Show there is a countable collection of closed balls $beta=B_n=overlineB_r_n(x_n):nin mathbbN$ such that $U=cup_nB_n^textint$'Equivalent' Exhaustion by compact setsShow every open $ U subseteq mathbbR^n$ is the union of a countable, locally finite collection of ballsa compact set $X$ has a countable set $S$ such that $overlineS = X$Give an example of a metric space $(X,d)$ and $Asubseteq X$ such that $textint(overlineA)notsubseteqoverlinetextint(A)$ and vice versa$mathbb R^n$ has countable basis of open balls? (Yes)How to show that the set of open balls with rational centres and rational radii form a countable base for $mathbbR^n$?$int_Uf(x)dx$ converge iff $int_U|f(x)|dx$ convergeProving the ''Countable Complement Extension Topology'' defines a topologyCountably Compact Equivalent to Nested Sequence PropertyShow that for each positive integer $n$, we can choose $x_n in U_n backslash mathbbN $ such that $x_n > n$.
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Show there is a countable collection of closed balls $beta=B_n=overlineB_r_n(x_n):nin mathbbN$ such that $U=cup_nB_n^textint$
'Equivalent' Exhaustion by compact setsShow every open $ U subseteq mathbbR^n$ is the union of a countable, locally finite collection of ballsa compact set $X$ has a countable set $S$ such that $overlineS = X$Give an example of a metric space $(X,d)$ and $Asubseteq X$ such that $textint(overlineA)notsubseteqoverlinetextint(A)$ and vice versa$mathbb R^n$ has countable basis of open balls? (Yes)How to show that the set of open balls with rational centres and rational radii form a countable base for $mathbbR^n$?$int_Uf(x)dx$ converge iff $int_U|f(x)|dx$ convergeProving the ''Countable Complement Extension Topology'' defines a topologyCountably Compact Equivalent to Nested Sequence PropertyShow that for each positive integer $n$, we can choose $x_n in U_n backslash mathbbN $ such that $x_n > n$.
$begingroup$
Let $U= cup C$ be an open set in $mathbbR^n$. Show there is a countable number of closed balls $beta=B_n=overlineB_r_n(x_n):nin mathbbN$ such that:
a.$U=cup_n B_n^textint$
b. For each $nin mathbbN$ there exists an open set $M$ such that $B_nsubseteq M$.
c. For every point $xin U$, there exists a neighbourhood $V$ of $x$ such that only finitely many elements of the cover $C$ intersect $V$ non trivially.
So I want to use compact exhaustion of the open set $U$.
Fix a compact exhaustion $K_n:ninmathbbN$ of $U$
Then since by definition for every $iinmathbbN$ $K_isubseteq K_i+1^textint$
I can make a compact set $L_n=K_nsetminus K_n-1^textint$ which will satisfy that each $L_i$ is compact as its an intersection of closed sets, which will be closed and is bounded by $K_i$.
From here I think what I want to do is to take balls $B_i(x)$ for each $xin L_i$ and take a radius which will not intersect more then $L_i+1$ or $L_i-1$. A set of all such balls $B_n$ in each $L_i$ will admit a finite subcover because each $L_i$ is compact. And the union of all such finite subcovers of each $L_i$ should still cover $U$ and because $L_n$ is countable this set $cup B_n$ will be countable.
I believe this would satisfy part a. but I don't really see part b and c. I also don't know what a non trivial intersection is.
real-analysis general-topology
asked Mar 29 at 22:28
AColoredReptileAColoredReptile
401210
$endgroup$
|
show 2 more comments
$begingroup$
Let $U= cup C$ be an open set in $mathbbR^n$. Show there is a countable number of closed balls $beta=B_n=overlineB_r_n(x_n):nin mathbbN$ such that:
a.$U=cup_n B_n^textint$
b. For each $nin mathbbN$ there exists an open set $M$ such that $B_nsubseteq M$.
c. For every point $xin U$, there exists a neighbourhood $V$ of $x$ such that only finitely many elements of the cover $C$ intersect $V$ non trivially.
