Prove that $K$ is a compact set by directly using the definition of compactness.
Let $X = mathbb R$ with usual distance metric. Let $K = {0}cup {1/n | n in N}$. Prove
that $K$ is a compact set by directly using the definition of compactness.
Definition: A set $C subset E$ is said to be compact if every collection of open sets that covers $C$ has a finite sub-collection that covers $C$. The metric space $(E,d)$ is said to be compact if E is so.
Can we start by constructing balls centered at $1/n$ with radius 1 and say it is an open cover of K. But since in the definition it says every open collection I do not know if it right to select a specific one.
real-analysis proof-verification metric-spaces compactness
edited Dec 28 '18 at 9:39
José Carlos Santos
152k22123225
asked Dec 28 '18 at 9:01
Pumpkin
5021417
add a comment |
Let $X = mathbb R$ with usual distance metric. Let $K = {0}cup {1/n | n in N}$. Prove
that $K$ is a compact set by directly using the definition of compactness.
Definition: A set $C subset E$ is said to be compact if every collection of open sets that covers $C$ has a finite sub-collection that covers $C$. The metric space $(E,d)$ is said to be compact if E is so.
Can we start by constructing balls centered at $1/n$ with radius 1 and say it is an open cover of K. But since in the definition it says every open collection I do not know if it right to select a specific one.
real-analysis proof-verification metric-spaces compactness
edited Dec 28 '18 at 9:39
José Carlos Santos
152k22123225
asked Dec 28 '18 at 9:01
Pumpkin
5021417
add a comment |
Let $X = mathbb R$ with usual distance metric. Let $K = {0}cup {1/n | n in N}$. Prove
that $K$ is a compact set by directly using the definition of compactness.
Definition: A set $C subset E$ is said to be compact if every collection of open sets that covers $C$ has a finite sub-collection that covers $C$. The metric space $(E,d)$ is said to be compact if E is so.
Can we start by constructing balls centered at $1/n$ with radius 1 and say it is an open cover of K. But since in the definition it says every open collection I do not know if it right to select a specific one.
real-analysis proof-verification metric-spaces compactness
edited Dec 28 '18 at 9:39
José Carlos Santos
152k22123225
asked Dec 28 '18 at 9:01
Pumpkin
5021417
Let $X = mathbb R$ with usual distance metric. Let $K = {0}cup {1/n | n in N}$. Prove
that $K$ is a compact set by directly using the definition of compactness.
Definition: A set $C subset E$ is said to be compact if every collection of open sets that covers $C$ has a finite sub-collection that covers $C$. The metric space $(E,d)$ is said to be compact if E is so.
Can we start by constructing balls centered at $1/n$ with radius 1 and say it is an open cover of K. But since in the definition it says every open collection I do not know if it right to select a specific one.
real-analysis proof-verification metric-spaces compactness
real-analysis proof-verification metric-spaces compactness
edited Dec 28 '18 at 9:39
José Carlos Santos
152k22123225
asked Dec 28 '18 at 9:01
Pumpkin
5021417
edited Dec 28 '18 at 9:39
José Carlos Santos
152k22123225
asked Dec 28 '18 at 9:01
Pumpkin
5021417
edited Dec 28 '18 at 9:39
José Carlos Santos
152k22123225
edited Dec 28 '18 at 9:39
José Carlos Santos
152k22123225
edited Dec 28 '18 at 9:39
José Carlos Santos
152k22123225
152k22123225
asked Dec 28 '18 at 9:01
Pumpkin
5021417
asked Dec 28 '18 at 9:01
Pumpkin
5021417
asked Dec 28 '18 at 9:01
Pumpkin
5021417
5021417
add a comment |
add a comment |
3 Answers
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Right: you cannot prove this starting from a specific cover.
If $mathcal U$ is an open cover of $K$, there is some $A_0inmathcal U$ such that $0in A_0$. Since $A_0$ is open and $lim_{ntoinfty}frac1n=0$, $frac1nin A_0$ is $n$ is large enough. So, there is some $Ninmathbb N$ such that $ngeqslant Nimpliesfrac1nin A_0$. For each $nin{1,2,ldots,N-1}$, let $A_ninmathcal U$ be such that $frac1nin A_n$. Then $bigl{A_n,|,nin{0,1,ldots,N-1}bigr}$ is a finite subcover of $mathcal U$.
answered Dec 28 '18 at 9:09
José Carlos Santos
152k22123225
add a comment |
Let $mathscr{U}$ be a cover of $K$; then, for each $n>0$, there is $U_ninmathscr{U}$ such that $1/nin U_n$.
