Mixing
Continuity of $textrm{argmin}$ set-valued mapping
0
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Let $X subset mathbb{R}^n$ be a finite set and $Phi : X mapsto 2^{X}$ be a set-valued mapping defined as follows:
$Phi(y) := underset{x in X}{textrm{argmin}} ; L(x, y)$.
I'm trying to figure out if, without any particular assumptions on $L: X times X to mathbb{R}$, $Phi(cdot)$ is continuous over the finite set $X$. I guess no - any hints or suggestions?
general-topology functional-analysis continuity operator-theory
asked Jan 23 at 11:23
GuidoLaremiGuidoLaremi
145
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add a comment |
0
$begingroup$
Let $X subset mathbb{R}^n$ be a finite set and $Phi : X mapsto 2^{X}$ be a set-valued mapping defined as follows:
$Phi(y) := underset{x in X}{textrm{argmin}} ; L(x, y)$.
I'm trying to figure out if, without any particular assumptions on $L: X times X to mathbb{R}$, $Phi(cdot)$ is continuous over the finite set $X$. I guess no - any hints or suggestions?
general-topology functional-analysis continuity operator-theory
asked Jan 23 at 11:23
GuidoLaremiGuidoLaremi
145
$endgroup$
$begingroup$
What would you have argmin do if there are multiple points $x$ for which $L$ attains the same (minimum) value for a given $y$? It seems natural to have $Phi$ be a set-valued function, in which case you might not have a natural topology.
$endgroup$
– or1426
Jan 23 at 12:26
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@BenW Thank you, I'm not really confident with discrete topology. To prove continuity I guess you should rely on the classic definition, right?
$endgroup$
– GuidoLaremi
Jan 23 at 12:52
$begingroup$
@or1426 Exactly, this is my point. You are right, $Phi$ is a set-valued mapping and, for any given $y$, there may be more than one minimizer.
$endgroup$
– GuidoLaremi
Jan 23 at 12:55
$begingroup$
@or1426 Oh right, I didn't even realize that. So, I guess $2^X$ would have to be endowed with a topology, too. In this case, $Phi$ might not be continuous after all given the trivial topology on $X$ (although it would always be continuous given the discrete topology on $X$).
$endgroup$
– Ben W
Jan 23 at 14:28
add a comment |
0
0
0
$begingroup$
Let $X subset mathbb{R}^n$ be a finite set and $Phi : X mapsto 2^{X}$ be a set-valued mapping defined as follows:
$Phi(y) := underset{x in X}{textrm{argmin}} ; L(x, y)$.
I'm trying to figure out if, without any particular assumptions on $L: X times X to mathbb{R}$, $Phi(cdot)$ is continuous over the finite set $X$. I guess no - any hints or suggestions?
general-topology functional-analysis continuity operator-theory
asked Jan 23 at 11:23
GuidoLaremiGuidoLaremi
145
$endgroup$
Let $X subset mathbb{R}^n$ be a finite set and $Phi : X mapsto 2^{X}$ be a set-valued mapping defined as follows:
$Phi(y) := underset{x in X}{textrm{argmin}} ; L(x, y)$.
I'm trying to figure out if, without any particular assumptions on $L: X times X to mathbb{R}$, $Phi(cdot)$ is continuous over the finite set $X$. I guess no - any hints or suggestions?
general-topology functional-analysis continuity operator-theory
general-topology functional-analysis continuity operator-theory
asked Jan 23 at 11:23
GuidoLaremiGuidoLaremi
145
asked Jan 23 at 11:23
GuidoLaremiGuidoLaremi
145
edited Jan 23 at 13:02
GuidoLaremi
asked Jan 23 at 11:23
GuidoLaremiGuidoLaremi
145
asked Jan 23 at 11:23
GuidoLaremiGuidoLaremi
145
asked Jan 23 at 11:23
GuidoLaremiGuidoLaremi
145
145
$begingroup$
What would you have argmin do if there are multiple points $x$ for which $L$ attains the same (minimum) value for a given $y$? It seems natural to have $Phi$ be a set-valued function, in which case you might not have a natural topology.
$endgroup$
– or1426
Jan 23 at 12:26
$begingroup$
@BenW Thank you, I'm not really confident with discrete topology. To prove continuity I guess you should rely on the classic definition, right?
$endgroup$
– GuidoLaremi
Jan 23 at 12:52
$begingroup$
@or1426 Exactly, this is my point. You are right, $Phi$ is a set-valued mapping and, for any given $y$, there may be more than one minimizer.