So I want to use compact exhaustion of the open set $U$.
Fix a compact exhaustion $K_n:ninmathbbN$ of $U$
Then since by definition for every $iinmathbbN$ $K_isubseteq K_i+1^textint$
I can make a compact set $L_n=K_nsetminus K_n-1^textint$ which will satisfy that each $L_i$ is compact as its an intersection of closed sets, which will be closed and is bounded by $K_i$.
From here I think what I want to do is to take balls $B_i(x)$ for each $xin L_i$ and take a radius which will not intersect more then $L_i+1$ or $L_i-1$. A set of all such balls $B_n$ in each $L_i$ will admit a finite subcover because each $L_i$ is compact. And the union of all such finite subcovers of each $L_i$ should still cover $U$ and because $L_n$ is countable this set $cup B_n$ will be countable.
I believe this would satisfy part a. but I don't really see part b and c. I also don't know what a non trivial intersection is.
real-analysis general-topology
asked Mar 29 at 22:28
AColoredReptileAColoredReptile
401210
$endgroup$
$begingroup$
I think you can do this by considering sets of dyadic cubes, and the corresponding inscribed/circumscribed balls.
$endgroup$
– Matematleta
Mar 29 at 22:52
$begingroup$
I've never seen dyadic cubes before.
$endgroup$
– AColoredReptile
Mar 29 at 22:52
1
$begingroup$
Is there an extra part of the condition in b), or what prevents us to take $M=mathbbR^n$ or $M=U$ for all $n$?
$endgroup$
– Henno Brandsma
Mar 30 at 13:10
$begingroup$
non-trivial intersection is onon-empty intersection, I think.
$endgroup$
– Henno Brandsma
Mar 30 at 13:11
$begingroup$
What is the cover $C$ in condition c)? Do you mean $beta$?
$endgroup$
– Paul Frost
Mar 30 at 17:24
|
show 2 more comments
$begingroup$
Let $U= cup C$ be an open set in $mathbbR^n$. Show there is a countable number of closed balls $beta=B_n=overlineB_r_n(x_n):nin mathbbN$ such that:
a.$U=cup_n B_n^textint$
b. For each $nin mathbbN$ there exists an open set $M$ such that $B_nsubseteq M$.
c. For every point $xin U$, there exists a neighbourhood $V$ of $x$ such that only finitely many elements of the cover $C$ intersect $V$ non trivially.
So I want to use compact exhaustion of the open set $U$.
Fix a compact exhaustion $K_n:ninmathbbN$ of $U$
Then since by definition for every $iinmathbbN$ $K_isubseteq K_i+1^textint$
I can make a compact set $L_n=K_nsetminus K_n-1^textint$ which will satisfy that each $L_i$ is compact as its an intersection of closed sets, which will be closed and is bounded by $K_i$.
From here I think what I want to do is to take balls $B_i(x)$ for each $xin L_i$ and take a radius which will not intersect more then $L_i+1$ or $L_i-1$. A set of all such balls $B_n$ in each $L_i$ will admit a finite subcover because each $L_i$ is compact. And the union of all such finite subcovers of each $L_i$ should still cover $U$ and because $L_n$ is countable this set $cup B_n$ will be countable.
I believe this would satisfy part a. but I don't really see part b and c. I also don't know what a non trivial intersection is.
real-analysis general-topology
asked Mar 29 at 22:28
AColoredReptileAColoredReptile
401210
$endgroup$
Let $U= cup C$ be an open set in $mathbbR^n$. Show there is a countable number of closed balls $beta=B_n=overlineB_r_n(x_n):nin mathbbN$ such that:
a.$U=cup_n B_n^textint$
b. For each $nin mathbbN$ there exists an open set $M$ such that $B_nsubseteq M$.
c. For every point $xin U$, there exists a neighbourhood $V$ of $x$ such that only finitely many elements of the cover $C$ intersect $V$ non trivially.
