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Strategy to prove a set is compact. Case: $S={0}cupbigcup_limits{i=1}^{infty}{frac{1}{n}}$ [duplicate]
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This question already has an answer here:
Prove that $ S={0}cupleft(bigcup_{n=0}^{infty} {frac{1}{n}}right)$ is a compact set in $mathbb{R}$.
3 answers
Verify that $S={0}cupbigcup_limits{i=1}^{infty}{frac{1}{n}}$ is compact subset of $mathbb{R}$ while $bigcup_limits{i=1}^{infty}{frac{1}{n}}$ is not.
If I take the sequences ${frac{1}{i}:iinmathbb{N}}$ there is no convergent subsequnce once ${0}notin bigcup_limits{i=1}^{infty}{frac{1}{n}}$ . So $bigcup_limits{i=1}^{infty}{frac{1}{n}}$ is not compact.
However when I examine $S={0}cupbigcup_limits{i=1}^{infty}{frac{1}{n}}$. I do not know how can I address all the coverings in order to prove it is compact.
Questions:
Is my proof right insofar? How should I prove $S$ is not compact? What is the strategy for that sort of cases?
Thanks in advance!
general-topology proof-verification compactness
edited Jan 21 at 19:13
José Carlos Santos
166k22132235
asked Jan 21 at 18:38
Pedro GomesPedro Gomes
1,9012721
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marked as duplicate by Math_QED, Lord Shark the Unknown, onurcanbektas, José Carlos Santos general-topology
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Jan 22 at 9:56
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.
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This question already has an answer here:
Prove that $ S={0}cupleft(bigcup_{n=0}^{infty} {frac{1}{n}}right)$ is a compact set in $mathbb{R}$.
3 answers
Verify that $S={0}cupbigcup_limits{i=1}^{infty}{frac{1}{n}}$ is compact subset of $mathbb{R}$ while $bigcup_limits{i=1}^{infty}{frac{1}{n}}$ is not.
If I take the sequences ${frac{1}{i}:iinmathbb{N}}$ there is no convergent subsequnce once ${0}notin bigcup_limits{i=1}^{infty}{frac{1}{n}}$ . So $bigcup_limits{i=1}^{infty}{frac{1}{n}}$ is not compact.
However when I examine $S={0}cupbigcup_limits{i=1}^{infty}{frac{1}{n}}$. I do not know how can I address all the coverings in order to prove it is compact.
Questions:
Is my proof right insofar? How should I prove $S$ is not compact? What is the strategy for that sort of cases?
Thanks in advance!
general-topology proof-verification compactness
edited Jan 21 at 19:13
José Carlos Santos
166k22132235
asked Jan 21 at 18:38
Pedro GomesPedro Gomes
1,9012721
$endgroup$
marked as duplicate by Math_QED, Lord Shark the Unknown, onurcanbektas, José Carlos Santos general-topology
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Jan 22 at 9:56
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.
3
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Possible duplicate of Prove that $ S={0}cupleft(bigcup_{n=0}^{infty} {frac{1}{n}}right)$ is a compact set in $mathbb{R}$. - It was until I added an answer that I saw this duplicate, so I kept it as it might be useful for other users.
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– Math_QED
Jan 21 at 19:10
add a comment |
$begingroup$
This question already has an answer here:
Prove that $ S={0}cupleft(bigcup_{n=0}^{infty} {frac{1}{n}}right)$ is a compact set in $mathbb{R}$.
3 answers
Verify that $S={0}cupbigcup_limits{i=1}^{infty}{frac{1}{n}}$ is compact subset of $mathbb{R}$ while $bigcup_limits{i=1}^{infty}{frac{1}{n}}$ is not.
If I take the sequences ${frac{1}{i}:iinmathbb{N}}$ there is no convergent subsequnce once ${0}notin bigcup_limits{i=1}^{infty}{frac{1}{n}}$ . So $bigcup_limits{i=1}^{infty}{frac{1}{n}}$ is not compact.
However when I examine $S={0}cupbigcup_limits{i=1}^{infty}{frac{1}{n}}$. I do not know how can I address all the coverings in order to prove it is compact.
Questions:
Is my proof right insofar? How should I prove $S$ is not compact? What is the strategy for that sort of cases?
