Let $f:Xrightarrow Bbb{R},cup,{+infty}$ be a map. Then, for $x_0in X,;f(x_0)geq suplimits_{Vin...
Let $f:Xrightarrow Bbb{R},cup,{+infty}$ be a map where $X$ is a real normed space. Then, for an arbitrary $x_0in X,$ I want to prove that the following always holds:
begin{align}f(x_0)geq suplimits_{Vin U(x_0)}inflimits_{xin V}f(x)
end{align}
MY TRIAL
Let $x_0in X,$ then $exists,Vin,U(x_0)$ such that $f(x_0)geq f(x),;forall;xin V$, where $U(x_0)$ is the set of all neighbourhoods of $x_0$. So,
begin{align}f(x_0)geq inflimits_{xin V}f(x),;;text{for some};;Vin U(x_0),
end{align}
which implies that begin{align}f(x_0)geq suplimits_{Vin U(x_0)}inflimits_{xin V}f(x)
end{align}
Kindly check if I am correct or wrong. If it happens that I'm wrong, kindly give an alternative proof. Thanks
functional-analysis functions supremum-and-infimum
asked Nov 24 at 13:28
Mike
1,194218
add a comment |
Let $f:Xrightarrow Bbb{R},cup,{+infty}$ be a map where $X$ is a real normed space. Then, for an arbitrary $x_0in X,$ I want to prove that the following always holds:
begin{align}f(x_0)geq suplimits_{Vin U(x_0)}inflimits_{xin V}f(x)
end{align}
MY TRIAL
Let $x_0in X,$ then $exists,Vin,U(x_0)$ such that $f(x_0)geq f(x),;forall;xin V$, where $U(x_0)$ is the set of all neighbourhoods of $x_0$. So,
begin{align}f(x_0)geq inflimits_{xin V}f(x),;;text{for some};;Vin U(x_0),
end{align}
which implies that begin{align}f(x_0)geq suplimits_{Vin U(x_0)}inflimits_{xin V}f(x)
end{align}
Kindly check if I am correct or wrong. If it happens that I'm wrong, kindly give an alternative proof. Thanks
functional-analysis functions supremum-and-infimum
asked Nov 24 at 13:28
Mike
1,194218
You need to give us a hint - what is $U(x_0)$?
– David C. Ullrich
Nov 24 at 13:32
@David C. Ullrich: Sorry! $U(x_0)$ is the set of all neighbourhoods of $x_0$
– Mike
Nov 24 at 13:34
@Mike Oh, so $X$ is a topological space? But why should that matter when $f$ is not necessarily continuos?
– Hagen von Eitzen
Nov 24 at 13:35
@Hagen von Eitzen: Yes! Infact, $X$ is a real normed space.
– Mike
Nov 24 at 13:36
add a comment |
Let $f:Xrightarrow Bbb{R},cup,{+infty}$ be a map where $X$ is a real normed space. Then, for an arbitrary $x_0in X,$ I want to prove that the following always holds:
begin{align}f(x_0)geq suplimits_{Vin U(x_0)}inflimits_{xin V}f(x)
end{align}
MY TRIAL
Let $x_0in X,$ then $exists,Vin,U(x_0)$ such that $f(x_0)geq f(x),;forall;xin V$, where $U(x_0)$ is the set of all neighbourhoods of $x_0$. So,
begin{align}f(x_0)geq inflimits_{xin V}f(x),;;text{for some};;Vin U(x_0),
end{align}
which implies that begin{align}f(x_0)geq suplimits_{Vin U(x_0)}inflimits_{xin V}f(x)
end{align}
Kindly check if I am correct or wrong. If it happens that I'm wrong, kindly give an alternative proof. Thanks
functional-analysis functions supremum-and-infimum
asked Nov 24 at 13:28
Mike
1,194218
Let $f:Xrightarrow Bbb{R},cup,{+infty}$ be a map where $X$ is a real normed space. Then, for an arbitrary $x_0in X,$ I want to prove that the following always holds:
begin{align}f(x_0)geq suplimits_{Vin U(x_0)}inflimits_{xin V}f(x)
end{align}
MY TRIAL
Let $x_0in X,$ then $exists,Vin,U(x_0)$ such that $f(x_0)geq f(x),;forall;xin V$, where $U(x_0)$ is the set of all neighbourhoods of $x_0$. So,
begin{align}f(x_0)geq inflimits_{xin V}f(x),;;text{for some};;Vin U(x_0),
end{align}
which implies that begin{align}f(x_0)geq suplimits_{Vin U(x_0)}inflimits_{xin V}f(x)
end{align}
Kindly check if I am correct or wrong. If it happens that I'm wrong, kindly give an alternative proof. Thanks
functional-analysis functions supremum-and-infimum
functional-analysis functions supremum-and-infimum
asked Nov 24 at 13:28
Mike
1,194218
asked Nov 24 at 13:28
Mike
1,194218
edited Nov 24 at 13:41
asked Nov 24 at 13:28
Mike
1,194218
asked Nov 24 at 13:28
Mike
1,194218
asked Nov 24 at 13:28
Mike
1,194218
1,194218
You need to give us a hint - what is $U(x_0)$?
