If $x$ is not in the closure of a convex set $A$, is there a point in $A$ that is closer than $x$ is to each point in $A$?Why does a convex set have the same interior points as its closure?Show that the set of points that are nearer $a$ than $b$ with respect to $lVert cdot rVert_2$ is convexOptimality conditions in convex programmingClosed convex set as the intersection of (tangent) half spacesFor any two disjoint convex open sets there is a hyperplane that strictly separates themQuestion about definition of separating hyperplanes (theorem)Approximating disjoint convex sets by subsets with positive separationFinding a convex set that approaches a boundary pointDistance of point, x from closed convex set C when x is not in C.Is the subspace of odd functions of $L^2(-1,1)$ convex?
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If $x$ is not in the closure of a convex set $A$, is there a point in $A$ that is closer than $x$ is to each point in $A$?
Why does a convex set have the same interior points as its closure?Show that the set of points that are nearer $a$ than $b$ with respect to $lVert cdot rVert_2$ is convexOptimality conditions in convex programmingClosed convex set as the intersection of (tangent) half spacesFor any two disjoint convex open sets there is a hyperplane that strictly separates themQuestion about definition of separating hyperplanes (theorem)Approximating disjoint convex sets by subsets with positive separationFinding a convex set that approaches a boundary pointDistance of point, x from closed convex set C when x is not in C.Is the subspace of odd functions of $L^2(-1,1)$ convex?
$begingroup$
I'm trying to assess the following conjecture:
$textbfConjecture.$ Suppose that $A$ is a non-empty convex subset of $mathbbR^N$ and that $x$ is not in the closure of $A$. Then there is a point $a' in A$ such that $forall a in A, , ||x - a|| > || a' - a ||.$
I am familiar with results that yield this if we assume that $A$ is closed (closest point theorem, strongly separating hyperplane theorem), but I do not want to assume that $A$ is closed.
Edit: What I have in mind here is that if $A$ is closed, then there is some point $a'$ that is closest to $x$, and the perpendicular bisector of $a'x$ strongly separates $x$ and $A$, showing that $a'$ is closer to any point in $A$ than $x$ is. I'm trying to see if the conjecture holds without assuming $A$ is closed.
convex-analysis convex-optimization
asked Mar 13 at 17:31
supergenericsupergeneric
394
New contributor
supergeneric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 1 more comment
$begingroup$
I'm trying to assess the following conjecture:
$textbfConjecture.$ Suppose that $A$ is a non-empty convex subset of $mathbbR^N$ and that $x$ is not in the closure of $A$. Then there is a point $a' in A$ such that $forall a in A, , ||x - a|| > || a' - a ||.$
I am familiar with results that yield this if we assume that $A$ is closed (closest point theorem, strongly separating hyperplane theorem), but I do not want to assume that $A$ is closed.
Edit: What I have in mind here is that if $A$ is closed, then there is some point $a'$ that is closest to $x$, and the perpendicular bisector of $a'x$ strongly separates $x$ and $A$, showing that $a'$ is closer to any point in $A$ than $x$ is. I'm trying to see if the conjecture holds without assuming $A$ is closed.
convex-analysis convex-optimization
asked Mar 13 at 17:31
supergenericsupergeneric
394
New contributor
supergeneric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
So the problem is that $a'$ might belong to $overline A setminus A$, right?
$endgroup$
– Giuseppe Negro
Mar 13 at 18:02
$begingroup$
@GiuseppeNegro Yes, that's exactly what I'm thinking.
$endgroup$
– supergeneric
Mar 13 at 18:17
$begingroup$
Is $a'$ unique if $A$ is closed? If $a'$ coincides with $P(x)$, the point of $A$ that is closest to $x$, then it is unique. Thus, it is easy to construct a counterexample; just fix $x$ and consider $AsetminusP(x)$. This set can still be convex (it is if $A$ is a circle, for example), but it is not closed, and no $a'$ can exist for the $x$ you fixed earlier.
