Continuity set of a difference of two upper semi-continuous real functions over a metric space
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The difference of two upper semi-continuous functions is in general neither upper- nor lower semi-continuous. But what can be said about the continuity set of such a functions, specifically its topological properties?
I know that the continuity set is a $G_delta$ set.
real-analysis general-topology metric-spaces continuity semicontinuous-functions
edited Sep 1 at 0:01
Martin Sleziak
43.7k6113260
asked Dec 14 '14 at 17:29
José
112
add a comment |Â
up vote
2
down vote
favorite
The difference of two upper semi-continuous functions is in general neither upper- nor lower semi-continuous. But what can be said about the continuity set of such a functions, specifically its topological properties?
I know that the continuity set is a $G_delta$ set.
real-analysis general-topology metric-spaces continuity semicontinuous-functions
edited Sep 1 at 0:01
Martin Sleziak
43.7k6113260
asked Dec 14 '14 at 17:29
José
112
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The difference of two upper semi-continuous functions is in general neither upper- nor lower semi-continuous. But what can be said about the continuity set of such a functions, specifically its topological properties?
I know that the continuity set is a $G_delta$ set.
real-analysis general-topology metric-spaces continuity semicontinuous-functions
edited Sep 1 at 0:01
Martin Sleziak
43.7k6113260
asked Dec 14 '14 at 17:29
José
112
The difference of two upper semi-continuous functions is in general neither upper- nor lower semi-continuous. But what can be said about the continuity set of such a functions, specifically its topological properties?
I know that the continuity set is a $G_delta$ set.
real-analysis general-topology metric-spaces continuity semicontinuous-functions
real-analysis general-topology metric-spaces continuity semicontinuous-functions
edited Sep 1 at 0:01
Martin Sleziak
43.7k6113260
asked Dec 14 '14 at 17:29
José
112
edited Sep 1 at 0:01
Martin Sleziak
43.7k6113260
asked Dec 14 '14 at 17:29
José
112
edited Sep 1 at 0:01
Martin Sleziak
43.7k6113260
edited Sep 1 at 0:01
Martin Sleziak
43.7k6113260
edited Sep 1 at 0:01
Martin Sleziak
43.7k6113260
43.7k6113260
asked Dec 14 '14 at 17:29
José
112
asked Dec 14 '14 at 17:29
José
112
asked Dec 14 '14 at 17:29
José
112
112
add a comment |Â
add a comment |Â
2 Answers
2
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oldest
votes
up vote
1
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Finally I found an useful result for my work. If $f$ is a real function over a polish space then $f$ is a Baire 1 function if and only if the restriction to a closed subset has a continuity point. That is the Baire Characterization Theorem. If a function is upper semi-continuous also it is Baire 1. The same for a difference of two upper semi-continuous real functions.
answered Dec 15 '14 at 11:04
Agudo
112
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up vote
0
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When you subtract, you throw upper semi-continuity properties out the window. This is because the negation could be very not-upper semi-continuous. For example,
$$f(x) = left{beginarraylr 1 & x in mathbbQ \ 0 & x notin mathbbQendarrayright.$$
is upper semi-continuous at each point in $mathbbQ$. So is
$$g(x) = 1$$
But $g - f$ is not upper semi-continuous at any point in $mathbbQ$.
Restriction of an upper semi-continuous function behaves better. The set of points where the restricted function is upper semi-continuous contains all points where the original function was upper semi-continuous, as well as possibly some new points on the boundary of the restricted domain.
answered Dec 14 '14 at 17:57
Chris Jones
717414
Thanks for your answer but I told about the restriction of a difference of two upper semicontinuous functions. I specially interesed in topological properties of the continuity set, I know that it is $G_delta$ but I need more.
– José
Dec 14 '14 at 18:58
Do you mean restriction in the sense of restricting the domain of a function?
– Chris Jones
Dec 14 '14 at 19:07
Yes, I do. Maybe restricted to a compact we can have something about the continuity set.
– José
Dec 14 '14 at 19:13
In general the points where $f-g$ is upper semicontinuous could be horrible. In the example above, if $f$ is the indicator function for the irrationals instead, $g-f$ is the indicator function for the rationals. But the rationals aren't $G_delta$. Any restriction of this function to a domain where continuity makes sense i.e. an interval will still have a non-$G_delta$ set of upper continuous points.