Thanks in advance!
general-topology proof-verification compactness
edited Jan 21 at 19:13
José Carlos Santos
166k22132235
asked Jan 21 at 18:38
Pedro GomesPedro Gomes
1,9012721
$endgroup$
This question already has an answer here:
Prove that $ S={0}cupleft(bigcup_{n=0}^{infty} {frac{1}{n}}right)$ is a compact set in $mathbb{R}$.
3 answers
Verify that $S={0}cupbigcup_limits{i=1}^{infty}{frac{1}{n}}$ is compact subset of $mathbb{R}$ while $bigcup_limits{i=1}^{infty}{frac{1}{n}}$ is not.
If I take the sequences ${frac{1}{i}:iinmathbb{N}}$ there is no convergent subsequnce once ${0}notin bigcup_limits{i=1}^{infty}{frac{1}{n}}$ . So $bigcup_limits{i=1}^{infty}{frac{1}{n}}$ is not compact.
However when I examine $S={0}cupbigcup_limits{i=1}^{infty}{frac{1}{n}}$. I do not know how can I address all the coverings in order to prove it is compact.
Questions:
Is my proof right insofar? How should I prove $S$ is not compact? What is the strategy for that sort of cases?
Thanks in advance!
This question already has an answer here:
Prove that $ S={0}cupleft(bigcup_{n=0}^{infty} {frac{1}{n}}right)$ is a compact set in $mathbb{R}$.
3 answers
general-topology proof-verification compactness
general-topology proof-verification compactness
edited Jan 21 at 19:13
José Carlos Santos
166k22132235
asked Jan 21 at 18:38
Pedro GomesPedro Gomes
1,9012721
edited Jan 21 at 19:13
José Carlos Santos
166k22132235
asked Jan 21 at 18:38
Pedro GomesPedro Gomes
1,9012721
edited Jan 21 at 19:13
José Carlos Santos
166k22132235
edited Jan 21 at 19:13
José Carlos Santos
166k22132235
edited Jan 21 at 19:13
José Carlos Santos
166k22132235
166k22132235
asked Jan 21 at 18:38
Pedro GomesPedro Gomes
1,9012721
asked Jan 21 at 18:38
Pedro GomesPedro Gomes
1,9012721
asked Jan 21 at 18:38
Pedro GomesPedro Gomes
1,9012721
1,9012721
marked as duplicate by Math_QED, Lord Shark the Unknown, onurcanbektas, José Carlos Santos general-topology
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Jan 22 at 9:56
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 Math_QED, Lord Shark the Unknown, onurcanbektas, José Carlos Santos general-topology
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Jan 22 at 9:56
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.
3
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Possible duplicate of Prove that $ S={0}cupleft(bigcup_{n=0}^{infty} {frac{1}{n}}right)$ is a compact set in $mathbb{R}$. - It was until I added an answer that I saw this duplicate, so I kept it as it might be useful for other users.
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– Math_QED
Jan 21 at 19:10
add a comment |
3
$begingroup$
Possible duplicate of Prove that $ S={0}cupleft(bigcup_{n=0}^{infty} {frac{1}{n}}right)$ is a compact set in $mathbb{R}$. - It was until I added an answer that I saw this duplicate, so I kept it as it might be useful for other users.
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– Math_QED
Jan 21 at 19:10
3
3
$begingroup$
Possible duplicate of Prove that $ S={0}cupleft(bigcup_{n=0}^{infty} {frac{1}{n}}right)$ is a compact set in $mathbb{R}$. - It was until I added an answer that I saw this duplicate, so I kept it as it might be useful for other users.
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– Math_QED
Jan 21 at 19:10
$begingroup$
Possible duplicate of Prove that $ S={0}cupleft(bigcup_{n=0}^{infty} {frac{1}{n}}right)$ is a compact set in $mathbb{R}$. - It was until I added an answer that I saw this duplicate, so I kept it as it might be useful for other users.
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Jan 21 at 19:10
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4 Answers
4
active
oldest
votes
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You proved correctly that the set $left{frac1n,middle|,ninmathbb{N}right}$ is not compact. If you add ${0}$, then it becomes compat because for any open cover $(A_lambda)_{lambdainLambda}$ of that set, $0in A_{lambda_0}$, for some $lambda_0inLambda$. And, since $lim_{ntoinfty}frac1n=0$, there is some $pinmathbb N$ such that $ngeqslant pimpliesfrac1nin A_{lambda_0}$. And if $n<p$, take some $lambda_ninLambda$ such that $frac1nin A_{lambda_n}$. So$${0}cupleft{frac1n,middle|,ninmathbb{N}right}subset A_{lambda_0}cup A_{lambda_1}cupcdotscup A_{lambda_{p-1}}.$$
answered Jan 21 at 18:46
José Carlos SantosJosé Carlos Santos
166k22132235
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The first part is right. The sequence you found has no convergent subsequence.