$endgroup$
– Giuseppe Negro
Mar 13 at 20:49
$begingroup$
@GiuseppeNegro I am not sure if your example with $A$ a circle works. I believe I've come up with an argument that covers this case, so please have a look to see if I've made a mistake in my reasoning.
$endgroup$
– K.Power
Mar 13 at 21:37
$begingroup$
What about taking the closest point to $x$ in $overlineA$ (which always exists) and moving just an infinitesimal quantity away from $x$ on the line passing through $x$ and that point to obtain $a'$? Under the small assumption that $A$ has non-empty interior, this $a'$ can be taken in $A$ and it should do the job, intuitively (then again, there might be some pathological cases...)
$endgroup$
– Daniel Robert-Nicoud
Mar 13 at 22:14
|
show 1 more comment
$begingroup$
I'm trying to assess the following conjecture:
$textbfConjecture.$ Suppose that $A$ is a non-empty convex subset of $mathbbR^N$ and that $x$ is not in the closure of $A$. Then there is a point $a' in A$ such that $forall a in A, , ||x - a|| > || a' - a ||.$
I am familiar with results that yield this if we assume that $A$ is closed (closest point theorem, strongly separating hyperplane theorem), but I do not want to assume that $A$ is closed.
Edit: What I have in mind here is that if $A$ is closed, then there is some point $a'$ that is closest to $x$, and the perpendicular bisector of $a'x$ strongly separates $x$ and $A$, showing that $a'$ is closer to any point in $A$ than $x$ is. I'm trying to see if the conjecture holds without assuming $A$ is closed.
convex-analysis convex-optimization
asked Mar 13 at 17:31
supergenericsupergeneric
394
New contributor
supergeneric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm trying to assess the following conjecture:
$textbfConjecture.$ Suppose that $A$ is a non-empty convex subset of $mathbbR^N$ and that $x$ is not in the closure of $A$. Then there is a point $a' in A$ such that $forall a in A, , ||x - a|| > || a' - a ||.$
I am familiar with results that yield this if we assume that $A$ is closed (closest point theorem, strongly separating hyperplane theorem), but I do not want to assume that $A$ is closed.
Edit: What I have in mind here is that if $A$ is closed, then there is some point $a'$ that is closest to $x$, and the perpendicular bisector of $a'x$ strongly separates $x$ and $A$, showing that $a'$ is closer to any point in $A$ than $x$ is. I'm trying to see if the conjecture holds without assuming $A$ is closed.
convex-analysis convex-optimization
convex-analysis convex-optimization
asked Mar 13 at 17:31
supergenericsupergeneric
394
New contributor
supergeneric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 13 at 17:31
supergenericsupergeneric
394
New contributor
supergeneric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 14 at 4:29
supergeneric
asked Mar 13 at 17:31
supergenericsupergeneric
394
New contributor
supergeneric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 13 at 17:31
supergenericsupergeneric
394
asked Mar 13 at 17:31
supergenericsupergeneric
394
394
New contributor
supergeneric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
supergeneric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
supergeneric is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
So the problem is that $a'$ might belong to $overline A setminus A$, right?
$endgroup$
– Giuseppe Negro
Mar 13 at 18:02
$begingroup$
@GiuseppeNegro Yes, that's exactly what I'm thinking.
$endgroup$
– supergeneric
Mar 13 at 18:17
$begingroup$
Is $a'$ unique if $A$ is closed? If $a'$ coincides with $P(x)$, the point of $A$ that is closest to $x$, then it is unique. Thus, it is easy to construct a counterexample; just fix $x$ and consider $AsetminusP(x)$. This set can still be convex (it is if $A$ is a circle, for example), but it is not closed, and no $a'$ can exist for the $x$ you fixed earlier.
$endgroup$
– Giuseppe Negro
Mar 13 at 20:49
$begingroup$
@GiuseppeNegro I am not sure if your example with $A$ a circle works. I believe I've come up with an argument that covers this case, so please have a look to see if I've made a mistake in my reasoning.
