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Suprenum of set of lower semicontinuous functions is lower semicontinuous.
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From Rudin, Real and Complex Analysis
We are given that a real valued function is lower semicontinuous if
$x: f(x)> alpha$ is open for every real $alpha$.
I'm thinking of proving the former as follows:
Take any family of lower semicontinuous functions
$f_a$ for some indexed $a$
Then we have that
$x: f_a (x)> alpha$ is open for all lower semicontinuous $f_a$.
Taking the sup of this set we know that it is still open.
Hence for $g=supf_a$ we have that the set
x: g(x) > $alpha$ must also be open...
I would like to know if there is a way to formalize this sketch?
I've seen other proofs of this statement but they seem to follow from a few more facts on lower semicontinuous functions, which are not immediately presented in this book.
real-analysis lebesgue-measure
asked Jul 27 at 8:41
Jaaziel
437
add a comment |Â
up vote
0
down vote
favorite
From Rudin, Real and Complex Analysis
We are given that a real valued function is lower semicontinuous if
$x: f(x)> alpha$ is open for every real $alpha$.
I'm thinking of proving the former as follows:
Take any family of lower semicontinuous functions
$f_a$ for some indexed $a$
Then we have that
$x: f_a (x)> alpha$ is open for all lower semicontinuous $f_a$.
Taking the sup of this set we know that it is still open.
Hence for $g=supf_a$ we have that the set
x: g(x) > $alpha$ must also be open...
I would like to know if there is a way to formalize this sketch?
I've seen other proofs of this statement but they seem to follow from a few more facts on lower semicontinuous functions, which are not immediately presented in this book.
real-analysis lebesgue-measure
asked Jul 27 at 8:41
Jaaziel
437
what's "the former"???
– mathworker21
Jul 27 at 8:43
The statement in the title..
– Jaaziel
Jul 27 at 8:43
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
From Rudin, Real and Complex Analysis
We are given that a real valued function is lower semicontinuous if
$x: f(x)> alpha$ is open for every real $alpha$.
I'm thinking of proving the former as follows:
Take any family of lower semicontinuous functions
$f_a$ for some indexed $a$
Then we have that
$x: f_a (x)> alpha$ is open for all lower semicontinuous $f_a$.
Taking the sup of this set we know that it is still open.
Hence for $g=supf_a$ we have that the set
x: g(x) > $alpha$ must also be open...
I would like to know if there is a way to formalize this sketch?
I've seen other proofs of this statement but they seem to follow from a few more facts on lower semicontinuous functions, which are not immediately presented in this book.
real-analysis lebesgue-measure
asked Jul 27 at 8:41
Jaaziel
437
From Rudin, Real and Complex Analysis
We are given that a real valued function is lower semicontinuous if
$x: f(x)> alpha$ is open for every real $alpha$.
I'm thinking of proving the former as follows:
Take any family of lower semicontinuous functions
$f_a$ for some indexed $a$
Then we have that
$x: f_a (x)> alpha$ is open for all lower semicontinuous $f_a$.
Taking the sup of this set we know that it is still open.
Hence for $g=supf_a$ we have that the set
x: g(x) > $alpha$ must also be open...
I would like to know if there is a way to formalize this sketch?
I've seen other proofs of this statement but they seem to follow from a few more facts on lower semicontinuous functions, which are not immediately presented in this book.
real-analysis lebesgue-measure
asked Jul 27 at 8:41
Jaaziel
437
asked Jul 27 at 8:41
Jaaziel
437
asked Jul 27 at 8:41
Jaaziel
437
asked Jul 27 at 8:41
Jaaziel
437
437
what's "the former"???
– mathworker21
Jul 27 at 8:43
The statement in the title..
– Jaaziel
Jul 27 at 8:43
add a comment |Â
what's "the former"???
– mathworker21
Jul 27 at 8:43
The statement in the title..
– Jaaziel
Jul 27 at 8:43
what's "the former"???
