Mixing
Rudin's functional analysis appendix A4 (a)
$begingroup$
Quick question about the following theorem:
If $K$ is a closed subset of a complete metric space $X$ then the following three properties are equivalent:
(a) $K$ is compact
(b) Every infinite subset of $K$ ha s a limit point in $K$
(c) $K$ is totally bounded
(a) Is proven by contradiction.
Assume (a). If $E subset K$ is infinite and no point of $K$ is a limit point of $E$, there's an open cover $left{ V_alpha right}$ of $K$ such that each $V_alpha$ contains at most one point of $E$.
Why is that? I was trying to get the same conclusion by using the definition limit point and compact, but this doesn't seem to lead me anywhere.
Therefore $left{ V_alpha right}$ has no finite subcover, a contradiction.
Can you elaborate on this bit as well?
functional-analysis proof-explanation compactness complete-spaces
asked Jan 17 at 10:02
user8469759user8469759
1,5381618
$endgroup$
add a comment |
$begingroup$
Quick question about the following theorem:
If $K$ is a closed subset of a complete metric space $X$ then the following three properties are equivalent:
(a) $K$ is compact
(b) Every infinite subset of $K$ ha s a limit point in $K$
(c) $K$ is totally bounded
(a) Is proven by contradiction.
Assume (a). If $E subset K$ is infinite and no point of $K$ is a limit point of $E$, there's an open cover $left{ V_alpha right}$ of $K$ such that each $V_alpha$ contains at most one point of $E$.
Why is that? I was trying to get the same conclusion by using the definition limit point and compact, but this doesn't seem to lead me anywhere.
Therefore $left{ V_alpha right}$ has no finite subcover, a contradiction.
Can you elaborate on this bit as well?
functional-analysis proof-explanation compactness complete-spaces
asked Jan 17 at 10:02
user8469759user8469759
1,5381618
$endgroup$
add a comment |
$begingroup$
Quick question about the following theorem:
If $K$ is a closed subset of a complete metric space $X$ then the following three properties are equivalent:
(a) $K$ is compact
(b) Every infinite subset of $K$ ha s a limit point in $K$
(c) $K$ is totally bounded
(a) Is proven by contradiction.
Assume (a). If $E subset K$ is infinite and no point of $K$ is a limit point of $E$, there's an open cover $left{ V_alpha right}$ of $K$ such that each $V_alpha$ contains at most one point of $E$.
Why is that? I was trying to get the same conclusion by using the definition limit point and compact, but this doesn't seem to lead me anywhere.
Therefore $left{ V_alpha right}$ has no finite subcover, a contradiction.
Can you elaborate on this bit as well?
functional-analysis proof-explanation compactness complete-spaces
asked Jan 17 at 10:02
user8469759user8469759
1,5381618
$endgroup$
Quick question about the following theorem:
If $K$ is a closed subset of a complete metric space $X$ then the following three properties are equivalent:
(a) $K$ is compact
(b) Every infinite subset of $K$ ha s a limit point in $K$
(c) $K$ is totally bounded
(a) Is proven by contradiction.
Assume (a). If $E subset K$ is infinite and no point of $K$ is a limit point of $E$, there's an open cover $left{ V_alpha right}$ of $K$ such that each $V_alpha$ contains at most one point of $E$.
Why is that? I was trying to get the same conclusion by using the definition limit point and compact, but this doesn't seem to lead me anywhere.
Therefore $left{ V_alpha right}$ has no finite subcover, a contradiction.
Can you elaborate on this bit as well?
functional-analysis proof-explanation compactness complete-spaces
functional-analysis proof-explanation compactness complete-spaces
asked Jan 17 at 10:02
user8469759user8469759
1,5381618
asked Jan 17 at 10:02
user8469759user8469759
1,5381618
asked Jan 17 at 10:02
user8469759user8469759
1,5381618
asked Jan 17 at 10:02
user8469759user8469759
1,5381618
asked Jan 17 at 10:02
user8469759user8469759
1,5381618
1,5381618
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
1) By definition of limit point. If no point is a limit point, you can find a neighborhood of each point not containing any other point of $E$. The union of all these neighborhoods cover $K$.
2) If you try to take a finite subcover, no matter how you choose them, you will only find finitely many points of E because every neighborhood you chose only contains at most one point of E. Hence you cannot cover E (and consequently K) with a finite subcover.
answered Jan 17 at 10:13
KlausKlaus
1,8179
$endgroup$
$begingroup$
I know I'm being annoying, but could you elaborate a bit more on the fact "you can find a neighborhood of each point not containing any other point of $E$"?