$endgroup$
– K.Power
Mar 13 at 21:37
$begingroup$
What about taking the closest point to $x$ in $overlineA$ (which always exists) and moving just an infinitesimal quantity away from $x$ on the line passing through $x$ and that point to obtain $a'$? Under the small assumption that $A$ has non-empty interior, this $a'$ can be taken in $A$ and it should do the job, intuitively (then again, there might be some pathological cases...)
$endgroup$
– Daniel Robert-Nicoud
Mar 13 at 22:14
|
show 1 more comment
1
$begingroup$
So the problem is that $a'$ might belong to $overline A setminus A$, right?
$endgroup$
– Giuseppe Negro
Mar 13 at 18:02
$begingroup$
@GiuseppeNegro Yes, that's exactly what I'm thinking.
$endgroup$
– supergeneric
Mar 13 at 18:17
$begingroup$
Is $a'$ unique if $A$ is closed? If $a'$ coincides with $P(x)$, the point of $A$ that is closest to $x$, then it is unique. Thus, it is easy to construct a counterexample; just fix $x$ and consider $AsetminusP(x)$. This set can still be convex (it is if $A$ is a circle, for example), but it is not closed, and no $a'$ can exist for the $x$ you fixed earlier.
$endgroup$
– Giuseppe Negro
Mar 13 at 20:49
$begingroup$
@GiuseppeNegro I am not sure if your example with $A$ a circle works. I believe I've come up with an argument that covers this case, so please have a look to see if I've made a mistake in my reasoning.
$endgroup$
– K.Power
Mar 13 at 21:37
$begingroup$
What about taking the closest point to $x$ in $overlineA$ (which always exists) and moving just an infinitesimal quantity away from $x$ on the line passing through $x$ and that point to obtain $a'$? Under the small assumption that $A$ has non-empty interior, this $a'$ can be taken in $A$ and it should do the job, intuitively (then again, there might be some pathological cases...)
$endgroup$
– Daniel Robert-Nicoud
Mar 13 at 22:14
1
1
$begingroup$
So the problem is that $a'$ might belong to $overline A setminus A$, right?
$endgroup$
– Giuseppe Negro
Mar 13 at 18:02
$begingroup$
So the problem is that $a'$ might belong to $overline A setminus A$, right?
$endgroup$
– Giuseppe Negro
Mar 13 at 18:02
$begingroup$
@GiuseppeNegro Yes, that's exactly what I'm thinking.
$endgroup$
– supergeneric
Mar 13 at 18:17
$begingroup$
@GiuseppeNegro Yes, that's exactly what I'm thinking.
$endgroup$
– supergeneric
Mar 13 at 18:17
$begingroup$
Is $a'$ unique if $A$ is closed? If $a'$ coincides with $P(x)$, the point of $A$ that is closest to $x$, then it is unique. Thus, it is easy to construct a counterexample; just fix $x$ and consider $AsetminusP(x)$. This set can still be convex (it is if $A$ is a circle, for example), but it is not closed, and no $a'$ can exist for the $x$ you fixed earlier.
$endgroup$
– Giuseppe Negro
Mar 13 at 20:49
$begingroup$
Is $a'$ unique if $A$ is closed? If $a'$ coincides with $P(x)$, the point of $A$ that is closest to $x$, then it is unique. Thus, it is easy to construct a counterexample; just fix $x$ and consider $AsetminusP(x)$. This set can still be convex (it is if $A$ is a circle, for example), but it is not closed, and no $a'$ can exist for the $x$ you fixed earlier.
$endgroup$
– Giuseppe Negro
Mar 13 at 20:49
$begingroup$
@GiuseppeNegro I am not sure if your example with $A$ a circle works. I believe I've come up with an argument that covers this case, so please have a look to see if I've made a mistake in my reasoning.
$endgroup$
– K.Power
Mar 13 at 21:37
$begingroup$
@GiuseppeNegro I am not sure if your example with $A$ a circle works. I believe I've come up with an argument that covers this case, so please have a look to see if I've made a mistake in my reasoning.
