Let $f:R^nrightarrow R$ be a lower semi-continuity function, how to show for any constant $r$ , $U={zin R^n :...
Let $f:R^nrightarrow R$ be a lower semi-continuity function, how to show for any constant $r$ $U={zin R^n : f(z)> r}$ is open ?
real-analysis general-topology continuity
asked Jun 5 '16 at 14:07
lanse7pty
1,7751823
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Let $f:R^nrightarrow R$ be a lower semi-continuity function, how to show for any constant $r$ $U={zin R^n : f(z)> r}$ is open ?
real-analysis general-topology continuity
asked Jun 5 '16 at 14:07
lanse7pty
1,7751823
add a comment |
Let $f:R^nrightarrow R$ be a lower semi-continuity function, how to show for any constant $r$ $U={zin R^n : f(z)> r}$ is open ?
real-analysis general-topology continuity
asked Jun 5 '16 at 14:07
lanse7pty
1,7751823
Let $f:R^nrightarrow R$ be a lower semi-continuity function, how to show for any constant $r$ $U={zin R^n : f(z)> r}$ is open ?
real-analysis general-topology continuity
real-analysis general-topology continuity
asked Jun 5 '16 at 14:07
lanse7pty
1,7751823
asked Jun 5 '16 at 14:07
lanse7pty
1,7751823
asked Jun 5 '16 at 14:07
lanse7pty
1,7751823
asked Jun 5 '16 at 14:07
lanse7pty
1,7751823
asked Jun 5 '16 at 14:07
lanse7pty
1,7751823
1,7751823
add a comment |
add a comment |
2 Answers
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Consider a sequence $x_n in mathbb{R}^nbackslash U = E$ and assume $x_nrightarrow x$. Then $f(x) le liminf f(x_n) le r$, so $xin E$ and $E$ is closed.
answered Jun 5 '16 at 14:11
Thomas
16.7k21631
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We have
begin{align*}
{ f le s } text{ closed for all } s in mathbb{R}
iff & { f > -s } text{ open for all } s in mathbb{R} \
iff & { f > s } text{ open for all } s in mathbb{R}
end{align*}
Now a subset $A subseteq X$ is called open in $X$, when
begin{equation} tag{$star$} label{eq:offen}
forall a in A exists delta > 0: B_{delta}(a) subset A.
end{equation}
Because $f$ is lower semi continuous on $X$, we have
begin{equation*}
forall c < f(x) exists delta > 0: f(y) > c forall y in B_{delta}(x).
end{equation*}
(follows from the standard $varepsilon-delta$-definition with $varepsilon := f(x) - c$.)
Now let $A := { f > s }$.
From definition eqref{eq:offen} the proposition follows, since for all $a in A { c in mathbb{R}: f(x) > c}$ exists a $delta > 0$, so that for all $y$ with $| x - y | < delta$, we have $y in A$ and therefore $f(y) > c$.
answered Nov 24 at 15:06
Viktor Glombik
513422
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
Consider a sequence $x_n in mathbb{R}^nbackslash U = E$ and assume $x_nrightarrow x$. Then $f(x) le liminf f(x_n) le r$, so $xin E$ and $E$ is closed.
answered Jun 5 '16 at 14:11
Thomas
16.7k21631
add a comment |
Consider a sequence $x_n in mathbb{R}^nbackslash U = E$ and assume $x_nrightarrow x$. Then $f(x) le liminf f(x_n) le r$, so $xin E$ and $E$ is closed.
answered Jun 5 '16 at 14:11
Thomas
16.7k21631
add a comment |
Consider a sequence $x_n in mathbb{R}^nbackslash U = E$ and assume $x_nrightarrow x$. Then $f(x) le liminf f(x_n) le r$, so $xin E$ and $E$ is closed.
answered Jun 5 '16 at 14:11
Thomas
16.7k21631
Consider a sequence $x_n in mathbb{R}^nbackslash U = E$ and assume $x_nrightarrow x$. Then $f(x) le liminf f(x_n) le r$, so $xin E$ and $E$ is closed.
answered Jun 5 '16 at 14:11
Thomas
16.7k21631
answered Jun 5 '16 at 14:11
Thomas
16.7k21631
answered Jun 5 '16 at 14:11
Thomas
16.7k21631
answered Jun 5 '16 at 14:11
Thomas
16.7k21631
16.7k21631
add a comment |
add a comment |
We have
begin{align*}
{ f le s } text{ closed for all } s in mathbb{R}
iff & { f > -s } text{ open for all } s in mathbb{R} \
iff & { f > s } text{ open for all } s in mathbb{R}
end{align*}
Now a subset $A subseteq X$ is called open in $X$, when
begin{equation} tag{$star$} label{eq:offen}
forall a in A exists delta > 0: B_{delta}(a) subset A.
end{equation}
Because $f$ is lower semi continuous on $X$, we have
begin{equation*}
forall c < f(x) exists delta > 0: f(y) > c forall y in B_{delta}(x).
end{equation*}
(follows from the standard $varepsilon-delta$-definition with $varepsilon := f(x) - c$.)