$endgroup$
– GuidoLaremi
Jan 23 at 12:55
$begingroup$
@or1426 Oh right, I didn't even realize that. So, I guess $2^X$ would have to be endowed with a topology, too. In this case, $Phi$ might not be continuous after all given the trivial topology on $X$ (although it would always be continuous given the discrete topology on $X$).
$endgroup$
– Ben W
Jan 23 at 14:28
add a comment |
$begingroup$
What would you have argmin do if there are multiple points $x$ for which $L$ attains the same (minimum) value for a given $y$? It seems natural to have $Phi$ be a set-valued function, in which case you might not have a natural topology.
$endgroup$
– or1426
Jan 23 at 12:26
$begingroup$
@BenW Thank you, I'm not really confident with discrete topology. To prove continuity I guess you should rely on the classic definition, right?
$endgroup$
– GuidoLaremi
Jan 23 at 12:52
$begingroup$
@or1426 Exactly, this is my point. You are right, $Phi$ is a set-valued mapping and, for any given $y$, there may be more than one minimizer.
$endgroup$
– GuidoLaremi
Jan 23 at 12:55
$begingroup$
@or1426 Oh right, I didn't even realize that. So, I guess $2^X$ would have to be endowed with a topology, too. In this case, $Phi$ might not be continuous after all given the trivial topology on $X$ (although it would always be continuous given the discrete topology on $X$).
$endgroup$
– Ben W
Jan 23 at 14:28
$begingroup$
What would you have argmin do if there are multiple points $x$ for which $L$ attains the same (minimum) value for a given $y$? It seems natural to have $Phi$ be a set-valued function, in which case you might not have a natural topology.
$endgroup$
– or1426
Jan 23 at 12:26
$begingroup$
What would you have argmin do if there are multiple points $x$ for which $L$ attains the same (minimum) value for a given $y$? It seems natural to have $Phi$ be a set-valued function, in which case you might not have a natural topology.
$endgroup$
– or1426
Jan 23 at 12:26
$begingroup$
@BenW Thank you, I'm not really confident with discrete topology. To prove continuity I guess you should rely on the classic definition, right?
$endgroup$
– GuidoLaremi
Jan 23 at 12:52
$begingroup$
@BenW Thank you, I'm not really confident with discrete topology. To prove continuity I guess you should rely on the classic definition, right?
$endgroup$
– GuidoLaremi
Jan 23 at 12:52
$begingroup$
@or1426 Exactly, this is my point. You are right, $Phi$ is a set-valued mapping and, for any given $y$, there may be more than one minimizer.
$endgroup$
– GuidoLaremi
Jan 23 at 12:55
$begingroup$
@or1426 Exactly, this is my point. You are right, $Phi$ is a set-valued mapping and, for any given $y$, there may be more than one minimizer.
$endgroup$
– GuidoLaremi
Jan 23 at 12:55
$begingroup$
@or1426 Oh right, I didn't even realize that. So, I guess $2^X$ would have to be endowed with a topology, too. In this case, $Phi$ might not be continuous after all given the trivial topology on $X$ (although it would always be continuous given the discrete topology on $X$).
$endgroup$
– Ben W
Jan 23 at 14:28
$begingroup$
@or1426 Oh right, I didn't even realize that. So, I guess $2^X$ would have to be endowed with a topology, too. In this case, $Phi$ might not be continuous after all given the trivial topology on $X$ (although it would always be continuous given the discrete topology on $X$).
$endgroup$
– Ben W
Jan 23 at 14:28
add a comment |
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$begingroup$
What would you have argmin do if there are multiple points $x$ for which $L$ attains the same (minimum) value for a given $y$? It seems natural to have $Phi$ be a set-valued function, in which case you might not have a natural topology.
$endgroup$
– or1426
Jan 23 at 12:26
$begingroup$
@BenW Thank you, I'm not really confident with discrete topology. To prove continuity I guess you should rely on the classic definition, right?
$endgroup$
– GuidoLaremi
Jan 23 at 12:52
$begingroup$
@or1426 Exactly, this is my point. You are right, $Phi$ is a set-valued mapping and, for any given $y$, there may be more than one minimizer.
$endgroup$
– GuidoLaremi
Jan 23 at 12:55
$begingroup$
@or1426 Oh right, I didn't even realize that. So, I guess $2^X$ would have to be endowed with a topology, too. In this case, $Phi$ might not be continuous after all given the trivial topology on $X$ (although it would always be continuous given the discrete topology on $X$).
$endgroup$
– Ben W
Jan 23 at 14:28