$endgroup$
– K.Power
Mar 13 at 21:37
$begingroup$
What about taking the closest point to $x$ in $overlineA$ (which always exists) and moving just an infinitesimal quantity away from $x$ on the line passing through $x$ and that point to obtain $a'$? Under the small assumption that $A$ has non-empty interior, this $a'$ can be taken in $A$ and it should do the job, intuitively (then again, there might be some pathological cases...)
$endgroup$
– Daniel Robert-Nicoud
Mar 13 at 22:14
$begingroup$
What about taking the closest point to $x$ in $overlineA$ (which always exists) and moving just an infinitesimal quantity away from $x$ on the line passing through $x$ and that point to obtain $a'$? Under the small assumption that $A$ has non-empty interior, this $a'$ can be taken in $A$ and it should do the job, intuitively (then again, there might be some pathological cases...)
$endgroup$
– Daniel Robert-Nicoud
Mar 13 at 22:14
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
This is a slightly heuristic argument, which I think may work. As you say there exists some $bin bar A$ such that $|x-a|>|b-a|$ for all $ain A$ (where $b=P_A(x)$), to avoid triviality we assume $bin bar Abackslash A$. Let us first assume that $overrightarrowxb$ intercepts $A$. So for all $varepsilon>0$ there exists an $a_varepsilonin Acapoverrightarrowxb$ such that $|b-a_varepsilon|<varepsilon$. Set $delta=|x-b|/2$. Convexity implies, for all $ain A$, that angle $overlineabangleoverlinebx$ cannot be acute. This in combination with the fact $x,b,a_delta$ all lie on the same line, and $|b-a_delta|<|b-x|$ implies that $|a-a_delta|<|a-x|$.
As is illustrated above we are using the fact that in a triangle $ABC$, if there exists a point $D$ on $AB$ such that $|AD|<|DB|$ and $angle ADC<pi/2$ we must have $|CA|<|CB|$.
If $overrightarrowxbcap A =emptyset$, I believe we can use a similar argument, but we will have to control the value of $delta$ more tightly (remembering that $a_delta$ is no longer on $overrightarrowxb$) depending on the angle made. If there was to be a counterexample I think it would have to be in this situation.
answered Mar 13 at 21:20
K.PowerK.Power
3,490926
$endgroup$
$begingroup$
Much thanks! I think that this strategy should work. I also think that any counterexamples would have to be in the second case. I’m still thinking about that case; I’ll post if I get somewhere.
$endgroup$
– supergeneric
Mar 14 at 3:07
$begingroup$
@supergeneric I hope you do. If I have time later I will try think about it again.
$endgroup$
– K.Power
Mar 14 at 10:28
add a comment |
$begingroup$
EDIT: I just realized that I misread the question, but I'll leave this here - maybe it helps anyway.
Since you referenced the closest point theorem, I thought you actually wanted to prove the following:
Suppose that $A$ is a non-empty convex subset of $mathbbR^N$ and that $x$ is not in the closure of $A$. Then there is a point $a' in A$ such that $|x - a'| le | x - a |$ for all $a in A$.
This statement is wrong. Take $A=(x,y)in Bbb R^2 : x^2+y^2le 1setminus (0,1)$ and $x=(0,2)$.
answered Mar 13 at 17:37
Mars PlasticMars Plastic
1,480121
$endgroup$
$begingroup$
I assume the counterexample is for your version of the conjecture? Because it does not work for the one in the OP. Also, apologies for the lack of clarity in the OP. I am trying to prove the original conjecture, but the reference to the closest point theorem made that confusing! I tried to clarify with an edit.
$endgroup$
– supergeneric
Mar 13 at 18:12
$begingroup$
Never mind - I apologize for not reading carefully enough. Yes, my example shows that the statement at the end of my answer is false, but it does not work for what you are actually interested in.
$endgroup$
– Mars Plastic
Mar 13 at 18:14
$begingroup$
No problem, thanks for looking!