Now let $A := { f > s }$.
From definition eqref{eq:offen} the proposition follows, since for all $a in A { c in mathbb{R}: f(x) > c}$ exists a $delta > 0$, so that for all $y$ with $| x - y | < delta$, we have $y in A$ and therefore $f(y) > c$.
answered Nov 24 at 15:06
Viktor Glombik
513422
add a comment |
We have
begin{align*}
{ f le s } text{ closed for all } s in mathbb{R}
iff & { f > -s } text{ open for all } s in mathbb{R} \
iff & { f > s } text{ open for all } s in mathbb{R}
end{align*}
Now a subset $A subseteq X$ is called open in $X$, when
begin{equation} tag{$star$} label{eq:offen}
forall a in A exists delta > 0: B_{delta}(a) subset A.
end{equation}
Because $f$ is lower semi continuous on $X$, we have
begin{equation*}
forall c < f(x) exists delta > 0: f(y) > c forall y in B_{delta}(x).
end{equation*}
(follows from the standard $varepsilon-delta$-definition with $varepsilon := f(x) - c$.)
Now let $A := { f > s }$.
From definition eqref{eq:offen} the proposition follows, since for all $a in A { c in mathbb{R}: f(x) > c}$ exists a $delta > 0$, so that for all $y$ with $| x - y | < delta$, we have $y in A$ and therefore $f(y) > c$.
answered Nov 24 at 15:06
Viktor Glombik
513422
add a comment |
We have
begin{align*}
{ f le s } text{ closed for all } s in mathbb{R}
iff & { f > -s } text{ open for all } s in mathbb{R} \
iff & { f > s } text{ open for all } s in mathbb{R}
end{align*}
Now a subset $A subseteq X$ is called open in $X$, when
begin{equation} tag{$star$} label{eq:offen}
forall a in A exists delta > 0: B_{delta}(a) subset A.
end{equation}
Because $f$ is lower semi continuous on $X$, we have
begin{equation*}
forall c < f(x) exists delta > 0: f(y) > c forall y in B_{delta}(x).
end{equation*}
(follows from the standard $varepsilon-delta$-definition with $varepsilon := f(x) - c$.)
Now let $A := { f > s }$.
From definition eqref{eq:offen} the proposition follows, since for all $a in A { c in mathbb{R}: f(x) > c}$ exists a $delta > 0$, so that for all $y$ with $| x - y | < delta$, we have $y in A$ and therefore $f(y) > c$.
answered Nov 24 at 15:06
Viktor Glombik
513422
We have
begin{align*}
{ f le s } text{ closed for all } s in mathbb{R}
iff & { f > -s } text{ open for all } s in mathbb{R} \
iff & { f > s } text{ open for all } s in mathbb{R}
end{align*}
Now a subset $A subseteq X$ is called open in $X$, when
begin{equation} tag{$star$} label{eq:offen}
forall a in A exists delta > 0: B_{delta}(a) subset A.
end{equation}
Because $f$ is lower semi continuous on $X$, we have
begin{equation*}
forall c < f(x) exists delta > 0: f(y) > c forall y in B_{delta}(x).
end{equation*}
(follows from the standard $varepsilon-delta$-definition with $varepsilon := f(x) - c$.)
Now let $A := { f > s }$.
From definition eqref{eq:offen} the proposition follows, since for all $a in A { c in mathbb{R}: f(x) > c}$ exists a $delta > 0$, so that for all $y$ with $| x - y | < delta$, we have $y in A$ and therefore $f(y) > c$.
answered Nov 24 at 15:06
Viktor Glombik
513422
answered Nov 24 at 15:06
Viktor Glombik
513422
answered Nov 24 at 15:06
Viktor Glombik
513422
answered Nov 24 at 15:06
Viktor Glombik
513422
513422
add a comment |
add a comment |
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