So I want to use compact exhaustion of the open set $U$.
Fix a compact exhaustion $K_n:ninmathbbN$ of $U$
Then since by definition for every $iinmathbbN$ $K_isubseteq K_i+1^textint$
I can make a compact set $L_n=K_nsetminus K_n-1^textint$ which will satisfy that each $L_i$ is compact as its an intersection of closed sets, which will be closed and is bounded by $K_i$.
From here I think what I want to do is to take balls $B_i(x)$ for each $xin L_i$ and take a radius which will not intersect more then $L_i+1$ or $L_i-1$. A set of all such balls $B_n$ in each $L_i$ will admit a finite subcover because each $L_i$ is compact. And the union of all such finite subcovers of each $L_i$ should still cover $U$ and because $L_n$ is countable this set $cup B_n$ will be countable.
I believe this would satisfy part a. but I don't really see part b and c. I also don't know what a non trivial intersection is.
real-analysis general-topology
real-analysis general-topology
asked Mar 29 at 22:28
AColoredReptileAColoredReptile
401210
asked Mar 29 at 22:28
AColoredReptileAColoredReptile
401210
edited Mar 30 at 17:58
AColoredReptile
asked Mar 29 at 22:28
AColoredReptileAColoredReptile
401210
asked Mar 29 at 22:28
AColoredReptileAColoredReptile
401210
asked Mar 29 at 22:28
AColoredReptileAColoredReptile
401210
401210
$begingroup$
I think you can do this by considering sets of dyadic cubes, and the corresponding inscribed/circumscribed balls.
$endgroup$
– Matematleta
Mar 29 at 22:52
$begingroup$
I've never seen dyadic cubes before.
$endgroup$
– AColoredReptile
Mar 29 at 22:52
1
$begingroup$
Is there an extra part of the condition in b), or what prevents us to take $M=mathbbR^n$ or $M=U$ for all $n$?
$endgroup$
– Henno Brandsma
Mar 30 at 13:10
$begingroup$
non-trivial intersection is onon-empty intersection, I think.
$endgroup$
– Henno Brandsma
Mar 30 at 13:11
$begingroup$
What is the cover $C$ in condition c)? Do you mean $beta$?
$endgroup$
– Paul Frost
Mar 30 at 17:24
|
show 2 more comments
$begingroup$
I think you can do this by considering sets of dyadic cubes, and the corresponding inscribed/circumscribed balls.
$endgroup$
– Matematleta
Mar 29 at 22:52
$begingroup$
I've never seen dyadic cubes before.
$endgroup$
– AColoredReptile
Mar 29 at 22:52
1
$begingroup$
Is there an extra part of the condition in b), or what prevents us to take $M=mathbbR^n$ or $M=U$ for all $n$?
$endgroup$
– Henno Brandsma
Mar 30 at 13:10
$begingroup$
non-trivial intersection is onon-empty intersection, I think.
$endgroup$
– Henno Brandsma
Mar 30 at 13:11
$begingroup$
What is the cover $C$ in condition c)? Do you mean $beta$?
$endgroup$
– Paul Frost
Mar 30 at 17:24
$begingroup$
I think you can do this by considering sets of dyadic cubes, and the corresponding inscribed/circumscribed balls.
$endgroup$
– Matematleta
Mar 29 at 22:52
$begingroup$
I think you can do this by considering sets of dyadic cubes, and the corresponding inscribed/circumscribed balls.
$endgroup$
– Matematleta
Mar 29 at 22:52
$begingroup$
I've never seen dyadic cubes before.
$endgroup$
– AColoredReptile
Mar 29 at 22:52
$begingroup$
I've never seen dyadic cubes before.
$endgroup$
– AColoredReptile
Mar 29 at 22:52
1
1
$begingroup$
Is there an extra part of the condition in b), or what prevents us to take $M=mathbbR^n$ or $M=U$ for all $n$?
