. Take $(0,1)subseteq [0,1]$ or the circle embedded as the equator of the sphere: $S^1subseteq S^2$.
If you are familiar with it, take the middle thirds cantor set $Csubseteq [0,1]$.
answered Dec 7 '18 at 5:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
$endgroup$
$begingroup$
Thank you for the examples. Would it also be correct to say that $[0,1] subseteq [0,1]$ works too? I know this seems trivial, but I guess I am just wondering if the infinite subset can be closed as well.
$endgroup$
– automattik
Dec 8 '18 at 5:38
1
$begingroup$
Yes that's also correct.
$endgroup$
– Antonios-Alexandros Robotis
Dec 8 '18 at 5:39
add a comment |
$begingroup$
There are many possibilities. Take $(0,1)subseteq [0,1]$ or the circle embedded as the equator of the sphere: $S^1subseteq S^2$.
If you are familiar with it, take the middle thirds cantor set $Csubseteq [0,1]$.
answered Dec 7 '18 at 5:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
$endgroup$
There are many possibilities. Take $(0,1)subseteq [0,1]$ or the circle embedded as the equator of the sphere: $S^1subseteq S^2$.
If you are familiar with it, take the middle thirds cantor set $Csubseteq [0,1]$.
answered Dec 7 '18 at 5:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
answered Dec 7 '18 at 5:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
answered Dec 7 '18 at 5:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
answered Dec 7 '18 at 5:08
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
9,91741640
$begingroup$
Thank you for the examples. Would it also be correct to say that $[0,1] subseteq [0,1]$ works too? I know this seems trivial, but I guess I am just wondering if the infinite subset can be closed as well.
$endgroup$
– automattik
Dec 8 '18 at 5:38
1
$begingroup$
Yes that's also correct.
$endgroup$
– Antonios-Alexandros Robotis
Dec 8 '18 at 5:39
add a comment |
$begingroup$
Thank you for the examples. Would it also be correct to say that $[0,1] subseteq [0,1]$ works too? I know this seems trivial, but I guess I am just wondering if the infinite subset can be closed as well.
$endgroup$
– automattik
Dec 8 '18 at 5:38
1
$begingroup$
Yes that's also correct.
$endgroup$
– Antonios-Alexandros Robotis
Dec 8 '18 at 5:39
$begingroup$
Thank you for the examples. Would it also be correct to say that $[0,1] subseteq [0,1]$ works too? I know this seems trivial, but I guess I am just wondering if the infinite subset can be closed as well.
$endgroup$
– automattik
Dec 8 '18 at 5:38
$begingroup$
Thank you for the examples. Would it also be correct to say that $[0,1] subseteq [0,1]$ works too? I know this seems trivial, but I guess I am just wondering if the infinite subset can be closed as well.
$endgroup$
– automattik
Dec 8 '18 at 5:38
1
1
$begingroup$
Yes that's also correct.
$endgroup$
– Antonios-Alexandros Robotis
Dec 8 '18 at 5:39
$begingroup$
Yes that's also correct.
$endgroup$
– Antonios-Alexandros Robotis
Dec 8 '18 at 5:39
add a comment |
$begingroup$
There are many examples as mentioned in the other answer. By Heine-Borel Theorem compactness is equivalent to the condition of being closed and bounded. Take the unit interval $I$ for example, you can pick the sequence $1,frac{1}{2},...,1/n...$ as an infinite subset, or any open(or closed) interval contained in $I$.
Although not mentioned in the description, a proof of the theorem can be found here:
Infinite subset of a compact set
answered Dec 7 '18 at 5:22
William SunWilliam Sun
471111
$endgroup$
add a comment |
$begingroup$
There are many examples as mentioned in the other answer. By Heine-Borel Theorem compactness is equivalent to the condition of being closed and bounded. Take the unit interval $I$ for example, you can pick the sequence $1,frac{1}{2},...,1/n...$ as an infinite subset, or any open(or closed) interval contained in $I$.
Although not mentioned in the description, a proof of the theorem can be found here:
Infinite subset of a compact set
answered Dec 7 '18 at 5:22
William SunWilliam Sun
471111
$endgroup$
add a comment |
$begingroup$
There are many examples as mentioned in the other answer. By Heine-Borel Theorem compactness is equivalent to the condition of being closed and bounded. Take the unit interval $I$ for example, you can pick the sequence $1,frac{1}{2},...,1/n...$ as an infinite subset, or any open(or closed) interval contained in $I$.
Although not mentioned in the description, a proof of the theorem can be found here:
Infinite subset of a compact set
answered Dec 7 '18 at 5:22
William SunWilliam Sun
471111
$endgroup$
There are many examples as mentioned in the other answer. By Heine-Borel Theorem compactness is equivalent to the condition of being closed and bounded. Take the unit interval $I$ for example, you can pick the sequence $1,frac{1}{2},...,1/n...$ as an infinite subset, or any open(or closed) interval contained in $I$.
Although not mentioned in the description, a proof of the theorem can be found here:
Infinite subset of a compact set
answered Dec 7 '18 at 5:22
William SunWilliam Sun
471111
answered Dec 7 '18 at 5:22
William SunWilliam Sun
471111
answered Dec 7 '18 at 5:22
William SunWilliam Sun
471111
answered Dec 7 '18 at 5:22
William SunWilliam Sun
471111
471111
add a comment |
add a comment |
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