For $S$ take some open covering $U_i$. Then some $U_j$ must contain $0$. Now observe that this $U_j$ already contains all but finitely many $frac{1}{n}$ and for the finitely many exceptions you can find finitely many other sets.
answered Jan 21 at 18:45
Paul KPaul K
2,775416
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add a comment |
$begingroup$
Intuitively $Ssetminus{0}$ is not compact because you can cover it by $bigcuplimits_{i=1}^{infty} B_n$ where $B_n=B(frac 1n,frac 1{4n})$ which is infinite and disjoint so you cannot extract a finite covering out of it.
But a covering of $S$ should at least contain a neighbourhood of $0$ or to simplify a ball $B(0,varepsilon)$.
Yet $varepsilon$ is fixed, so it will contain all balls $B_n$ for $n>n_0$ large enough (i.e. $n_0>frac 1varepsilon$). Thus you are left with $n_0+1$ balls which is a finite covering.
If instead of taking balls, you take random open sets, just consider the intersections of all sets not containing $0$ with the open set containing $0$. Similarly the infinity of sets vanishes in the set containing $0$ and it remains only a finite number of them for the rest of the points away for $0$.
answered Jan 21 at 18:58
zwimzwim
12.5k831
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add a comment |
$begingroup$
You can prove in general that if $(X,d)$ is a metric space and $x_n to x$ in $X$ for the metric $d$, then ${x_n mid n in mathbb{N}}cup {x}$ is compact. Your question is a special case of this.
answered Jan 21 at 19:08
Math_QEDMath_QED
7,69231453
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add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You proved correctly that the set $left{frac1n,middle|,ninmathbb{N}right}$ is not compact. If you add ${0}$, then it becomes compat because for any open cover $(A_lambda)_{lambdainLambda}$ of that set, $0in A_{lambda_0}$, for some $lambda_0inLambda$. And, since $lim_{ntoinfty}frac1n=0$, there is some $pinmathbb N$ such that $ngeqslant pimpliesfrac1nin A_{lambda_0}$. And if $n<p$, take some $lambda_ninLambda$ such that $frac1nin A_{lambda_n}$. So$${0}cupleft{frac1n,middle|,ninmathbb{N}right}subset A_{lambda_0}cup A_{lambda_1}cupcdotscup A_{lambda_{p-1}}.$$
answered Jan 21 at 18:46
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
add a comment |
$begingroup$
You proved correctly that the set $left{frac1n,middle|,ninmathbb{N}right}$ is not compact. If you add ${0}$, then it becomes compat because for any open cover $(A_lambda)_{lambdainLambda}$ of that set, $0in A_{lambda_0}$, for some $lambda_0inLambda$. And, since $lim_{ntoinfty}frac1n=0$, there is some $pinmathbb N$ such that $ngeqslant pimpliesfrac1nin A_{lambda_0}$. And if $n<p$, take some $lambda_ninLambda$ such that $frac1nin A_{lambda_n}$. So$${0}cupleft{frac1n,middle|,ninmathbb{N}right}subset A_{lambda_0}cup A_{lambda_1}cupcdotscup A_{lambda_{p-1}}.$$
answered Jan 21 at 18:46
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
add a comment |
$begingroup$
You proved correctly that the set $left{frac1n,middle|,ninmathbb{N}right}$ is not compact. If you add ${0}$, then it becomes compat because for any open cover $(A_lambda)_{lambdainLambda}$ of that set, $0in A_{lambda_0}$, for some $lambda_0inLambda$. And, since $lim_{ntoinfty}frac1n=0$, there is some $pinmathbb N$ such that $ngeqslant pimpliesfrac1nin A_{lambda_0}$. And if $n<p$, take some $lambda_ninLambda$ such that $frac1nin A_{lambda_n}$. So$${0}cupleft{frac1n,middle|,ninmathbb{N}right}subset A_{lambda_0}cup A_{lambda_1}cupcdotscup A_{lambda_{p-1}}.$$
answered Jan 21 at 18:46
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
You proved correctly that the set $left{frac1n,middle|,ninmathbb{N}right}$ is not compact. If you add ${0}$, then it becomes compat because for any open cover $(A_lambda)_{lambdainLambda}$ of that set, $0in A_{lambda_0}$, for some $lambda_0inLambda$. And, since $lim_{ntoinfty}frac1n=0$, there is some $pinmathbb N$ such that $ngeqslant pimpliesfrac1nin A_{lambda_0}$. And if $n<p$, take some $lambda_ninLambda$ such that $frac1nin A_{lambda_n}$. So$${0}cupleft{frac1n,middle|,ninmathbb{N}right}subset A_{lambda_0}cup A_{lambda_1}cupcdotscup A_{lambda_{p-1}}.$$
answered Jan 21 at 18:46
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 21 at 18:46
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 21 at 18:46
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 21 at 18:46
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
add a comment |
add a comment |
$begingroup$
The first part is right. The sequence you found has no convergent subsequence.