– Chris Jones
Dec 14 '14 at 19:44
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
Finally I found an useful result for my work. If $f$ is a real function over a polish space then $f$ is a Baire 1 function if and only if the restriction to a closed subset has a continuity point. That is the Baire Characterization Theorem. If a function is upper semi-continuous also it is Baire 1. The same for a difference of two upper semi-continuous real functions.
answered Dec 15 '14 at 11:04
Agudo
112
add a comment |Â
up vote
1
down vote
Finally I found an useful result for my work. If $f$ is a real function over a polish space then $f$ is a Baire 1 function if and only if the restriction to a closed subset has a continuity point. That is the Baire Characterization Theorem. If a function is upper semi-continuous also it is Baire 1. The same for a difference of two upper semi-continuous real functions.
answered Dec 15 '14 at 11:04
Agudo
112
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Finally I found an useful result for my work. If $f$ is a real function over a polish space then $f$ is a Baire 1 function if and only if the restriction to a closed subset has a continuity point. That is the Baire Characterization Theorem. If a function is upper semi-continuous also it is Baire 1. The same for a difference of two upper semi-continuous real functions.
answered Dec 15 '14 at 11:04
Agudo
112
Finally I found an useful result for my work. If $f$ is a real function over a polish space then $f$ is a Baire 1 function if and only if the restriction to a closed subset has a continuity point. That is the Baire Characterization Theorem. If a function is upper semi-continuous also it is Baire 1. The same for a difference of two upper semi-continuous real functions.
answered Dec 15 '14 at 11:04
Agudo
112
answered Dec 15 '14 at 11:04
Agudo
112
answered Dec 15 '14 at 11:04
Agudo
112
answered Dec 15 '14 at 11:04
Agudo
112
112
add a comment |Â
add a comment |Â
up vote
0
down vote
When you subtract, you throw upper semi-continuity properties out the window. This is because the negation could be very not-upper semi-continuous. For example,
$$f(x) = left{beginarraylr 1 & x in mathbbQ \ 0 & x notin mathbbQendarrayright.$$
is upper semi-continuous at each point in $mathbbQ$. So is
$$g(x) = 1$$
But $g - f$ is not upper semi-continuous at any point in $mathbbQ$.
Restriction of an upper semi-continuous function behaves better. The set of points where the restricted function is upper semi-continuous contains all points where the original function was upper semi-continuous, as well as possibly some new points on the boundary of the restricted domain.
answered Dec 14 '14 at 17:57
Chris Jones
717414
Thanks for your answer but I told about the restriction of a difference of two upper semicontinuous functions. I specially interesed in topological properties of the continuity set, I know that it is $G_delta$ but I need more.
– José
Dec 14 '14 at 18:58
Do you mean restriction in the sense of restricting the domain of a function?
– Chris Jones
Dec 14 '14 at 19:07
Yes, I do. Maybe restricted to a compact we can have something about the continuity set.
– José
Dec 14 '14 at 19:13
In general the points where $f-g$ is upper semicontinuous could be horrible. In the example above, if $f$ is the indicator function for the irrationals instead, $g-f$ is the indicator function for the rationals. But the rationals aren't $G_delta$. Any restriction of this function to a domain where continuity makes sense i.e. an interval will still have a non-$G_delta$ set of upper continuous points.
– Chris Jones
Dec 14 '14 at 19:44
add a comment |Â
up vote
0
down vote
When you subtract, you throw upper semi-continuity properties out the window. This is because the negation could be very not-upper semi-continuous. For example,
$$f(x) = left{beginarraylr 1 & x in mathbbQ \ 0 & x notin mathbbQendarrayright.$$
is upper semi-continuous at each point in $mathbbQ$. So is
$$g(x) = 1$$
But $g - f$ is not upper semi-continuous at any point in $mathbbQ$.
Restriction of an upper semi-continuous function behaves better. The set of points where the restricted function is upper semi-continuous contains all points where the original function was upper semi-continuous, as well as possibly some new points on the boundary of the restricted domain.
answered Dec 14 '14 at 17:57
Chris Jones
717414
Thanks for your answer but I told about the restriction of a difference of two upper semicontinuous functions. I specially interesed in topological properties of the continuity set, I know that it is $G_delta$ but I need more.
– José
Dec 14 '14 at 18:58
Do you mean restriction in the sense of restricting the domain of a function?
– Chris Jones
Dec 14 '14 at 19:07
Yes, I do. Maybe restricted to a compact we can have something about the continuity set.
– José
Dec 14 '14 at 19:13
In general the points where $f-g$ is upper semicontinuous could be horrible. In the example above, if $f$ is the indicator function for the irrationals instead, $g-f$ is the indicator function for the rationals. But the rationals aren't $G_delta$. Any restriction of this function to a domain where continuity makes sense i.e. an interval will still have a non-$G_delta$ set of upper continuous points.
– Chris Jones
Dec 14 '14 at 19:44
add a comment |Â
up vote
0
down vote
up vote
0
down vote
When you subtract, you throw upper semi-continuity properties out the window. This is because the negation could be very not-upper semi-continuous. For example,
$$f(x) = left{beginarraylr 1 & x in mathbbQ \ 0 & x notin mathbbQendarrayright.$$
is upper semi-continuous at each point in $mathbbQ$. So is
$$g(x) = 1$$
But $g - f$ is not upper semi-continuous at any point in $mathbbQ$.