$endgroup$
– supergeneric
Mar 13 at 18:17
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
This is a slightly heuristic argument, which I think may work. As you say there exists some $bin bar A$ such that $|x-a|>|b-a|$ for all $ain A$ (where $b=P_A(x)$), to avoid triviality we assume $bin bar Abackslash A$. Let us first assume that $overrightarrowxb$ intercepts $A$. So for all $varepsilon>0$ there exists an $a_varepsilonin Acapoverrightarrowxb$ such that $|b-a_varepsilon|<varepsilon$. Set $delta=|x-b|/2$. Convexity implies, for all $ain A$, that angle $overlineabangleoverlinebx$ cannot be acute. This in combination with the fact $x,b,a_delta$ all lie on the same line, and $|b-a_delta|<|b-x|$ implies that $|a-a_delta|<|a-x|$.
As is illustrated above we are using the fact that in a triangle $ABC$, if there exists a point $D$ on $AB$ such that $|AD|<|DB|$ and $angle ADC<pi/2$ we must have $|CA|<|CB|$.
If $overrightarrowxbcap A =emptyset$, I believe we can use a similar argument, but we will have to control the value of $delta$ more tightly (remembering that $a_delta$ is no longer on $overrightarrowxb$) depending on the angle made. If there was to be a counterexample I think it would have to be in this situation.
answered Mar 13 at 21:20
K.PowerK.Power
3,490926
$endgroup$
$begingroup$
Much thanks! I think that this strategy should work. I also think that any counterexamples would have to be in the second case. I’m still thinking about that case; I’ll post if I get somewhere.
$endgroup$
– supergeneric
Mar 14 at 3:07
$begingroup$
@supergeneric I hope you do. If I have time later I will try think about it again.
$endgroup$
– K.Power
Mar 14 at 10:28
add a comment |
$begingroup$
This is a slightly heuristic argument, which I think may work. As you say there exists some $bin bar A$ such that $|x-a|>|b-a|$ for all $ain A$ (where $b=P_A(x)$), to avoid triviality we assume $bin bar Abackslash A$. Let us first assume that $overrightarrowxb$ intercepts $A$. So for all $varepsilon>0$ there exists an $a_varepsilonin Acapoverrightarrowxb$ such that $|b-a_varepsilon|<varepsilon$. Set $delta=|x-b|/2$. Convexity implies, for all $ain A$, that angle $overlineabangleoverlinebx$ cannot be acute. This in combination with the fact $x,b,a_delta$ all lie on the same line, and $|b-a_delta|<|b-x|$ implies that $|a-a_delta|<|a-x|$.
As is illustrated above we are using the fact that in a triangle $ABC$, if there exists a point $D$ on $AB$ such that $|AD|<|DB|$ and $angle ADC<pi/2$ we must have $|CA|<|CB|$.
If $overrightarrowxbcap A =emptyset$, I believe we can use a similar argument, but we will have to control the value of $delta$ more tightly (remembering that $a_delta$ is no longer on $overrightarrowxb$) depending on the angle made. If there was to be a counterexample I think it would have to be in this situation.
answered Mar 13 at 21:20
K.PowerK.Power
3,490926
$endgroup$
$begingroup$
Much thanks! I think that this strategy should work. I also think that any counterexamples would have to be in the second case. I’m still thinking about that case; I’ll post if I get somewhere.
$endgroup$
– supergeneric
Mar 14 at 3:07
$begingroup$
@supergeneric I hope you do. If I have time later I will try think about it again.
$endgroup$
– K.Power
Mar 14 at 10:28
add a comment |
$begingroup$
This is a slightly heuristic argument, which I think may work. As you say there exists some $bin bar A$ such that $|x-a|>|b-a|$ for all $ain A$ (where $b=P_A(x)$), to avoid triviality we assume $bin bar Abackslash A$. Let us first assume that $overrightarrowxb$ intercepts $A$. So for all $varepsilon>0$ there exists an $a_varepsilonin Acapoverrightarrowxb$ such that $|b-a_varepsilon|<varepsilon$. Set $delta=|x-b|/2$. Convexity implies, for all $ain A$, that angle $overlineabangleoverlinebx$ cannot be acute. This in combination with the fact $x,b,a_delta$ all lie on the same line, and $|b-a_delta|<|b-x|$ implies that $|a-a_delta|<|a-x|$.