There is also $U_0inmathscr{U}$ such that $0in U_0$.
Can you go on?
answered Dec 28 '18 at 9:04
egreg
179k1484201
add a comment |
Hint:Let ${U_{alpha}}$ be an open cover of $K$. Let $0in U_{alpha_0}$. Since $U_{alpha_0}$ is open, there exists $rgt 0$ such that $B(0,r)subseteq U_{alpha_0}$. As $0$ is the limit point of the set$Ksetminus {0}= {1/n | n in Bbb N}$,there are only finitely many points of $Ksetminus{0}$ outside of $B(0,r)$. Can you proceed from here?
answered Dec 28 '18 at 9:10
Thomas Shelby
1,887217
add a comment |
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3 Answers
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oldest
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3 Answers
3
active
oldest
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active
oldest
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active
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votes
Right: you cannot prove this starting from a specific cover.
If $mathcal U$ is an open cover of $K$, there is some $A_0inmathcal U$ such that $0in A_0$. Since $A_0$ is open and $lim_{ntoinfty}frac1n=0$, $frac1nin A_0$ is $n$ is large enough. So, there is some $Ninmathbb N$ such that $ngeqslant Nimpliesfrac1nin A_0$. For each $nin{1,2,ldots,N-1}$, let $A_ninmathcal U$ be such that $frac1nin A_n$. Then $bigl{A_n,|,nin{0,1,ldots,N-1}bigr}$ is a finite subcover of $mathcal U$.
answered Dec 28 '18 at 9:09
José Carlos Santos
152k22123225
add a comment |
Right: you cannot prove this starting from a specific cover.
If $mathcal U$ is an open cover of $K$, there is some $A_0inmathcal U$ such that $0in A_0$. Since $A_0$ is open and $lim_{ntoinfty}frac1n=0$, $frac1nin A_0$ is $n$ is large enough. So, there is some $Ninmathbb N$ such that $ngeqslant Nimpliesfrac1nin A_0$. For each $nin{1,2,ldots,N-1}$, let $A_ninmathcal U$ be such that $frac1nin A_n$. Then $bigl{A_n,|,nin{0,1,ldots,N-1}bigr}$ is a finite subcover of $mathcal U$.
answered Dec 28 '18 at 9:09
José Carlos Santos
152k22123225
add a comment |
Right: you cannot prove this starting from a specific cover.
If $mathcal U$ is an open cover of $K$, there is some $A_0inmathcal U$ such that $0in A_0$. Since $A_0$ is open and $lim_{ntoinfty}frac1n=0$, $frac1nin A_0$ is $n$ is large enough. So, there is some $Ninmathbb N$ such that $ngeqslant Nimpliesfrac1nin A_0$. For each $nin{1,2,ldots,N-1}$, let $A_ninmathcal U$ be such that $frac1nin A_n$. Then $bigl{A_n,|,nin{0,1,ldots,N-1}bigr}$ is a finite subcover of $mathcal U$.
answered Dec 28 '18 at 9:09
José Carlos Santos
152k22123225
Right: you cannot prove this starting from a specific cover.
If $mathcal U$ is an open cover of $K$, there is some $A_0inmathcal U$ such that $0in A_0$. Since $A_0$ is open and $lim_{ntoinfty}frac1n=0$, $frac1nin A_0$ is $n$ is large enough. So, there is some $Ninmathbb N$ such that $ngeqslant Nimpliesfrac1nin A_0$. For each $nin{1,2,ldots,N-1}$, let $A_ninmathcal U$ be such that $frac1nin A_n$. Then $bigl{A_n,|,nin{0,1,ldots,N-1}bigr}$ is a finite subcover of $mathcal U$.
answered Dec 28 '18 at 9:09
José Carlos Santos
152k22123225
answered Dec 28 '18 at 9:09
José Carlos Santos
152k22123225
answered Dec 28 '18 at 9:09
José Carlos Santos
152k22123225
answered Dec 28 '18 at 9:09
José Carlos Santos
152k22123225
152k22123225
add a comment |
add a comment |
Let $mathscr{U}$ be a cover of $K$; then, for each $n>0$, there is $U_ninmathscr{U}$ such that $1/nin U_n$.