For $S$ take some open covering $U_i$. Then some $U_j$ must contain $0$. Now observe that this $U_j$ already contains all but finitely many $frac{1}{n}$ and for the finitely many exceptions you can find finitely many other sets.
answered Jan 21 at 18:45
Paul KPaul K
2,775416
$endgroup$
add a comment |
$begingroup$
The first part is right. The sequence you found has no convergent subsequence.
For $S$ take some open covering $U_i$. Then some $U_j$ must contain $0$. Now observe that this $U_j$ already contains all but finitely many $frac{1}{n}$ and for the finitely many exceptions you can find finitely many other sets.
answered Jan 21 at 18:45
Paul KPaul K
2,775416
$endgroup$
add a comment |
$begingroup$
The first part is right. The sequence you found has no convergent subsequence.
For $S$ take some open covering $U_i$. Then some $U_j$ must contain $0$. Now observe that this $U_j$ already contains all but finitely many $frac{1}{n}$ and for the finitely many exceptions you can find finitely many other sets.
answered Jan 21 at 18:45
Paul KPaul K
2,775416
$endgroup$
The first part is right. The sequence you found has no convergent subsequence.
For $S$ take some open covering $U_i$. Then some $U_j$ must contain $0$. Now observe that this $U_j$ already contains all but finitely many $frac{1}{n}$ and for the finitely many exceptions you can find finitely many other sets.
answered Jan 21 at 18:45
Paul KPaul K
2,775416
answered Jan 21 at 18:45
Paul KPaul K
2,775416
answered Jan 21 at 18:45
Paul KPaul K
2,775416
answered Jan 21 at 18:45
Paul KPaul K
2,775416
2,775416
add a comment |
add a comment |
$begingroup$
Intuitively $Ssetminus{0}$ is not compact because you can cover it by $bigcuplimits_{i=1}^{infty} B_n$ where $B_n=B(frac 1n,frac 1{4n})$ which is infinite and disjoint so you cannot extract a finite covering out of it.
But a covering of $S$ should at least contain a neighbourhood of $0$ or to simplify a ball $B(0,varepsilon)$.
Yet $varepsilon$ is fixed, so it will contain all balls $B_n$ for $n>n_0$ large enough (i.e. $n_0>frac 1varepsilon$). Thus you are left with $n_0+1$ balls which is a finite covering.
If instead of taking balls, you take random open sets, just consider the intersections of all sets not containing $0$ with the open set containing $0$. Similarly the infinity of sets vanishes in the set containing $0$ and it remains only a finite number of them for the rest of the points away for $0$.
answered Jan 21 at 18:58
zwimzwim
12.5k831
$endgroup$
add a comment |
$begingroup$
Intuitively $Ssetminus{0}$ is not compact because you can cover it by $bigcuplimits_{i=1}^{infty} B_n$ where $B_n=B(frac 1n,frac 1{4n})$ which is infinite and disjoint so you cannot extract a finite covering out of it.
But a covering of $S$ should at least contain a neighbourhood of $0$ or to simplify a ball $B(0,varepsilon)$.
Yet $varepsilon$ is fixed, so it will contain all balls $B_n$ for $n>n_0$ large enough (i.e. $n_0>frac 1varepsilon$). Thus you are left with $n_0+1$ balls which is a finite covering.