$endgroup$
– Henno Brandsma
Mar 30 at 13:10
$begingroup$
Is there an extra part of the condition in b), or what prevents us to take $M=mathbbR^n$ or $M=U$ for all $n$?
$endgroup$
– Henno Brandsma
Mar 30 at 13:10
$begingroup$
non-trivial intersection is onon-empty intersection, I think.
$endgroup$
– Henno Brandsma
Mar 30 at 13:11
$begingroup$
non-trivial intersection is onon-empty intersection, I think.
$endgroup$
– Henno Brandsma
Mar 30 at 13:11
$begingroup$
What is the cover $C$ in condition c)? Do you mean $beta$?
$endgroup$
– Paul Frost
Mar 30 at 17:24
$begingroup$
What is the cover $C$ in condition c)? Do you mean $beta$?
$endgroup$
– Paul Frost
Mar 30 at 17:24
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Here's an answer to part a: This is just the same as, or a slight modification of the proof that was posted earlier.
For any set $U$ the set of balls $$mathcalB=~qin mathbbQ^n, p in mathbbQ_ge 0$$ form a countable cover of $U$.
Let $x$ be an arbitrary point in $U$. Then $x$ is contained in an open ball $B_r_x(x) subseteq U$. Let $q$ be a rational point contained within the ball $B_r_x/3(x)$. Then we can choose a ball $B_p(q) in mathcalB$ with $r_x/3 < p <2r_x/3$. Then $$x in B_p(q) subset B_r_x(x) subseteq U.$$
Hence $U$ is the union of a subset of $mathcalB$.
answered Mar 31 at 2:00
AutumnAutumn
361
$endgroup$
add a comment |
$begingroup$
Take a family $V=V_n:nin Bbb N$ of bounded open sets such that $U=cup V$ and such that $overline V_n subset V_n+1$ for each $n.$ (To get $V,$ see APPENDIX below.)
Let $K_1=overline V_1.$ For $n>1$ let $K_n=bar V_n V_n-1.$ Each $K_n$ is compact.
Let $C_1$ be a finite cover of $K_1$ by open balls, such that $overline cup C_1subset V_2.$ This is possible because $K_1$ is compact, $V_2$ is open, and $K_1subset V_2.$
For $n>1$ let $C_n$ be a finite cover of $K_n$ by open balls, such that $overline cup C_nsubset V_n+1$ and also such that $cup C_n$ is disjoint from $cup_jle n-2(overline cup C_j).$
The last condition above is vacuous when $n=2.$ When $n>2$ it is crucial, and possible because $S_n=cup_jle n-2(overline cup C_j)$ is a subset of $cup_jle n-1V_j=V_n-1,$ and $V_n-1$ is disjoint from $K_n,$ so the closed set $S_n$ and the compact set $K_n$ are disjoint.
Now $C=cup_nin Bbb nC_n$ is a countable family of open balls, and $U=cup C=cup _bin C, int(bar b)=cup_bin C,bar b.$
If $pin U$ then for some $n$ we have $pin V_nsubset overline V_n=cup_jle nK_j$ so for some $jle n$ and some $b in C_j$ we will have $pin bin C_j.$ Now this $b$ is disjoint from $cup C_m$ for every $mge n+2.$ So $b'in C: b'cap bne phi$ is a subset of the finite set $cup_jle n+1C_j.$
APPENDIX. Let $bar 0$ be the origin in $Bbb R^n.$
(i). If $U=Bbb R^n$ let $V_n=B_n(bar 0)$ for each $nin Bbb N.$
(ii). If $Une Bbb R^n,$ let $U^c=Bbb R^n$ $U$ and let $d$ be a metric for the topology on $Bbb R^n,$ and for $pin U$ let $d(p,U^c)=inf d(p,q):qin U^c.$
Now let $V_1=pin Ucap B_1(bar 0): d(p,U^c)>1.$ And for $n>1$ let $V_n= pin Ucap B_n(bar 0): d(p,U^c)>1/n.$ I will leave it to the reader to confirm that $V_n:nin Bbb N$ has the required properties.
answered Mar 31 at 18:24
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
$begingroup$
This works if you replace $Bbb R^n$ with any metric space in which closed bounded subsets are compact.