As is illustrated above we are using the fact that in a triangle $ABC$, if there exists a point $D$ on $AB$ such that $|AD|<|DB|$ and $angle ADC<pi/2$ we must have $|CA|<|CB|$.
If $overrightarrowxbcap A =emptyset$, I believe we can use a similar argument, but we will have to control the value of $delta$ more tightly (remembering that $a_delta$ is no longer on $overrightarrowxb$) depending on the angle made. If there was to be a counterexample I think it would have to be in this situation.
answered Mar 13 at 21:20
K.PowerK.Power
3,490926
$endgroup$
This is a slightly heuristic argument, which I think may work. As you say there exists some $bin bar A$ such that $|x-a|>|b-a|$ for all $ain A$ (where $b=P_A(x)$), to avoid triviality we assume $bin bar Abackslash A$. Let us first assume that $overrightarrowxb$ intercepts $A$. So for all $varepsilon>0$ there exists an $a_varepsilonin Acapoverrightarrowxb$ such that $|b-a_varepsilon|<varepsilon$. Set $delta=|x-b|/2$. Convexity implies, for all $ain A$, that angle $overlineabangleoverlinebx$ cannot be acute. This in combination with the fact $x,b,a_delta$ all lie on the same line, and $|b-a_delta|<|b-x|$ implies that $|a-a_delta|<|a-x|$.
As is illustrated above we are using the fact that in a triangle $ABC$, if there exists a point $D$ on $AB$ such that $|AD|<|DB|$ and $angle ADC<pi/2$ we must have $|CA|<|CB|$.
If $overrightarrowxbcap A =emptyset$, I believe we can use a similar argument, but we will have to control the value of $delta$ more tightly (remembering that $a_delta$ is no longer on $overrightarrowxb$) depending on the angle made. If there was to be a counterexample I think it would have to be in this situation.
answered Mar 13 at 21:20
K.PowerK.Power
3,490926
edited Mar 14 at 10:27
answered Mar 13 at 21:20
K.PowerK.Power
3,490926
answered Mar 13 at 21:20
K.PowerK.Power
3,490926
answered Mar 13 at 21:20
K.PowerK.Power
3,490926
3,490926
$begingroup$
Much thanks! I think that this strategy should work. I also think that any counterexamples would have to be in the second case. I’m still thinking about that case; I’ll post if I get somewhere.
$endgroup$
– supergeneric
Mar 14 at 3:07
$begingroup$
@supergeneric I hope you do. If I have time later I will try think about it again.
$endgroup$
– K.Power
Mar 14 at 10:28
add a comment |
$begingroup$
Much thanks! I think that this strategy should work. I also think that any counterexamples would have to be in the second case. I’m still thinking about that case; I’ll post if I get somewhere.
$endgroup$
– supergeneric
Mar 14 at 3:07
$begingroup$
@supergeneric I hope you do. If I have time later I will try think about it again.
$endgroup$
– K.Power
Mar 14 at 10:28
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Much thanks! I think that this strategy should work. I also think that any counterexamples would have to be in the second case. I’m still thinking about that case; I’ll post if I get somewhere.
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– supergeneric
Mar 14 at 3:07
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Much thanks! I think that this strategy should work. I also think that any counterexamples would have to be in the second case. I’m still thinking about that case; I’ll post if I get somewhere.
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– supergeneric
Mar 14 at 3:07
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@supergeneric I hope you do. If I have time later I will try think about it again.
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– K.Power
Mar 14 at 10:28
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@supergeneric I hope you do. If I have time later I will try think about it again.
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– K.Power
Mar 14 at 10:28
add a comment |
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EDIT: I just realized that I misread the question, but I'll leave this here - maybe it helps anyway.
Since you referenced the closest point theorem, I thought you actually wanted to prove the following:
Suppose that $A$ is a non-empty convex subset of $mathbbR^N$ and that $x$ is not in the closure of $A$. Then there is a point $a' in A$ such that $|x - a'| le | x - a |$ for all $a in A$.
