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The Lusternik–Schnirelmann Theorem For Open and Closed Sets
Clash Royale CLAN TAG#URR8PPP
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The generalized Lusternik-Schnirelmann Theorem states that
If $S^n$ is covered by $n+1$ sets $A_1, A_2, ... ,A_n+1$ such that each $A_i$ is either open or closed, then there exists an $i$ such that $A_i$ has a pair of antipodal points.
I'm having trouble proving this theorem for "$A_i$ is either open or closed". I can prove the theorem given the hypothesis that all $A_i$ are closed.
Define $d_i : S^n to mathbbR$ by $d_i(x) = inf : y in A_i $. Clearly, this function is continuous. Now, consider the map
beginalign*
Psi: S^n & longrightarrow mathbbR^n \
x & longrightarrow (d_1(x), ..., d_n(x)).
endalign*
From the Borsuk-Ulam theorem, there exists an $x in S^n$ such that $Psi(x) = Psi(-x) = Omega$. If any of the coordinates of $Omega$ are $0$, then, since the $A_i$ are closed, $x, -x$ must be limit points of some $A_i$ and hence are in that $A_i$. If none of the coordinates are $0$, then $x, -x$ are both in $A_n+1$.
If the $A_i$'s could also be open, the above argument does not work.
Any suggestions on how to prove the theorem would be appreciated.
real-analysis general-topology
asked Aug 1 at 18:11
Alan Yan
605311
add a comment |Â
up vote
2
down vote
favorite
The generalized Lusternik-Schnirelmann Theorem states that
If $S^n$ is covered by $n+1$ sets $A_1, A_2, ... ,A_n+1$ such that each $A_i$ is either open or closed, then there exists an $i$ such that $A_i$ has a pair of antipodal points.
I'm having trouble proving this theorem for "$A_i$ is either open or closed". I can prove the theorem given the hypothesis that all $A_i$ are closed.
Define $d_i : S^n to mathbbR$ by $d_i(x) = inf : y in A_i $. Clearly, this function is continuous. Now, consider the map
beginalign*
Psi: S^n & longrightarrow mathbbR^n \
x & longrightarrow (d_1(x), ..., d_n(x)).
endalign*
From the Borsuk-Ulam theorem, there exists an $x in S^n$ such that $Psi(x) = Psi(-x) = Omega$. If any of the coordinates of $Omega$ are $0$, then, since the $A_i$ are closed, $x, -x$ must be limit points of some $A_i$ and hence are in that $A_i$. If none of the coordinates are $0$, then $x, -x$ are both in $A_n+1$.
If the $A_i$'s could also be open, the above argument does not work.
Any suggestions on how to prove the theorem would be appreciated.
real-analysis general-topology
asked Aug 1 at 18:11
Alan Yan
605311
If some $A_i$ is open, then can you shrink it to a suitable closed subset?
– Lord Shark the Unknown
Aug 1 at 18:33
@LordSharktheUnknown what if $A_1,...,A_n$ are distinct points and $A_n+1$ their complement?
– Akababa
Aug 1 at 21:32
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The generalized Lusternik-Schnirelmann Theorem states that
If $S^n$ is covered by $n+1$ sets $A_1, A_2, ... ,A_n+1$ such that each $A_i$ is either open or closed, then there exists an $i$ such that $A_i$ has a pair of antipodal points.
I'm having trouble proving this theorem for "$A_i$ is either open or closed". I can prove the theorem given the hypothesis that all $A_i$ are closed.
Define $d_i : S^n to mathbbR$ by $d_i(x) = inf : y in A_i $. Clearly, this function is continuous. Now, consider the map
beginalign*
Psi: S^n & longrightarrow mathbbR^n \
x & longrightarrow (d_1(x), ..., d_n(x)).
endalign*
From the Borsuk-Ulam theorem, there exists an $x in S^n$ such that $Psi(x) = Psi(-x) = Omega$. If any of the coordinates of $Omega$ are $0$, then, since the $A_i$ are closed, $x, -x$ must be limit points of some $A_i$ and hence are in that $A_i$. If none of the coordinates are $0$, then $x, -x$ are both in $A_n+1$.