$endgroup$
– DanielWainfleet
Mar 31 at 18:37
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Here's an answer to part a: This is just the same as, or a slight modification of the proof that was posted earlier.
For any set $U$ the set of balls $$mathcalB=~qin mathbbQ^n, p in mathbbQ_ge 0$$ form a countable cover of $U$.
Let $x$ be an arbitrary point in $U$. Then $x$ is contained in an open ball $B_r_x(x) subseteq U$. Let $q$ be a rational point contained within the ball $B_r_x/3(x)$. Then we can choose a ball $B_p(q) in mathcalB$ with $r_x/3 < p <2r_x/3$. Then $$x in B_p(q) subset B_r_x(x) subseteq U.$$
Hence $U$ is the union of a subset of $mathcalB$.
answered Mar 31 at 2:00
AutumnAutumn
361
$endgroup$
add a comment |
$begingroup$
Here's an answer to part a: This is just the same as, or a slight modification of the proof that was posted earlier.
For any set $U$ the set of balls $$mathcalB=~qin mathbbQ^n, p in mathbbQ_ge 0$$ form a countable cover of $U$.
Let $x$ be an arbitrary point in $U$. Then $x$ is contained in an open ball $B_r_x(x) subseteq U$. Let $q$ be a rational point contained within the ball $B_r_x/3(x)$. Then we can choose a ball $B_p(q) in mathcalB$ with $r_x/3 < p <2r_x/3$. Then $$x in B_p(q) subset B_r_x(x) subseteq U.$$
Hence $U$ is the union of a subset of $mathcalB$.
answered Mar 31 at 2:00
AutumnAutumn
361
$endgroup$
add a comment |
$begingroup$
Here's an answer to part a: This is just the same as, or a slight modification of the proof that was posted earlier.
For any set $U$ the set of balls $$mathcalB=~qin mathbbQ^n, p in mathbbQ_ge 0$$ form a countable cover of $U$.
Let $x$ be an arbitrary point in $U$. Then $x$ is contained in an open ball $B_r_x(x) subseteq U$. Let $q$ be a rational point contained within the ball $B_r_x/3(x)$. Then we can choose a ball $B_p(q) in mathcalB$ with $r_x/3 < p <2r_x/3$. Then $$x in B_p(q) subset B_r_x(x) subseteq U.$$
Hence $U$ is the union of a subset of $mathcalB$.
answered Mar 31 at 2:00
AutumnAutumn
361
$endgroup$
Here's an answer to part a: This is just the same as, or a slight modification of the proof that was posted earlier.
For any set $U$ the set of balls $$mathcalB=~qin mathbbQ^n, p in mathbbQ_ge 0$$ form a countable cover of $U$.
Let $x$ be an arbitrary point in $U$. Then $x$ is contained in an open ball $B_r_x(x) subseteq U$. Let $q$ be a rational point contained within the ball $B_r_x/3(x)$. Then we can choose a ball $B_p(q) in mathcalB$ with $r_x/3 < p <2r_x/3$. Then $$x in B_p(q) subset B_r_x(x) subseteq U.$$
Hence $U$ is the union of a subset of $mathcalB$.
answered Mar 31 at 2:00
AutumnAutumn
361
answered Mar 31 at 2:00
AutumnAutumn
361
answered Mar 31 at 2:00
AutumnAutumn
361
answered Mar 31 at 2:00
AutumnAutumn
361
361
add a comment |
add a comment |
$begingroup$
Take a family $V=V_n:nin Bbb N$ of bounded open sets such that $U=cup V$ and such that $overline V_n subset V_n+1$ for each $n.$ (To get $V,$ see APPENDIX below.)