– David C. Ullrich
Nov 24 at 13:32
@David C. Ullrich: Sorry! $U(x_0)$ is the set of all neighbourhoods of $x_0$
– Mike
Nov 24 at 13:34
@Mike Oh, so $X$ is a topological space? But why should that matter when $f$ is not necessarily continuos?
– Hagen von Eitzen
Nov 24 at 13:35
@Hagen von Eitzen: Yes! Infact, $X$ is a real normed space.
– Mike
Nov 24 at 13:36
add a comment |
You need to give us a hint - what is $U(x_0)$?
– David C. Ullrich
Nov 24 at 13:32
@David C. Ullrich: Sorry! $U(x_0)$ is the set of all neighbourhoods of $x_0$
– Mike
Nov 24 at 13:34
@Mike Oh, so $X$ is a topological space? But why should that matter when $f$ is not necessarily continuos?
– Hagen von Eitzen
Nov 24 at 13:35
@Hagen von Eitzen: Yes! Infact, $X$ is a real normed space.
– Mike
Nov 24 at 13:36
You need to give us a hint - what is $U(x_0)$?
– David C. Ullrich
Nov 24 at 13:32
You need to give us a hint - what is $U(x_0)$?
– David C. Ullrich
Nov 24 at 13:32
@David C. Ullrich: Sorry! $U(x_0)$ is the set of all neighbourhoods of $x_0$
– Mike
Nov 24 at 13:34
@David C. Ullrich: Sorry! $U(x_0)$ is the set of all neighbourhoods of $x_0$
– Mike
Nov 24 at 13:34
@Mike Oh, so $X$ is a topological space? But why should that matter when $f$ is not necessarily continuos?
– Hagen von Eitzen
Nov 24 at 13:35
@Mike Oh, so $X$ is a topological space? But why should that matter when $f$ is not necessarily continuos?
– Hagen von Eitzen
Nov 24 at 13:35
@Hagen von Eitzen: Yes! Infact, $X$ is a real normed space.
– Mike
Nov 24 at 13:36
@Hagen von Eitzen: Yes! Infact, $X$ is a real normed space.
– Mike
Nov 24 at 13:36
add a comment |
1 Answer
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There is not the slightest reason why $Vin U(x_0)$ with $f(x_0)ge f(x),forall xin V$ should exist.
However, if $Vin U(x_0)$ then $x_0in V$ (here, I make a wild guess that $U(x_0)$ denotes some non-empty subset of the power set of $X$ and each of its elements contains $x_0$) and hence $inf_{xin V} f(x)le f(x_0)$.
answered Nov 24 at 13:32
Hagen von Eitzen
276k21268495
add a comment |
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There is not the slightest reason why $Vin U(x_0)$ with $f(x_0)ge f(x),forall xin V$ should exist.
However, if $Vin U(x_0)$ then $x_0in V$ (here, I make a wild guess that $U(x_0)$ denotes some non-empty subset of the power set of $X$ and each of its elements contains $x_0$) and hence $inf_{xin V} f(x)le f(x_0)$.
answered Nov 24 at 13:32
Hagen von Eitzen
276k21268495
add a comment |
There is not the slightest reason why $Vin U(x_0)$ with $f(x_0)ge f(x),forall xin V$ should exist.
However, if $Vin U(x_0)$ then $x_0in V$ (here, I make a wild guess that $U(x_0)$ denotes some non-empty subset of the power set of $X$ and each of its elements contains $x_0$) and hence $inf_{xin V} f(x)le f(x_0)$.
answered Nov 24 at 13:32
Hagen von Eitzen
276k21268495
add a comment |
There is not the slightest reason why $Vin U(x_0)$ with $f(x_0)ge f(x),forall xin V$ should exist.
However, if $Vin U(x_0)$ then $x_0in V$ (here, I make a wild guess that $U(x_0)$ denotes some non-empty subset of the power set of $X$ and each of its elements contains $x_0$) and hence $inf_{xin V} f(x)le f(x_0)$.
answered Nov 24 at 13:32
Hagen von Eitzen
276k21268495
There is not the slightest reason why $Vin U(x_0)$ with $f(x_0)ge f(x),forall xin V$ should exist.
However, if $Vin U(x_0)$ then $x_0in V$ (here, I make a wild guess that $U(x_0)$ denotes some non-empty subset of the power set of $X$ and each of its elements contains $x_0$) and hence $inf_{xin V} f(x)le f(x_0)$.
answered Nov 24 at 13:32
Hagen von Eitzen
276k21268495
answered Nov 24 at 13:32
Hagen von Eitzen
276k21268495
answered Nov 24 at 13:32
Hagen von Eitzen
276k21268495
answered Nov 24 at 13:32
Hagen von Eitzen
276k21268495
276k21268495
add a comment |
add a comment |
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You need to give us a hint - what is $U(x_0)$?
– David C. Ullrich
Nov 24 at 13:32
@David C. Ullrich: Sorry! $U(x_0)$ is the set of all neighbourhoods of $x_0$
– Mike
Nov 24 at 13:34
@Mike Oh, so $X$ is a topological space? But why should that matter when $f$ is not necessarily continuos?
– Hagen von Eitzen
Nov 24 at 13:35
@Hagen von Eitzen: Yes! Infact, $X$ is a real normed space.
– Mike
Nov 24 at 13:36