This statement is wrong. Take $A=(x,y)in Bbb R^2 : x^2+y^2le 1setminus (0,1)$ and $x=(0,2)$.
answered Mar 13 at 17:37
Mars PlasticMars Plastic
1,480121
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$begingroup$
I assume the counterexample is for your version of the conjecture? Because it does not work for the one in the OP. Also, apologies for the lack of clarity in the OP. I am trying to prove the original conjecture, but the reference to the closest point theorem made that confusing! I tried to clarify with an edit.
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– supergeneric
Mar 13 at 18:12
$begingroup$
Never mind - I apologize for not reading carefully enough. Yes, my example shows that the statement at the end of my answer is false, but it does not work for what you are actually interested in.
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– Mars Plastic
Mar 13 at 18:14
$begingroup$
No problem, thanks for looking!
$endgroup$
– supergeneric
Mar 13 at 18:17
add a comment |
$begingroup$
EDIT: I just realized that I misread the question, but I'll leave this here - maybe it helps anyway.
Since you referenced the closest point theorem, I thought you actually wanted to prove the following:
Suppose that $A$ is a non-empty convex subset of $mathbbR^N$ and that $x$ is not in the closure of $A$. Then there is a point $a' in A$ such that $|x - a'| le | x - a |$ for all $a in A$.
This statement is wrong. Take $A=(x,y)in Bbb R^2 : x^2+y^2le 1setminus (0,1)$ and $x=(0,2)$.
answered Mar 13 at 17:37
Mars PlasticMars Plastic
1,480121
$endgroup$
$begingroup$
I assume the counterexample is for your version of the conjecture? Because it does not work for the one in the OP. Also, apologies for the lack of clarity in the OP. I am trying to prove the original conjecture, but the reference to the closest point theorem made that confusing! I tried to clarify with an edit.
$endgroup$
– supergeneric
Mar 13 at 18:12
$begingroup$
Never mind - I apologize for not reading carefully enough. Yes, my example shows that the statement at the end of my answer is false, but it does not work for what you are actually interested in.
$endgroup$
– Mars Plastic
Mar 13 at 18:14
$begingroup$
No problem, thanks for looking!
$endgroup$
– supergeneric
Mar 13 at 18:17
add a comment |
$begingroup$
EDIT: I just realized that I misread the question, but I'll leave this here - maybe it helps anyway.
Since you referenced the closest point theorem, I thought you actually wanted to prove the following:
Suppose that $A$ is a non-empty convex subset of $mathbbR^N$ and that $x$ is not in the closure of $A$. Then there is a point $a' in A$ such that $|x - a'| le | x - a |$ for all $a in A$.
This statement is wrong. Take $A=(x,y)in Bbb R^2 : x^2+y^2le 1setminus (0,1)$ and $x=(0,2)$.
answered Mar 13 at 17:37
Mars PlasticMars Plastic
1,480121
$endgroup$
EDIT: I just realized that I misread the question, but I'll leave this here - maybe it helps anyway.
Since you referenced the closest point theorem, I thought you actually wanted to prove the following:
Suppose that $A$ is a non-empty convex subset of $mathbbR^N$ and that $x$ is not in the closure of $A$. Then there is a point $a' in A$ such that $|x - a'| le | x - a |$ for all $a in A$.
This statement is wrong. Take $A=(x,y)in Bbb R^2 : x^2+y^2le 1setminus (0,1)$ and $x=(0,2)$.
answered Mar 13 at 17:37
Mars PlasticMars Plastic
1,480121
edited Mar 13 at 18:49
answered Mar 13 at 17:37
Mars PlasticMars Plastic
1,480121
answered Mar 13 at 17:37
Mars PlasticMars Plastic
1,480121
answered Mar 13 at 17:37
Mars PlasticMars Plastic
1,480121
1,480121
$begingroup$
I assume the counterexample is for your version of the conjecture? Because it does not work for the one in the OP. Also, apologies for the lack of clarity in the OP. I am trying to prove the original conjecture, but the reference to the closest point theorem made that confusing! I tried to clarify with an edit.