Let $K_1=overline V_1.$ For $n>1$ let $K_n=bar V_n V_n-1.$ Each $K_n$ is compact.
Let $C_1$ be a finite cover of $K_1$ by open balls, such that $overline cup C_1subset V_2.$ This is possible because $K_1$ is compact, $V_2$ is open, and $K_1subset V_2.$
For $n>1$ let $C_n$ be a finite cover of $K_n$ by open balls, such that $overline cup C_nsubset V_n+1$ and also such that $cup C_n$ is disjoint from $cup_jle n-2(overline cup C_j).$
The last condition above is vacuous when $n=2.$ When $n>2$ it is crucial, and possible because $S_n=cup_jle n-2(overline cup C_j)$ is a subset of $cup_jle n-1V_j=V_n-1,$ and $V_n-1$ is disjoint from $K_n,$ so the closed set $S_n$ and the compact set $K_n$ are disjoint.
Now $C=cup_nin Bbb nC_n$ is a countable family of open balls, and $U=cup C=cup _bin C, int(bar b)=cup_bin C,bar b.$
If $pin U$ then for some $n$ we have $pin V_nsubset overline V_n=cup_jle nK_j$ so for some $jle n$ and some $b in C_j$ we will have $pin bin C_j.$ Now this $b$ is disjoint from $cup C_m$ for every $mge n+2.$ So $b'in C: b'cap bne phi$ is a subset of the finite set $cup_jle n+1C_j.$
APPENDIX. Let $bar 0$ be the origin in $Bbb R^n.$
(i). If $U=Bbb R^n$ let $V_n=B_n(bar 0)$ for each $nin Bbb N.$
(ii). If $Une Bbb R^n,$ let $U^c=Bbb R^n$ $U$ and let $d$ be a metric for the topology on $Bbb R^n,$ and for $pin U$ let $d(p,U^c)=inf d(p,q):qin U^c.$
Now let $V_1=pin Ucap B_1(bar 0): d(p,U^c)>1.$ And for $n>1$ let $V_n= pin Ucap B_n(bar 0): d(p,U^c)>1/n.$ I will leave it to the reader to confirm that $V_n:nin Bbb N$ has the required properties.
answered Mar 31 at 18:24
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
$begingroup$
This works if you replace $Bbb R^n$ with any metric space in which closed bounded subsets are compact.
$endgroup$
– DanielWainfleet
Mar 31 at 18:37
add a comment |
$begingroup$
Take a family $V=V_n:nin Bbb N$ of bounded open sets such that $U=cup V$ and such that $overline V_n subset V_n+1$ for each $n.$ (To get $V,$ see APPENDIX below.)
Let $K_1=overline V_1.$ For $n>1$ let $K_n=bar V_n V_n-1.$ Each $K_n$ is compact.
Let $C_1$ be a finite cover of $K_1$ by open balls, such that $overline cup C_1subset V_2.$ This is possible because $K_1$ is compact, $V_2$ is open, and $K_1subset V_2.$
For $n>1$ let $C_n$ be a finite cover of $K_n$ by open balls, such that $overline cup C_nsubset V_n+1$ and also such that $cup C_n$ is disjoint from $cup_jle n-2(overline cup C_j).$
The last condition above is vacuous when $n=2.$ When $n>2$ it is crucial, and possible because $S_n=cup_jle n-2(overline cup C_j)$ is a subset of $cup_jle n-1V_j=V_n-1,$ and $V_n-1$ is disjoint from $K_n,$ so the closed set $S_n$ and the compact set $K_n$ are disjoint.