$endgroup$
– user8469759
Jan 17 at 10:21
$begingroup$
Well, what is your definition of limit point?
$endgroup$
– Klaus
Jan 17 at 10:30
$begingroup$
x is a limit point for E if every neighborhood intersects E (except for x)
$endgroup$
– user8469759
Jan 17 at 10:33
$begingroup$
Hence if $x$ is not a limit point, there is a neighborhood that does not intersect $E$ (except for $x$). ;-)
$endgroup$
– Klaus
Jan 17 at 10:36
$begingroup$
Ok, I struggled with the negation... shame on me. I'll give a second read to the proof given your explanation as well.
$endgroup$
– user8469759
Jan 17 at 10:38
add a comment |
$begingroup$
For every $xin E$ you can choose an open set $V_x$ such that $Ecap V_x={x}$.
For every $xin K-E$ you can choose an open set $V_x$ such that $xin V_x$ and $Ecap V_x=varnothing$.
This because every $xin K$ is not a limit point of $E$.
The collection ${V_xmid xin K}$ is an open cover of $K$.
answered Jan 17 at 10:13
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
Attempt (Credit to drhab).
Assume $E subset K$ is infinite and no point of $K$ is a limit point.
1) Let $e in E$. Then , since $e in K$, $e$ is not a limit point of $E$.
There is an open $U_e$ s.t. $U_ecap E =$ {$e$}.
2) Let $y in K-E_y$. then since $y$ is not a limit point of $E$, there is an open $V_y$ s.t. $V_y cap E = emptyset$.
3) $displaystyle{cup_{e in E}} U_e displaystyle{cup_{y in K-E}} V_y$ is an open cover of $K$ that does not have a finite subcover(Why?).
Contradiction to $K$ compact.
answered Jan 17 at 17:01
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
$begingroup$
Your point 3) isn't clear.
$endgroup$
– user8469759
Jan 17 at 17:35
$begingroup$
User8469759.Please explain. $E subset cup_{e in E} U_e$, likewise the other union : $K -E subset cup_{y in K-E} V_y$.
$endgroup$
– Peter Szilas
Jan 17 at 18:07
$begingroup$
Just an issue with the latex, it's ok now.
$endgroup$
– user8469759
Jan 17 at 18:20
$begingroup$
User8469759.Thanks.
$endgroup$
– Peter Szilas
Jan 17 at 18:29
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
1) By definition of limit point. If no point is a limit point, you can find a neighborhood of each point not containing any other point of $E$. The union of all these neighborhoods cover $K$.
2) If you try to take a finite subcover, no matter how you choose them, you will only find finitely many points of E because every neighborhood you chose only contains at most one point of E. Hence you cannot cover E (and consequently K) with a finite subcover.
answered Jan 17 at 10:13
KlausKlaus
1,8179
$endgroup$
$begingroup$
I know I'm being annoying, but could you elaborate a bit more on the fact "you can find a neighborhood of each point not containing any other point of $E$"?
$endgroup$
– user8469759
Jan 17 at 10:21
$begingroup$
Well, what is your definition of limit point?
$endgroup$
– Klaus
Jan 17 at 10:30
$begingroup$
x is a limit point for E if every neighborhood intersects E (except for x)
$endgroup$
– user8469759
Jan 17 at 10:33
$begingroup$
Hence if $x$ is not a limit point, there is a neighborhood that does not intersect $E$ (except for $x$). ;-)
$endgroup$
– Klaus
Jan 17 at 10:36
$begingroup$
Ok, I struggled with the negation... shame on me. I'll give a second read to the proof given your explanation as well.
$endgroup$
– user8469759
Jan 17 at 10:38
add a comment |
$begingroup$
1) By definition of limit point. If no point is a limit point, you can find a neighborhood of each point not containing any other point of $E$. The union of all these neighborhoods cover $K$.
2) If you try to take a finite subcover, no matter how you choose them, you will only find finitely many points of E because every neighborhood you chose only contains at most one point of E. Hence you cannot cover E (and consequently K) with a finite subcover.
answered Jan 17 at 10:13
KlausKlaus
1,8179
$endgroup$
$begingroup$
I know I'm being annoying, but could you elaborate a bit more on the fact "you can find a neighborhood of each point not containing any other point of $E$"?
$endgroup$
– user8469759
Jan 17 at 10:21
$begingroup$
Well, what is your definition of limit point?