Restriction of an upper semi-continuous function behaves better. The set of points where the restricted function is upper semi-continuous contains all points where the original function was upper semi-continuous, as well as possibly some new points on the boundary of the restricted domain.
answered Dec 14 '14 at 17:57
Chris Jones
717414
When you subtract, you throw upper semi-continuity properties out the window. This is because the negation could be very not-upper semi-continuous. For example,
$$f(x) = left{beginarraylr 1 & x in mathbbQ \ 0 & x notin mathbbQendarrayright.$$
is upper semi-continuous at each point in $mathbbQ$. So is
$$g(x) = 1$$
But $g - f$ is not upper semi-continuous at any point in $mathbbQ$.
Restriction of an upper semi-continuous function behaves better. The set of points where the restricted function is upper semi-continuous contains all points where the original function was upper semi-continuous, as well as possibly some new points on the boundary of the restricted domain.
answered Dec 14 '14 at 17:57
Chris Jones
717414
edited Dec 14 '14 at 19:40
answered Dec 14 '14 at 17:57
Chris Jones
717414
answered Dec 14 '14 at 17:57
Chris Jones
717414
answered Dec 14 '14 at 17:57
Chris Jones
717414
717414
Thanks for your answer but I told about the restriction of a difference of two upper semicontinuous functions. I specially interesed in topological properties of the continuity set, I know that it is $G_delta$ but I need more.
– José
Dec 14 '14 at 18:58
Do you mean restriction in the sense of restricting the domain of a function?
– Chris Jones
Dec 14 '14 at 19:07
Yes, I do. Maybe restricted to a compact we can have something about the continuity set.
– José
Dec 14 '14 at 19:13
In general the points where $f-g$ is upper semicontinuous could be horrible. In the example above, if $f$ is the indicator function for the irrationals instead, $g-f$ is the indicator function for the rationals. But the rationals aren't $G_delta$. Any restriction of this function to a domain where continuity makes sense i.e. an interval will still have a non-$G_delta$ set of upper continuous points.
– Chris Jones
Dec 14 '14 at 19:44
add a comment |Â
Thanks for your answer but I told about the restriction of a difference of two upper semicontinuous functions. I specially interesed in topological properties of the continuity set, I know that it is $G_delta$ but I need more.
– José
Dec 14 '14 at 18:58
Do you mean restriction in the sense of restricting the domain of a function?
– Chris Jones
Dec 14 '14 at 19:07
Yes, I do. Maybe restricted to a compact we can have something about the continuity set.
– José
Dec 14 '14 at 19:13
In general the points where $f-g$ is upper semicontinuous could be horrible. In the example above, if $f$ is the indicator function for the irrationals instead, $g-f$ is the indicator function for the rationals. But the rationals aren't $G_delta$. Any restriction of this function to a domain where continuity makes sense i.e. an interval will still have a non-$G_delta$ set of upper continuous points.
– Chris Jones
Dec 14 '14 at 19:44
Thanks for your answer but I told about the restriction of a difference of two upper semicontinuous functions. I specially interesed in topological properties of the continuity set, I know that it is $G_delta$ but I need more.
– José
Dec 14 '14 at 18:58
Thanks for your answer but I told about the restriction of a difference of two upper semicontinuous functions. I specially interesed in topological properties of the continuity set, I know that it is $G_delta$ but I need more.
– José
Dec 14 '14 at 18:58
Do you mean restriction in the sense of restricting the domain of a function?
– Chris Jones
Dec 14 '14 at 19:07
Do you mean restriction in the sense of restricting the domain of a function?
– Chris Jones
Dec 14 '14 at 19:07
Yes, I do. Maybe restricted to a compact we can have something about the continuity set.
– José
Dec 14 '14 at 19:13
Yes, I do. Maybe restricted to a compact we can have something about the continuity set.
– José
Dec 14 '14 at 19:13
In general the points where $f-g$ is upper semicontinuous could be horrible. In the example above, if $f$ is the indicator function for the irrationals instead, $g-f$ is the indicator function for the rationals. But the rationals aren't $G_delta$. Any restriction of this function to a domain where continuity makes sense i.e. an interval will still have a non-$G_delta$ set of upper continuous points.
– Chris Jones
Dec 14 '14 at 19:44
In general the points where $f-g$ is upper semicontinuous could be horrible. In the example above, if $f$ is the indicator function for the irrationals instead, $g-f$ is the indicator function for the rationals. But the rationals aren't $G_delta$. Any restriction of this function to a domain where continuity makes sense i.e. an interval will still have a non-$G_delta$ set of upper continuous points.
– Chris Jones
Dec 14 '14 at 19:44
add a comment |Â
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