Now $C=cup_nin Bbb nC_n$ is a countable family of open balls, and $U=cup C=cup _bin C, int(bar b)=cup_bin C,bar b.$
If $pin U$ then for some $n$ we have $pin V_nsubset overline V_n=cup_jle nK_j$ so for some $jle n$ and some $b in C_j$ we will have $pin bin C_j.$ Now this $b$ is disjoint from $cup C_m$ for every $mge n+2.$ So $b'in C: b'cap bne phi$ is a subset of the finite set $cup_jle n+1C_j.$
APPENDIX. Let $bar 0$ be the origin in $Bbb R^n.$
(i). If $U=Bbb R^n$ let $V_n=B_n(bar 0)$ for each $nin Bbb N.$
(ii). If $Une Bbb R^n,$ let $U^c=Bbb R^n$ $U$ and let $d$ be a metric for the topology on $Bbb R^n,$ and for $pin U$ let $d(p,U^c)=inf d(p,q):qin U^c.$
Now let $V_1=pin Ucap B_1(bar 0): d(p,U^c)>1.$ And for $n>1$ let $V_n= pin Ucap B_n(bar 0): d(p,U^c)>1/n.$ I will leave it to the reader to confirm that $V_n:nin Bbb N$ has the required properties.
answered Mar 31 at 18:24
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
$begingroup$
This works if you replace $Bbb R^n$ with any metric space in which closed bounded subsets are compact.
$endgroup$
– DanielWainfleet
Mar 31 at 18:37
add a comment |
$begingroup$
Take a family $V=V_n:nin Bbb N$ of bounded open sets such that $U=cup V$ and such that $overline V_n subset V_n+1$ for each $n.$ (To get $V,$ see APPENDIX below.)
Let $K_1=overline V_1.$ For $n>1$ let $K_n=bar V_n V_n-1.$ Each $K_n$ is compact.
Let $C_1$ be a finite cover of $K_1$ by open balls, such that $overline cup C_1subset V_2.$ This is possible because $K_1$ is compact, $V_2$ is open, and $K_1subset V_2.$
For $n>1$ let $C_n$ be a finite cover of $K_n$ by open balls, such that $overline cup C_nsubset V_n+1$ and also such that $cup C_n$ is disjoint from $cup_jle n-2(overline cup C_j).$
The last condition above is vacuous when $n=2.$ When $n>2$ it is crucial, and possible because $S_n=cup_jle n-2(overline cup C_j)$ is a subset of $cup_jle n-1V_j=V_n-1,$ and $V_n-1$ is disjoint from $K_n,$ so the closed set $S_n$ and the compact set $K_n$ are disjoint.
Now $C=cup_nin Bbb nC_n$ is a countable family of open balls, and $U=cup C=cup _bin C, int(bar b)=cup_bin C,bar b.$
If $pin U$ then for some $n$ we have $pin V_nsubset overline V_n=cup_jle nK_j$ so for some $jle n$ and some $b in C_j$ we will have $pin bin C_j.$ Now this $b$ is disjoint from $cup C_m$ for every $mge n+2.$ So $b'in C: b'cap bne phi$ is a subset of the finite set $cup_jle n+1C_j.$
APPENDIX. Let $bar 0$ be the origin in $Bbb R^n.$
(i). If $U=Bbb R^n$ let $V_n=B_n(bar 0)$ for each $nin Bbb N.$
(ii). If $Une Bbb R^n,$ let $U^c=Bbb R^n$ $U$ and let $d$ be a metric for the topology on $Bbb R^n,$ and for $pin U$ let $d(p,U^c)=inf d(p,q):qin U^c.$
Now let $V_1=pin Ucap B_1(bar 0): d(p,U^c)>1.$ And for $n>1$ let $V_n= pin Ucap B_n(bar 0): d(p,U^c)>1/n.$ I will leave it to the reader to confirm that $V_n:nin Bbb N$ has the required properties.
answered Mar 31 at 18:24
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
Take a family $V=V_n:nin Bbb N$ of bounded open sets such that $U=cup V$ and such that $overline V_n subset V_n+1$ for each $n.$ (To get $V,$ see APPENDIX below.)
Let $K_1=overline V_1.$ For $n>1$ let $K_n=bar V_n V_n-1.$ Each $K_n$ is compact.