$endgroup$
– Klaus
Jan 17 at 10:30
$begingroup$
x is a limit point for E if every neighborhood intersects E (except for x)
$endgroup$
– user8469759
Jan 17 at 10:33
$begingroup$
Hence if $x$ is not a limit point, there is a neighborhood that does not intersect $E$ (except for $x$). ;-)
$endgroup$
– Klaus
Jan 17 at 10:36
$begingroup$
Ok, I struggled with the negation... shame on me. I'll give a second read to the proof given your explanation as well.
$endgroup$
– user8469759
Jan 17 at 10:38
add a comment |
$begingroup$
1) By definition of limit point. If no point is a limit point, you can find a neighborhood of each point not containing any other point of $E$. The union of all these neighborhoods cover $K$.
2) If you try to take a finite subcover, no matter how you choose them, you will only find finitely many points of E because every neighborhood you chose only contains at most one point of E. Hence you cannot cover E (and consequently K) with a finite subcover.
answered Jan 17 at 10:13
KlausKlaus
1,8179
$endgroup$
1) By definition of limit point. If no point is a limit point, you can find a neighborhood of each point not containing any other point of $E$. The union of all these neighborhoods cover $K$.
2) If you try to take a finite subcover, no matter how you choose them, you will only find finitely many points of E because every neighborhood you chose only contains at most one point of E. Hence you cannot cover E (and consequently K) with a finite subcover.
answered Jan 17 at 10:13
KlausKlaus
1,8179
answered Jan 17 at 10:13
KlausKlaus
1,8179
answered Jan 17 at 10:13
KlausKlaus
1,8179
answered Jan 17 at 10:13
KlausKlaus
1,8179
1,8179
$begingroup$
I know I'm being annoying, but could you elaborate a bit more on the fact "you can find a neighborhood of each point not containing any other point of $E$"?
$endgroup$
– user8469759
Jan 17 at 10:21
$begingroup$
Well, what is your definition of limit point?
$endgroup$
– Klaus
Jan 17 at 10:30
$begingroup$
x is a limit point for E if every neighborhood intersects E (except for x)
$endgroup$
– user8469759
Jan 17 at 10:33
$begingroup$
Hence if $x$ is not a limit point, there is a neighborhood that does not intersect $E$ (except for $x$). ;-)
$endgroup$
– Klaus
Jan 17 at 10:36
$begingroup$
Ok, I struggled with the negation... shame on me. I'll give a second read to the proof given your explanation as well.
$endgroup$
– user8469759
Jan 17 at 10:38
add a comment |
$begingroup$
I know I'm being annoying, but could you elaborate a bit more on the fact "you can find a neighborhood of each point not containing any other point of $E$"?
$endgroup$
– user8469759
Jan 17 at 10:21
$begingroup$
Well, what is your definition of limit point?
$endgroup$
– Klaus
Jan 17 at 10:30
$begingroup$
x is a limit point for E if every neighborhood intersects E (except for x)
$endgroup$
– user8469759
Jan 17 at 10:33
$begingroup$
Hence if $x$ is not a limit point, there is a neighborhood that does not intersect $E$ (except for $x$). ;-)
$endgroup$
– Klaus
Jan 17 at 10:36
$begingroup$
Ok, I struggled with the negation... shame on me. I'll give a second read to the proof given your explanation as well.
$endgroup$
– user8469759
Jan 17 at 10:38
$begingroup$
I know I'm being annoying, but could you elaborate a bit more on the fact "you can find a neighborhood of each point not containing any other point of $E$"?
$endgroup$
– user8469759
Jan 17 at 10:21
$begingroup$
I know I'm being annoying, but could you elaborate a bit more on the fact "you can find a neighborhood of each point not containing any other point of $E$"?
$endgroup$
– user8469759
Jan 17 at 10:21
$begingroup$
Well, what is your definition of limit point?
$endgroup$
– Klaus
Jan 17 at 10:30
$begingroup$
Well, what is your definition of limit point?
$endgroup$
– Klaus
Jan 17 at 10:30
$begingroup$
x is a limit point for E if every neighborhood intersects E (except for x)
$endgroup$
– user8469759
Jan 17 at 10:33
$begingroup$
x is a limit point for E if every neighborhood intersects E (except for x)
$endgroup$
– user8469759
Jan 17 at 10:33
$begingroup$
Hence if $x$ is not a limit point, there is a neighborhood that does not intersect $E$ (except for $x$). ;-)
$endgroup$
– Klaus
Jan 17 at 10:36
$begingroup$
Hence if $x$ is not a limit point, there is a neighborhood that does not intersect $E$ (except for $x$). ;-)
$endgroup$
– Klaus
Jan 17 at 10:36
$begingroup$
Ok, I struggled with the negation... shame on me. I'll give a second read to the proof given your explanation as well.
