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Continuity at a point in terms of closure
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If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverline{A} implies f(x_0)inoverline{f(A)}$.
I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!
Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^{-1}(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverline{A}$, we have $Acap f^{-1}(V)neqvarnothing$. Let $xin Acap f^{-1}(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverline{f(A)}$.
general-topology continuity
asked 2 days ago
BlondCaféBlondCafé
364
$endgroup$
add a comment |
$begingroup$
If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverline{A} implies f(x_0)inoverline{f(A)}$.
I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!
Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^{-1}(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverline{A}$, we have $Acap f^{-1}(V)neqvarnothing$. Let $xin Acap f^{-1}(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverline{f(A)}$.
general-topology continuity
asked 2 days ago
BlondCaféBlondCafé
364
$endgroup$
add a comment |
$begingroup$
If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverline{A} implies f(x_0)inoverline{f(A)}$.
I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!
Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^{-1}(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverline{A}$, we have $Acap f^{-1}(V)neqvarnothing$. Let $xin Acap f^{-1}(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverline{f(A)}$.
general-topology continuity
asked 2 days ago
BlondCaféBlondCafé
364
$endgroup$
If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverline{A} implies f(x_0)inoverline{f(A)}$.
I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!
Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^{-1}(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverline{A}$, we have $Acap f^{-1}(V)neqvarnothing$. Let $xin Acap f^{-1}(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverline{f(A)}$.
general-topology continuity
general-topology continuity
asked 2 days ago
BlondCaféBlondCafé
364
asked 2 days ago
BlondCaféBlondCafé
364
asked 2 days ago
BlondCaféBlondCafé
364
asked 2 days ago
BlondCaféBlondCafé
364
asked 2 days ago
BlondCaféBlondCafé
364
364
add a comment |
add a comment |
2 Answers
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Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overline{f^{-1}(Ysetminus V)}$, then $f(x_o)in overline{f(f^{-1}(Ysetminus V))}subset overline{Ysetminus V}= Ysetminus V$, a contradiction, so $x_0 notin overline{f^{-1}(Ysetminus V)}$. Then, if $U = Xsetminus overline{f^{-1}(Ysetminus V)}$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.
answered 2 days ago
guchiheguchihe
21919
$endgroup$
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$begingroup$
It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.
Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^{-1}[Ysetminus V] neq emptyset$.
It follows that then $x_0 in overline{f^{-1}[Ysetminus V]}$ and so the assumption on $f$ would imply that $y=f(x_0) in overline{f[f^{-1}[Ysetminus V]]}$. But $f[f^{-1}[B]] subseteq B$ for any $B$ so we'd deduce that $y in overline{Ysetminus V} = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.
answered 2 days ago
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
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$begingroup$
Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overline{f^{-1}(Ysetminus V)}$, then $f(x_o)in overline{f(f^{-1}(Ysetminus V))}subset overline{Ysetminus V}= Ysetminus V$, a contradiction, so $x_0 notin overline{f^{-1}(Ysetminus V)}$. Then, if $U = Xsetminus overline{f^{-1}(Ysetminus V)}$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.
answered 2 days ago
guchiheguchihe
21919
$endgroup$
add a comment |
$begingroup$
Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overline{f^{-1}(Ysetminus V)}$, then $f(x_o)in overline{f(f^{-1}(Ysetminus V))}subset overline{Ysetminus V}= Ysetminus V$, a contradiction, so $x_0 notin overline{f^{-1}(Ysetminus V)}$. Then, if $U = Xsetminus overline{f^{-1}(Ysetminus V)}$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.
answered 2 days ago
guchiheguchihe
21919
$endgroup$
add a comment |
$begingroup$
Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overline{f^{-1}(Ysetminus V)}$, then $f(x_o)in overline{f(f^{-1}(Ysetminus V))}subset overline{Ysetminus V}= Ysetminus V$, a contradiction, so $x_0 notin overline{f^{-1}(Ysetminus V)}$. Then, if $U = Xsetminus overline{f^{-1}(Ysetminus V)}$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.
answered 2 days ago
guchiheguchihe
21919
$endgroup$
Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overline{f^{-1}(Ysetminus V)}$, then $f(x_o)in overline{f(f^{-1}(Ysetminus V))}subset overline{Ysetminus V}= Ysetminus V$, a contradiction, so $x_0 notin overline{f^{-1}(Ysetminus V)}$. Then, if $U = Xsetminus overline{f^{-1}(Ysetminus V)}$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.
answered 2 days ago
guchiheguchihe
21919
answered 2 days ago
guchiheguchihe
21919
answered 2 days ago
guchiheguchihe
21919
answered 2 days ago
guchiheguchihe
21919
21919
add a comment |
add a comment |
$begingroup$
It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.
Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^{-1}[Ysetminus V] neq emptyset$.
It follows that then $x_0 in overline{f^{-1}[Ysetminus V]}$ and so the assumption on $f$ would imply that $y=f(x_0) in overline{f[f^{-1}[Ysetminus V]]}$. But $f[f^{-1}[B]] subseteq B$ for any $B$ so we'd deduce that $y in overline{Ysetminus V} = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.
answered 2 days ago
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.
Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^{-1}[Ysetminus V] neq emptyset$.
It follows that then $x_0 in overline{f^{-1}[Ysetminus V]}$ and so the assumption on $f$ would imply that $y=f(x_0) in overline{f[f^{-1}[Ysetminus V]]}$. But $f[f^{-1}[B]] subseteq B$ for any $B$ so we'd deduce that $y in overline{Ysetminus V} = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.
answered 2 days ago
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.
Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^{-1}[Ysetminus V] neq emptyset$.
It follows that then $x_0 in overline{f^{-1}[Ysetminus V]}$ and so the assumption on $f$ would imply that $y=f(x_0) in overline{f[f^{-1}[Ysetminus V]]}$. But $f[f^{-1}[B]] subseteq B$ for any $B$ so we'd deduce that $y in overline{Ysetminus V} = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.
answered 2 days ago
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.
Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^{-1}[Ysetminus V] neq emptyset$.
It follows that then $x_0 in overline{f^{-1}[Ysetminus V]}$ and so the assumption on $f$ would imply that $y=f(x_0) in overline{f[f^{-1}[Ysetminus V]]}$. But $f[f^{-1}[B]] subseteq B$ for any $B$ so we'd deduce that $y in overline{Ysetminus V} = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.
answered 2 days ago
Henno BrandsmaHenno Brandsma
116k349127
answered 2 days ago
Henno BrandsmaHenno Brandsma
116k349127
answered 2 days ago
Henno BrandsmaHenno Brandsma
116k349127
answered 2 days ago
Henno BrandsmaHenno Brandsma
116k349127
116k349127
add a comment |
add a comment |
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