$endgroup$
– supergeneric
Mar 13 at 18:12
$begingroup$
Never mind - I apologize for not reading carefully enough. Yes, my example shows that the statement at the end of my answer is false, but it does not work for what you are actually interested in.
$endgroup$
– Mars Plastic
Mar 13 at 18:14
$begingroup$
No problem, thanks for looking!
$endgroup$
– supergeneric
Mar 13 at 18:17
add a comment |
$begingroup$
I assume the counterexample is for your version of the conjecture? Because it does not work for the one in the OP. Also, apologies for the lack of clarity in the OP. I am trying to prove the original conjecture, but the reference to the closest point theorem made that confusing! I tried to clarify with an edit.
$endgroup$
– supergeneric
Mar 13 at 18:12
$begingroup$
Never mind - I apologize for not reading carefully enough. Yes, my example shows that the statement at the end of my answer is false, but it does not work for what you are actually interested in.
$endgroup$
– Mars Plastic
Mar 13 at 18:14
$begingroup$
No problem, thanks for looking!
$endgroup$
– supergeneric
Mar 13 at 18:17
$begingroup$
I assume the counterexample is for your version of the conjecture? Because it does not work for the one in the OP. Also, apologies for the lack of clarity in the OP. I am trying to prove the original conjecture, but the reference to the closest point theorem made that confusing! I tried to clarify with an edit.
$endgroup$
– supergeneric
Mar 13 at 18:12
$begingroup$
I assume the counterexample is for your version of the conjecture? Because it does not work for the one in the OP. Also, apologies for the lack of clarity in the OP. I am trying to prove the original conjecture, but the reference to the closest point theorem made that confusing! I tried to clarify with an edit.
$endgroup$
– supergeneric
Mar 13 at 18:12
$begingroup$
Never mind - I apologize for not reading carefully enough. Yes, my example shows that the statement at the end of my answer is false, but it does not work for what you are actually interested in.
$endgroup$
– Mars Plastic
Mar 13 at 18:14
$begingroup$
Never mind - I apologize for not reading carefully enough. Yes, my example shows that the statement at the end of my answer is false, but it does not work for what you are actually interested in.
$endgroup$
– Mars Plastic
Mar 13 at 18:14
$begingroup$
No problem, thanks for looking!
$endgroup$
– supergeneric
Mar 13 at 18:17
$begingroup$
No problem, thanks for looking!
$endgroup$
– supergeneric
Mar 13 at 18:17
add a comment |
supergeneric is a new contributor. Be nice, and check out our Code of Conduct.
supergeneric is a new contributor. Be nice, and check out our Code of Conduct.
supergeneric is a new contributor. Be nice, and check out our Code of Conduct.
supergeneric is a new contributor. Be nice, and check out our Code of Conduct.
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So the problem is that $a'$ might belong to $overline A setminus A$, right?
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– Giuseppe Negro
Mar 13 at 18:02
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@GiuseppeNegro Yes, that's exactly what I'm thinking.
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– supergeneric
Mar 13 at 18:17
$begingroup$
Is $a'$ unique if $A$ is closed? If $a'$ coincides with $P(x)$, the point of $A$ that is closest to $x$, then it is unique. Thus, it is easy to construct a counterexample; just fix $x$ and consider $AsetminusP(x)$. This set can still be convex (it is if $A$ is a circle, for example), but it is not closed, and no $a'$ can exist for the $x$ you fixed earlier.
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– Giuseppe Negro
Mar 13 at 20:49
$begingroup$
@GiuseppeNegro I am not sure if your example with $A$ a circle works. I believe I've come up with an argument that covers this case, so please have a look to see if I've made a mistake in my reasoning.
$endgroup$
– K.Power
Mar 13 at 21:37
$begingroup$
What about taking the closest point to $x$ in $overlineA$ (which always exists) and moving just an infinitesimal quantity away from $x$ on the line passing through $x$ and that point to obtain $a'$? Under the small assumption that $A$ has non-empty interior, this $a'$ can be taken in $A$ and it should do the job, intuitively (then again, there might be some pathological cases...)
$endgroup$
– Daniel Robert-Nicoud
Mar 13 at 22:14