– mathworker21
Jul 27 at 8:43
what's "the former"???
– mathworker21
Jul 27 at 8:43
The statement in the title..
– Jaaziel
Jul 27 at 8:43
The statement in the title..
– Jaaziel
Jul 27 at 8:43
add a comment |Â
2 Answers
2
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oldest
votes
up vote
1
down vote
I'm a bit confused what you're asking. It seems all you need to do is show that if $x : f_a(x) > alpha$ is open for each $alpha$, then $x : sup_a f_a(x) > alpha$ is open. But this is easy. If $sup_a f_a(x) > alpha$, then $f_a(x) > alpha$ for some $a$ and so by assumption, there's some neighborhood around $x$ for which $f_a > alpha$, and then on this neighborhood, $sup_a f_a > alpha$, as desired.
answered Jul 27 at 8:45
mathworker21
6,4301727
add a comment |Â
up vote
1
down vote
$x:sup_a f_a(x)>alpha =cup_a x:f_a(x)>alpha $ and union of open sets is open.
answered Jul 27 at 9:06
Kavi Rama Murthy
20k2829
Actually that was one of my first thoughts.. how do we know it's certain that it's a union?
– Jaaziel
Jul 27 at 9:07
1
Just show that each side is contained in the other. If a number is less than the supremum of certain numbers then it must be less than one of those numbers; this shows that LHS is contained in RHS. To show that RHS is contained in LHS just observe that each of the sets $x:f_a(x)>alpha $ is contained in LHS.
– Kavi Rama Murthy
Jul 27 at 9:21
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I'm a bit confused what you're asking. It seems all you need to do is show that if $x : f_a(x) > alpha$ is open for each $alpha$, then $x : sup_a f_a(x) > alpha$ is open. But this is easy. If $sup_a f_a(x) > alpha$, then $f_a(x) > alpha$ for some $a$ and so by assumption, there's some neighborhood around $x$ for which $f_a > alpha$, and then on this neighborhood, $sup_a f_a > alpha$, as desired.
answered Jul 27 at 8:45
mathworker21
6,4301727
add a comment |Â
up vote
1
down vote
I'm a bit confused what you're asking. It seems all you need to do is show that if $x : f_a(x) > alpha$ is open for each $alpha$, then $x : sup_a f_a(x) > alpha$ is open. But this is easy. If $sup_a f_a(x) > alpha$, then $f_a(x) > alpha$ for some $a$ and so by assumption, there's some neighborhood around $x$ for which $f_a > alpha$, and then on this neighborhood, $sup_a f_a > alpha$, as desired.
answered Jul 27 at 8:45
mathworker21
6,4301727
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I'm a bit confused what you're asking. It seems all you need to do is show that if $x : f_a(x) > alpha$ is open for each $alpha$, then $x : sup_a f_a(x) > alpha$ is open. But this is easy. If $sup_a f_a(x) > alpha$, then $f_a(x) > alpha$ for some $a$ and so by assumption, there's some neighborhood around $x$ for which $f_a > alpha$, and then on this neighborhood, $sup_a f_a > alpha$, as desired.
answered Jul 27 at 8:45
mathworker21
6,4301727
I'm a bit confused what you're asking. It seems all you need to do is show that if $x : f_a(x) > alpha$ is open for each $alpha$, then $x : sup_a f_a(x) > alpha$ is open. But this is easy. If $sup_a f_a(x) > alpha$, then $f_a(x) > alpha$ for some $a$ and so by assumption, there's some neighborhood around $x$ for which $f_a > alpha$, and then on this neighborhood, $sup_a f_a > alpha$, as desired.
answered Jul 27 at 8:45
mathworker21
6,4301727
answered Jul 27 at 8:45
mathworker21
6,4301727
answered Jul 27 at 8:45
mathworker21
6,4301727
answered Jul 27 at 8:45
mathworker21
6,4301727
6,4301727
add a comment |Â
add a comment |Â
up vote
1
down vote
$x:sup_a f_a(x)>alpha =cup_a x:f_a(x)>alpha $ and union of open sets is open.
answered Jul 27 at 9:06
Kavi Rama Murthy
20k2829
Actually that was one of my first thoughts.. how do we know it's certain that it's a union?