If the $A_i$'s could also be open, the above argument does not work.
Any suggestions on how to prove the theorem would be appreciated.
real-analysis general-topology
asked Aug 1 at 18:11
Alan Yan
605311
The generalized Lusternik-Schnirelmann Theorem states that
If $S^n$ is covered by $n+1$ sets $A_1, A_2, ... ,A_n+1$ such that each $A_i$ is either open or closed, then there exists an $i$ such that $A_i$ has a pair of antipodal points.
I'm having trouble proving this theorem for "$A_i$ is either open or closed". I can prove the theorem given the hypothesis that all $A_i$ are closed.
Define $d_i : S^n to mathbbR$ by $d_i(x) = inf : y in A_i $. Clearly, this function is continuous. Now, consider the map
beginalign*
Psi: S^n & longrightarrow mathbbR^n \
x & longrightarrow (d_1(x), ..., d_n(x)).
endalign*
From the Borsuk-Ulam theorem, there exists an $x in S^n$ such that $Psi(x) = Psi(-x) = Omega$. If any of the coordinates of $Omega$ are $0$, then, since the $A_i$ are closed, $x, -x$ must be limit points of some $A_i$ and hence are in that $A_i$. If none of the coordinates are $0$, then $x, -x$ are both in $A_n+1$.
If the $A_i$'s could also be open, the above argument does not work.
Any suggestions on how to prove the theorem would be appreciated.
real-analysis general-topology
asked Aug 1 at 18:11
Alan Yan
605311
asked Aug 1 at 18:11
Alan Yan
605311
asked Aug 1 at 18:11
Alan Yan
605311
asked Aug 1 at 18:11
Alan Yan
605311
605311
If some $A_i$ is open, then can you shrink it to a suitable closed subset?
– Lord Shark the Unknown
Aug 1 at 18:33
@LordSharktheUnknown what if $A_1,...,A_n$ are distinct points and $A_n+1$ their complement?
– Akababa
Aug 1 at 21:32
add a comment |Â
If some $A_i$ is open, then can you shrink it to a suitable closed subset?
– Lord Shark the Unknown
Aug 1 at 18:33
@LordSharktheUnknown what if $A_1,...,A_n$ are distinct points and $A_n+1$ their complement?
– Akababa
Aug 1 at 21:32
If some $A_i$ is open, then can you shrink it to a suitable closed subset?
– Lord Shark the Unknown
Aug 1 at 18:33
If some $A_i$ is open, then can you shrink it to a suitable closed subset?
– Lord Shark the Unknown
Aug 1 at 18:33
@LordSharktheUnknown what if $A_1,...,A_n$ are distinct points and $A_n+1$ their complement?
– Akababa
Aug 1 at 21:32
@LordSharktheUnknown what if $A_1,...,A_n$ are distinct points and $A_n+1$ their complement?
– Akababa
Aug 1 at 21:32
add a comment |Â
1 Answer
1
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up vote
2
down vote
accepted
You can extend your argument as follows: let $x$ be defined as above. If $x,-xin A_n+1$ we're done, so WLOG assume $xnotin A_n+1implies xin A_i$ for some $ineq n+1$. Thus the $i$th coordinate of $Omega$ is $0$, so if $A_i$ is closed we're done by your proof. If $A_i$ is open, then $-xin bar A_i$ so $B(x,delta)subseteq A_i$ for some $delta>0$ and $B(-x,delta)cap A_ineqemptyset$, so if we pick some $y$ from the latter then $y,-yin A_i$.
answered Aug 1 at 22:01
Akababa
2,557922
Thanks! That was a nice argument!