$endgroup$
– user8469759
Jan 17 at 10:38
$begingroup$
Ok, I struggled with the negation... shame on me. I'll give a second read to the proof given your explanation as well.
$endgroup$
– user8469759
Jan 17 at 10:38
add a comment |
$begingroup$
For every $xin E$ you can choose an open set $V_x$ such that $Ecap V_x={x}$.
For every $xin K-E$ you can choose an open set $V_x$ such that $xin V_x$ and $Ecap V_x=varnothing$.
This because every $xin K$ is not a limit point of $E$.
The collection ${V_xmid xin K}$ is an open cover of $K$.
answered Jan 17 at 10:13
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
For every $xin E$ you can choose an open set $V_x$ such that $Ecap V_x={x}$.
For every $xin K-E$ you can choose an open set $V_x$ such that $xin V_x$ and $Ecap V_x=varnothing$.
This because every $xin K$ is not a limit point of $E$.
The collection ${V_xmid xin K}$ is an open cover of $K$.
answered Jan 17 at 10:13
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
For every $xin E$ you can choose an open set $V_x$ such that $Ecap V_x={x}$.
For every $xin K-E$ you can choose an open set $V_x$ such that $xin V_x$ and $Ecap V_x=varnothing$.
This because every $xin K$ is not a limit point of $E$.
The collection ${V_xmid xin K}$ is an open cover of $K$.
answered Jan 17 at 10:13
drhabdrhab
102k545136
$endgroup$
For every $xin E$ you can choose an open set $V_x$ such that $Ecap V_x={x}$.
For every $xin K-E$ you can choose an open set $V_x$ such that $xin V_x$ and $Ecap V_x=varnothing$.
This because every $xin K$ is not a limit point of $E$.
The collection ${V_xmid xin K}$ is an open cover of $K$.
answered Jan 17 at 10:13
drhabdrhab
102k545136
answered Jan 17 at 10:13
drhabdrhab
102k545136
answered Jan 17 at 10:13
drhabdrhab
102k545136
answered Jan 17 at 10:13
drhabdrhab
102k545136
102k545136
add a comment |
add a comment |
$begingroup$
Attempt (Credit to drhab).
Assume $E subset K$ is infinite and no point of $K$ is a limit point.
1) Let $e in E$. Then , since $e in K$, $e$ is not a limit point of $E$.
There is an open $U_e$ s.t. $U_ecap E =$ {$e$}.
2) Let $y in K-E_y$. then since $y$ is not a limit point of $E$, there is an open $V_y$ s.t. $V_y cap E = emptyset$.
3) $displaystyle{cup_{e in E}} U_e displaystyle{cup_{y in K-E}} V_y$ is an open cover of $K$ that does not have a finite subcover(Why?).
Contradiction to $K$ compact.
answered Jan 17 at 17:01
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
$begingroup$
Your point 3) isn't clear.
$endgroup$
– user8469759
Jan 17 at 17:35
$begingroup$
User8469759.Please explain. $E subset cup_{e in E} U_e$, likewise the other union : $K -E subset cup_{y in K-E} V_y$.
$endgroup$
– Peter Szilas
Jan 17 at 18:07
$begingroup$
Just an issue with the latex, it's ok now.
$endgroup$
– user8469759
Jan 17 at 18:20
$begingroup$
User8469759.Thanks.
$endgroup$
– Peter Szilas
Jan 17 at 18:29
add a comment |
$begingroup$
Attempt (Credit to drhab).
Assume $E subset K$ is infinite and no point of $K$ is a limit point.
1) Let $e in E$. Then , since $e in K$, $e$ is not a limit point of $E$.
There is an open $U_e$ s.t. $U_ecap E =$ {$e$}.
2) Let $y in K-E_y$. then since $y$ is not a limit point of $E$, there is an open $V_y$ s.t. $V_y cap E = emptyset$.
3) $displaystyle{cup_{e in E}} U_e displaystyle{cup_{y in K-E}} V_y$ is an open cover of $K$ that does not have a finite subcover(Why?).
Contradiction to $K$ compact.
answered Jan 17 at 17:01
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
$begingroup$
Your point 3) isn't clear.
$endgroup$
– user8469759
Jan 17 at 17:35
$begingroup$
User8469759.Please explain. $E subset cup_{e in E} U_e$, likewise the other union : $K -E subset cup_{y in K-E} V_y$.