If instead of taking balls, you take random open sets, just consider the intersections of all sets not containing $0$ with the open set containing $0$. Similarly the infinity of sets vanishes in the set containing $0$ and it remains only a finite number of them for the rest of the points away for $0$.
answered Jan 21 at 18:58
zwimzwim
12.5k831
$endgroup$
add a comment |
$begingroup$
Intuitively $Ssetminus{0}$ is not compact because you can cover it by $bigcuplimits_{i=1}^{infty} B_n$ where $B_n=B(frac 1n,frac 1{4n})$ which is infinite and disjoint so you cannot extract a finite covering out of it.
But a covering of $S$ should at least contain a neighbourhood of $0$ or to simplify a ball $B(0,varepsilon)$.
Yet $varepsilon$ is fixed, so it will contain all balls $B_n$ for $n>n_0$ large enough (i.e. $n_0>frac 1varepsilon$). Thus you are left with $n_0+1$ balls which is a finite covering.
If instead of taking balls, you take random open sets, just consider the intersections of all sets not containing $0$ with the open set containing $0$. Similarly the infinity of sets vanishes in the set containing $0$ and it remains only a finite number of them for the rest of the points away for $0$.
answered Jan 21 at 18:58
zwimzwim
12.5k831
$endgroup$
Intuitively $Ssetminus{0}$ is not compact because you can cover it by $bigcuplimits_{i=1}^{infty} B_n$ where $B_n=B(frac 1n,frac 1{4n})$ which is infinite and disjoint so you cannot extract a finite covering out of it.
But a covering of $S$ should at least contain a neighbourhood of $0$ or to simplify a ball $B(0,varepsilon)$.
Yet $varepsilon$ is fixed, so it will contain all balls $B_n$ for $n>n_0$ large enough (i.e. $n_0>frac 1varepsilon$). Thus you are left with $n_0+1$ balls which is a finite covering.
If instead of taking balls, you take random open sets, just consider the intersections of all sets not containing $0$ with the open set containing $0$. Similarly the infinity of sets vanishes in the set containing $0$ and it remains only a finite number of them for the rest of the points away for $0$.
answered Jan 21 at 18:58
zwimzwim
12.5k831
answered Jan 21 at 18:58
zwimzwim
12.5k831
answered Jan 21 at 18:58
zwimzwim
12.5k831
answered Jan 21 at 18:58
zwimzwim
12.5k831
12.5k831
add a comment |
add a comment |
$begingroup$
You can prove in general that if $(X,d)$ is a metric space and $x_n to x$ in $X$ for the metric $d$, then ${x_n mid n in mathbb{N}}cup {x}$ is compact. Your question is a special case of this.
answered Jan 21 at 19:08
Math_QEDMath_QED
7,69231453
$endgroup$
add a comment |
$begingroup$
You can prove in general that if $(X,d)$ is a metric space and $x_n to x$ in $X$ for the metric $d$, then ${x_n mid n in mathbb{N}}cup {x}$ is compact. Your question is a special case of this.
answered Jan 21 at 19:08
Math_QEDMath_QED
7,69231453
$endgroup$
add a comment |
$begingroup$
You can prove in general that if $(X,d)$ is a metric space and $x_n to x$ in $X$ for the metric $d$, then ${x_n mid n in mathbb{N}}cup {x}$ is compact. Your question is a special case of this.
answered Jan 21 at 19:08
Math_QEDMath_QED
7,69231453
$endgroup$
You can prove in general that if $(X,d)$ is a metric space and $x_n to x$ in $X$ for the metric $d$, then ${x_n mid n in mathbb{N}}cup {x}$ is compact. Your question is a special case of this.
answered Jan 21 at 19:08
Math_QEDMath_QED
7,69231453
answered Jan 21 at 19:08
Math_QEDMath_QED
7,69231453
answered Jan 21 at 19:08
Math_QEDMath_QED
7,69231453
answered Jan 21 at 19:08
Math_QEDMath_QED
7,69231453
7,69231453
add a comment |
add a comment |
3
$begingroup$
Possible duplicate of Prove that $ S={0}cupleft(bigcup_{n=0}^{infty} {frac{1}{n}}right)$ is a compact set in $mathbb{R}$. - It was until I added an answer that I saw this duplicate, so I kept it as it might be useful for other users.
$endgroup$
– Math_QED
Jan 21 at 19:10