– Alan Yan
Aug 2 at 0:27
Thanks, yours was too :)
– Akababa
Aug 2 at 20:35
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You can extend your argument as follows: let $x$ be defined as above. If $x,-xin A_n+1$ we're done, so WLOG assume $xnotin A_n+1implies xin A_i$ for some $ineq n+1$. Thus the $i$th coordinate of $Omega$ is $0$, so if $A_i$ is closed we're done by your proof. If $A_i$ is open, then $-xin bar A_i$ so $B(x,delta)subseteq A_i$ for some $delta>0$ and $B(-x,delta)cap A_ineqemptyset$, so if we pick some $y$ from the latter then $y,-yin A_i$.
answered Aug 1 at 22:01
Akababa
2,557922
Thanks! That was a nice argument!
– Alan Yan
Aug 2 at 0:27
Thanks, yours was too :)
– Akababa
Aug 2 at 20:35
add a comment |Â
up vote
2
down vote
accepted
You can extend your argument as follows: let $x$ be defined as above. If $x,-xin A_n+1$ we're done, so WLOG assume $xnotin A_n+1implies xin A_i$ for some $ineq n+1$. Thus the $i$th coordinate of $Omega$ is $0$, so if $A_i$ is closed we're done by your proof. If $A_i$ is open, then $-xin bar A_i$ so $B(x,delta)subseteq A_i$ for some $delta>0$ and $B(-x,delta)cap A_ineqemptyset$, so if we pick some $y$ from the latter then $y,-yin A_i$.
answered Aug 1 at 22:01
Akababa
2,557922
Thanks! That was a nice argument!
– Alan Yan
Aug 2 at 0:27
Thanks, yours was too :)
– Akababa
Aug 2 at 20:35
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You can extend your argument as follows: let $x$ be defined as above. If $x,-xin A_n+1$ we're done, so WLOG assume $xnotin A_n+1implies xin A_i$ for some $ineq n+1$. Thus the $i$th coordinate of $Omega$ is $0$, so if $A_i$ is closed we're done by your proof. If $A_i$ is open, then $-xin bar A_i$ so $B(x,delta)subseteq A_i$ for some $delta>0$ and $B(-x,delta)cap A_ineqemptyset$, so if we pick some $y$ from the latter then $y,-yin A_i$.
answered Aug 1 at 22:01
Akababa
2,557922
You can extend your argument as follows: let $x$ be defined as above. If $x,-xin A_n+1$ we're done, so WLOG assume $xnotin A_n+1implies xin A_i$ for some $ineq n+1$. Thus the $i$th coordinate of $Omega$ is $0$, so if $A_i$ is closed we're done by your proof. If $A_i$ is open, then $-xin bar A_i$ so $B(x,delta)subseteq A_i$ for some $delta>0$ and $B(-x,delta)cap A_ineqemptyset$, so if we pick some $y$ from the latter then $y,-yin A_i$.
answered Aug 1 at 22:01
Akababa
2,557922
answered Aug 1 at 22:01
Akababa
2,557922
answered Aug 1 at 22:01
Akababa
2,557922
answered Aug 1 at 22:01
Akababa
2,557922
2,557922
Thanks! That was a nice argument!
– Alan Yan
Aug 2 at 0:27
Thanks, yours was too :)
– Akababa
Aug 2 at 20:35
add a comment |Â
Thanks! That was a nice argument!
– Alan Yan
Aug 2 at 0:27
Thanks, yours was too :)
– Akababa
Aug 2 at 20:35
Thanks! That was a nice argument!
– Alan Yan
Aug 2 at 0:27
Thanks! That was a nice argument!
– Alan Yan
Aug 2 at 0:27
Thanks, yours was too :)
– Akababa
Aug 2 at 20:35
Thanks, yours was too :)
– Akababa
Aug 2 at 20:35
add a comment |Â
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If some $A_i$ is open, then can you shrink it to a suitable closed subset?
– Lord Shark the Unknown
Aug 1 at 18:33
@LordSharktheUnknown what if $A_1,...,A_n$ are distinct points and $A_n+1$ their complement?
– Akababa
Aug 1 at 21:32