$endgroup$
– Peter Szilas
Jan 17 at 18:07
$begingroup$
Just an issue with the latex, it's ok now.
$endgroup$
– user8469759
Jan 17 at 18:20
$begingroup$
User8469759.Thanks.
$endgroup$
– Peter Szilas
Jan 17 at 18:29
add a comment |
$begingroup$
Attempt (Credit to drhab).
Assume $E subset K$ is infinite and no point of $K$ is a limit point.
1) Let $e in E$. Then , since $e in K$, $e$ is not a limit point of $E$.
There is an open $U_e$ s.t. $U_ecap E =$ {$e$}.
2) Let $y in K-E_y$. then since $y$ is not a limit point of $E$, there is an open $V_y$ s.t. $V_y cap E = emptyset$.
3) $displaystyle{cup_{e in E}} U_e displaystyle{cup_{y in K-E}} V_y$ is an open cover of $K$ that does not have a finite subcover(Why?).
Contradiction to $K$ compact.
answered Jan 17 at 17:01
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
Attempt (Credit to drhab).
Assume $E subset K$ is infinite and no point of $K$ is a limit point.
1) Let $e in E$. Then , since $e in K$, $e$ is not a limit point of $E$.
There is an open $U_e$ s.t. $U_ecap E =$ {$e$}.
2) Let $y in K-E_y$. then since $y$ is not a limit point of $E$, there is an open $V_y$ s.t. $V_y cap E = emptyset$.
3) $displaystyle{cup_{e in E}} U_e displaystyle{cup_{y in K-E}} V_y$ is an open cover of $K$ that does not have a finite subcover(Why?).
Contradiction to $K$ compact.
answered Jan 17 at 17:01
Peter SzilasPeter Szilas
11.4k2822
edited Jan 17 at 18:12
answered Jan 17 at 17:01
Peter SzilasPeter Szilas
11.4k2822
answered Jan 17 at 17:01
Peter SzilasPeter Szilas
11.4k2822
answered Jan 17 at 17:01
Peter SzilasPeter Szilas
11.4k2822
11.4k2822
$begingroup$
Your point 3) isn't clear.
$endgroup$
– user8469759
Jan 17 at 17:35
$begingroup$
User8469759.Please explain. $E subset cup_{e in E} U_e$, likewise the other union : $K -E subset cup_{y in K-E} V_y$.
$endgroup$
– Peter Szilas
Jan 17 at 18:07
$begingroup$
Just an issue with the latex, it's ok now.
$endgroup$
– user8469759
Jan 17 at 18:20
$begingroup$
User8469759.Thanks.
$endgroup$
– Peter Szilas
Jan 17 at 18:29
add a comment |
$begingroup$
Your point 3) isn't clear.
$endgroup$
– user8469759
Jan 17 at 17:35
$begingroup$
User8469759.Please explain. $E subset cup_{e in E} U_e$, likewise the other union : $K -E subset cup_{y in K-E} V_y$.
$endgroup$
– Peter Szilas
Jan 17 at 18:07
$begingroup$
Just an issue with the latex, it's ok now.
$endgroup$
– user8469759
Jan 17 at 18:20
$begingroup$
User8469759.Thanks.
$endgroup$
– Peter Szilas
Jan 17 at 18:29
$begingroup$
Your point 3) isn't clear.
$endgroup$
– user8469759
Jan 17 at 17:35
$begingroup$
Your point 3) isn't clear.
$endgroup$
– user8469759
Jan 17 at 17:35
$begingroup$
User8469759.Please explain. $E subset cup_{e in E} U_e$, likewise the other union : $K -E subset cup_{y in K-E} V_y$.
$endgroup$
– Peter Szilas
Jan 17 at 18:07
$begingroup$
User8469759.Please explain. $E subset cup_{e in E} U_e$, likewise the other union : $K -E subset cup_{y in K-E} V_y$.
$endgroup$
– Peter Szilas
Jan 17 at 18:07
$begingroup$
Just an issue with the latex, it's ok now.
$endgroup$
– user8469759
Jan 17 at 18:20
$begingroup$
Just an issue with the latex, it's ok now.
$endgroup$
– user8469759
Jan 17 at 18:20
$begingroup$
User8469759.Thanks.
$endgroup$
– Peter Szilas
Jan 17 at 18:29
$begingroup$
User8469759.Thanks.
$endgroup$
– Peter Szilas
Jan 17 at 18:29
add a comment |
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