Let $C_1$ be a finite cover of $K_1$ by open balls, such that $overline cup C_1subset V_2.$ This is possible because $K_1$ is compact, $V_2$ is open, and $K_1subset V_2.$
For $n>1$ let $C_n$ be a finite cover of $K_n$ by open balls, such that $overline cup C_nsubset V_n+1$ and also such that $cup C_n$ is disjoint from $cup_jle n-2(overline cup C_j).$
The last condition above is vacuous when $n=2.$ When $n>2$ it is crucial, and possible because $S_n=cup_jle n-2(overline cup C_j)$ is a subset of $cup_jle n-1V_j=V_n-1,$ and $V_n-1$ is disjoint from $K_n,$ so the closed set $S_n$ and the compact set $K_n$ are disjoint.
Now $C=cup_nin Bbb nC_n$ is a countable family of open balls, and $U=cup C=cup _bin C, int(bar b)=cup_bin C,bar b.$
If $pin U$ then for some $n$ we have $pin V_nsubset overline V_n=cup_jle nK_j$ so for some $jle n$ and some $b in C_j$ we will have $pin bin C_j.$ Now this $b$ is disjoint from $cup C_m$ for every $mge n+2.$ So $b'in C: b'cap bne phi$ is a subset of the finite set $cup_jle n+1C_j.$
APPENDIX. Let $bar 0$ be the origin in $Bbb R^n.$
(i). If $U=Bbb R^n$ let $V_n=B_n(bar 0)$ for each $nin Bbb N.$
(ii). If $Une Bbb R^n,$ let $U^c=Bbb R^n$ $U$ and let $d$ be a metric for the topology on $Bbb R^n,$ and for $pin U$ let $d(p,U^c)=inf d(p,q):qin U^c.$
Now let $V_1=pin Ucap B_1(bar 0): d(p,U^c)>1.$ And for $n>1$ let $V_n= pin Ucap B_n(bar 0): d(p,U^c)>1/n.$ I will leave it to the reader to confirm that $V_n:nin Bbb N$ has the required properties.
answered Mar 31 at 18:24
DanielWainfleetDanielWainfleet
35.8k31648
edited Mar 31 at 18:34
answered Mar 31 at 18:24
DanielWainfleetDanielWainfleet
35.8k31648
answered Mar 31 at 18:24
DanielWainfleetDanielWainfleet
35.8k31648
answered Mar 31 at 18:24
DanielWainfleetDanielWainfleet
35.8k31648
35.8k31648
$begingroup$
This works if you replace $Bbb R^n$ with any metric space in which closed bounded subsets are compact.
$endgroup$
– DanielWainfleet
Mar 31 at 18:37
add a comment |
$begingroup$
This works if you replace $Bbb R^n$ with any metric space in which closed bounded subsets are compact.
$endgroup$
– DanielWainfleet
Mar 31 at 18:37
$begingroup$
This works if you replace $Bbb R^n$ with any metric space in which closed bounded subsets are compact.
$endgroup$
– DanielWainfleet
Mar 31 at 18:37
$begingroup$
This works if you replace $Bbb R^n$ with any metric space in which closed bounded subsets are compact.
$endgroup$
– DanielWainfleet
Mar 31 at 18:37
add a comment |
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$begingroup$
I think you can do this by considering sets of dyadic cubes, and the corresponding inscribed/circumscribed balls.
$endgroup$
– Matematleta
Mar 29 at 22:52
$begingroup$
I've never seen dyadic cubes before.
$endgroup$
– AColoredReptile
Mar 29 at 22:52
1
$begingroup$
Is there an extra part of the condition in b), or what prevents us to take $M=mathbbR^n$ or $M=U$ for all $n$?
$endgroup$
– Henno Brandsma
Mar 30 at 13:10
$begingroup$
non-trivial intersection is onon-empty intersection, I think.
$endgroup$
– Henno Brandsma
Mar 30 at 13:11
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What is the cover $C$ in condition c)? Do you mean $beta$?
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– Paul Frost
Mar 30 at 17:24