There is also $U_0inmathscr{U}$ such that $0in U_0$.
Can you go on?
answered Dec 28 '18 at 9:04
egreg
179k1484201
add a comment |
Let $mathscr{U}$ be a cover of $K$; then, for each $n>0$, there is $U_ninmathscr{U}$ such that $1/nin U_n$.
There is also $U_0inmathscr{U}$ such that $0in U_0$.
Can you go on?
answered Dec 28 '18 at 9:04
egreg
179k1484201
add a comment |
Let $mathscr{U}$ be a cover of $K$; then, for each $n>0$, there is $U_ninmathscr{U}$ such that $1/nin U_n$.
There is also $U_0inmathscr{U}$ such that $0in U_0$.
Can you go on?
answered Dec 28 '18 at 9:04
egreg
179k1484201
Let $mathscr{U}$ be a cover of $K$; then, for each $n>0$, there is $U_ninmathscr{U}$ such that $1/nin U_n$.
There is also $U_0inmathscr{U}$ such that $0in U_0$.
Can you go on?
answered Dec 28 '18 at 9:04
egreg
179k1484201
answered Dec 28 '18 at 9:04
egreg
179k1484201
answered Dec 28 '18 at 9:04
egreg
179k1484201
answered Dec 28 '18 at 9:04
egreg
179k1484201
179k1484201
add a comment |
add a comment |
Hint:Let ${U_{alpha}}$ be an open cover of $K$. Let $0in U_{alpha_0}$. Since $U_{alpha_0}$ is open, there exists $rgt 0$ such that $B(0,r)subseteq U_{alpha_0}$. As $0$ is the limit point of the set$Ksetminus {0}= {1/n | n in Bbb N}$,there are only finitely many points of $Ksetminus{0}$ outside of $B(0,r)$. Can you proceed from here?
answered Dec 28 '18 at 9:10
Thomas Shelby
1,887217
add a comment |
Hint:Let ${U_{alpha}}$ be an open cover of $K$. Let $0in U_{alpha_0}$. Since $U_{alpha_0}$ is open, there exists $rgt 0$ such that $B(0,r)subseteq U_{alpha_0}$. As $0$ is the limit point of the set$Ksetminus {0}= {1/n | n in Bbb N}$,there are only finitely many points of $Ksetminus{0}$ outside of $B(0,r)$. Can you proceed from here?
answered Dec 28 '18 at 9:10
Thomas Shelby
1,887217
add a comment |
Hint:Let ${U_{alpha}}$ be an open cover of $K$. Let $0in U_{alpha_0}$. Since $U_{alpha_0}$ is open, there exists $rgt 0$ such that $B(0,r)subseteq U_{alpha_0}$. As $0$ is the limit point of the set$Ksetminus {0}= {1/n | n in Bbb N}$,there are only finitely many points of $Ksetminus{0}$ outside of $B(0,r)$. Can you proceed from here?
answered Dec 28 '18 at 9:10
Thomas Shelby
1,887217
Hint:Let ${U_{alpha}}$ be an open cover of $K$. Let $0in U_{alpha_0}$. Since $U_{alpha_0}$ is open, there exists $rgt 0$ such that $B(0,r)subseteq U_{alpha_0}$. As $0$ is the limit point of the set$Ksetminus {0}= {1/n | n in Bbb N}$,there are only finitely many points of $Ksetminus{0}$ outside of $B(0,r)$. Can you proceed from here?
answered Dec 28 '18 at 9:10
Thomas Shelby
1,887217
edited Dec 28 '18 at 9:29
answered Dec 28 '18 at 9:10
Thomas Shelby
1,887217
answered Dec 28 '18 at 9:10
Thomas Shelby
1,887217
answered Dec 28 '18 at 9:10
Thomas Shelby
1,887217
1,887217
add a comment |
add a comment |
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