$U={xin mathbb{R^n} :g_i(x)leq 0 : text{for} quad i=1,…,m }$ is closed
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Let $U={xin mathbb{R^n} :g_i(x)leq 0 : forall i=1,...,m }$ $subset$ $mathbb{R^n}$.
with $(g_i)_{1leq i leq n}$ are convex functions from $mathbb{R^n}$ in $mathbb{R}$.
we want to prove that $U$ is a closed set.
general-topology convex-analysis
$endgroup$
add a comment |
$begingroup$
Let $U={xin mathbb{R^n} :g_i(x)leq 0 : forall i=1,...,m }$ $subset$ $mathbb{R^n}$.
with $(g_i)_{1leq i leq n}$ are convex functions from $mathbb{R^n}$ in $mathbb{R}$.
we want to prove that $U$ is a closed set.
general-topology convex-analysis
$endgroup$
$begingroup$
Should it be $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$?
$endgroup$
– TrostAft
Dec 19 '18 at 7:23
add a comment |
$begingroup$
Let $U={xin mathbb{R^n} :g_i(x)leq 0 : forall i=1,...,m }$ $subset$ $mathbb{R^n}$.
with $(g_i)_{1leq i leq n}$ are convex functions from $mathbb{R^n}$ in $mathbb{R}$.
we want to prove that $U$ is a closed set.
general-topology convex-analysis
$endgroup$
Let $U={xin mathbb{R^n} :g_i(x)leq 0 : forall i=1,...,m }$ $subset$ $mathbb{R^n}$.
with $(g_i)_{1leq i leq n}$ are convex functions from $mathbb{R^n}$ in $mathbb{R}$.
we want to prove that $U$ is a closed set.
general-topology convex-analysis
general-topology convex-analysis
edited Dec 19 '18 at 7:46
Hossien Sahebjame
817
817
asked Dec 19 '18 at 7:21
Abderrahmane OuachouachAbderrahmane Ouachouach
61
61
$begingroup$
Should it be $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$?
$endgroup$
– TrostAft
Dec 19 '18 at 7:23
add a comment |
$begingroup$
Should it be $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$?
$endgroup$
– TrostAft
Dec 19 '18 at 7:23
$begingroup$
Should it be $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$?
$endgroup$
– TrostAft
Dec 19 '18 at 7:23
$begingroup$
Should it be $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$?
$endgroup$
– TrostAft
Dec 19 '18 at 7:23
add a comment |
2 Answers
2
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oldest
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$begingroup$
Any convex function from $mathbb R^{n}$ to $mathbb R$ is continuous. It follows that $U$ is the intersection of a $m$ closed sets, hence closed.
$endgroup$
add a comment |
$begingroup$
I think we have $g=(g_1,...,g_m): mathbb R^n to mathbb R^m$ and $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$ with each $g_i$ convex.
Let $U_i := {x in mathbb{R}^n : g_i(x) leq 0}$.
Since $g_i$ is continuous, we have that $U_i$ is closed. Hence $U= bigcap_{i=1}^n U_i$ is closed.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
oldest
votes
$begingroup$
Any convex function from $mathbb R^{n}$ to $mathbb R$ is continuous. It follows that $U$ is the intersection of a $m$ closed sets, hence closed.
$endgroup$
add a comment |
$begingroup$
Any convex function from $mathbb R^{n}$ to $mathbb R$ is continuous. It follows that $U$ is the intersection of a $m$ closed sets, hence closed.
$endgroup$
add a comment |
$begingroup$
Any convex function from $mathbb R^{n}$ to $mathbb R$ is continuous. It follows that $U$ is the intersection of a $m$ closed sets, hence closed.
$endgroup$
Any convex function from $mathbb R^{n}$ to $mathbb R$ is continuous. It follows that $U$ is the intersection of a $m$ closed sets, hence closed.
answered Dec 19 '18 at 7:29
Kavi Rama MurthyKavi Rama Murthy
56k42158
56k42158
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$begingroup$
I think we have $g=(g_1,...,g_m): mathbb R^n to mathbb R^m$ and $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$ with each $g_i$ convex.
Let $U_i := {x in mathbb{R}^n : g_i(x) leq 0}$.
Since $g_i$ is continuous, we have that $U_i$ is closed. Hence $U= bigcap_{i=1}^n U_i$ is closed.
$endgroup$
add a comment |
$begingroup$
I think we have $g=(g_1,...,g_m): mathbb R^n to mathbb R^m$ and $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$ with each $g_i$ convex.
Let $U_i := {x in mathbb{R}^n : g_i(x) leq 0}$.
Since $g_i$ is continuous, we have that $U_i$ is closed. Hence $U= bigcap_{i=1}^n U_i$ is closed.
$endgroup$
add a comment |
$begingroup$
I think we have $g=(g_1,...,g_m): mathbb R^n to mathbb R^m$ and $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$ with each $g_i$ convex.
Let $U_i := {x in mathbb{R}^n : g_i(x) leq 0}$.
Since $g_i$ is continuous, we have that $U_i$ is closed. Hence $U= bigcap_{i=1}^n U_i$ is closed.
$endgroup$
I think we have $g=(g_1,...,g_m): mathbb R^n to mathbb R^m$ and $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$ with each $g_i$ convex.
Let $U_i := {x in mathbb{R}^n : g_i(x) leq 0}$.
Since $g_i$ is continuous, we have that $U_i$ is closed. Hence $U= bigcap_{i=1}^n U_i$ is closed.
answered Dec 19 '18 at 7:30
FredFred
45.2k1847
45.2k1847
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$begingroup$
Should it be $U = {x in mathbb{R}^n : g_i(x) leq 0, forall i = 1, dots, m}$?
$endgroup$
– TrostAft
Dec 19 '18 at 7:23