– Jaaziel
Jul 27 at 9:07
1
Just show that each side is contained in the other. If a number is less than the supremum of certain numbers then it must be less than one of those numbers; this shows that LHS is contained in RHS. To show that RHS is contained in LHS just observe that each of the sets $x:f_a(x)>alpha $ is contained in LHS.
– Kavi Rama Murthy
Jul 27 at 9:21
add a comment |Â
up vote
1
down vote
$x:sup_a f_a(x)>alpha =cup_a x:f_a(x)>alpha $ and union of open sets is open.
answered Jul 27 at 9:06
Kavi Rama Murthy
20k2829
Actually that was one of my first thoughts.. how do we know it's certain that it's a union?
– Jaaziel
Jul 27 at 9:07
1
Just show that each side is contained in the other. If a number is less than the supremum of certain numbers then it must be less than one of those numbers; this shows that LHS is contained in RHS. To show that RHS is contained in LHS just observe that each of the sets $x:f_a(x)>alpha $ is contained in LHS.
– Kavi Rama Murthy
Jul 27 at 9:21
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$x:sup_a f_a(x)>alpha =cup_a x:f_a(x)>alpha $ and union of open sets is open.
answered Jul 27 at 9:06
Kavi Rama Murthy
20k2829
$x:sup_a f_a(x)>alpha =cup_a x:f_a(x)>alpha $ and union of open sets is open.
answered Jul 27 at 9:06
Kavi Rama Murthy
20k2829
answered Jul 27 at 9:06
Kavi Rama Murthy
20k2829
answered Jul 27 at 9:06
Kavi Rama Murthy
20k2829
answered Jul 27 at 9:06
Kavi Rama Murthy
20k2829
20k2829
Actually that was one of my first thoughts.. how do we know it's certain that it's a union?
– Jaaziel
Jul 27 at 9:07
1
Just show that each side is contained in the other. If a number is less than the supremum of certain numbers then it must be less than one of those numbers; this shows that LHS is contained in RHS. To show that RHS is contained in LHS just observe that each of the sets $x:f_a(x)>alpha $ is contained in LHS.
– Kavi Rama Murthy
Jul 27 at 9:21
add a comment |Â
Actually that was one of my first thoughts.. how do we know it's certain that it's a union?
– Jaaziel
Jul 27 at 9:07
1
Just show that each side is contained in the other. If a number is less than the supremum of certain numbers then it must be less than one of those numbers; this shows that LHS is contained in RHS. To show that RHS is contained in LHS just observe that each of the sets $x:f_a(x)>alpha $ is contained in LHS.
– Kavi Rama Murthy
Jul 27 at 9:21
Actually that was one of my first thoughts.. how do we know it's certain that it's a union?
– Jaaziel
Jul 27 at 9:07
Actually that was one of my first thoughts.. how do we know it's certain that it's a union?
– Jaaziel
Jul 27 at 9:07
1
1
Just show that each side is contained in the other. If a number is less than the supremum of certain numbers then it must be less than one of those numbers; this shows that LHS is contained in RHS. To show that RHS is contained in LHS just observe that each of the sets $x:f_a(x)>alpha $ is contained in LHS.
– Kavi Rama Murthy
Jul 27 at 9:21
Just show that each side is contained in the other. If a number is less than the supremum of certain numbers then it must be less than one of those numbers; this shows that LHS is contained in RHS. To show that RHS is contained in LHS just observe that each of the sets $x:f_a(x)>alpha $ is contained in LHS.
– Kavi Rama Murthy
Jul 27 at 9:21
add a comment |Â
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what's "the former"???
– mathworker21
Jul 27 at 8:43
The statement in the title..
– Jaaziel
Jul 27 at 8:43