Open sets in continuous function space
$begingroup$
Let $Y,Z$ Hausdorff spaces, $Y$ compact, let $Fsubseteq Z$ closed, $G subseteq Y$ open.
Then ${ f | f^{-1} (F)subseteq G } $ is open in $mathbb{C} (Y,Z) $ with the open-compact topology.
I don't know how to prove this.
general-topology
edited Dec 4 '18 at 0:41
Bernard
119k740113
asked Dec 4 '18 at 0:35
Brau Morales MtzBrau Morales Mtz
61
$endgroup$
add a comment |
$begingroup$
Let $Y,Z$ Hausdorff spaces, $Y$ compact, let $Fsubseteq Z$ closed, $G subseteq Y$ open.
Then ${ f | f^{-1} (F)subseteq G } $ is open in $mathbb{C} (Y,Z) $ with the open-compact topology.
I don't know how to prove this.
general-topology
edited Dec 4 '18 at 0:41
Bernard
119k740113
asked Dec 4 '18 at 0:35
Brau Morales MtzBrau Morales Mtz
61
$endgroup$
add a comment |
$begingroup$
Let $Y,Z$ Hausdorff spaces, $Y$ compact, let $Fsubseteq Z$ closed, $G subseteq Y$ open.
Then ${ f | f^{-1} (F)subseteq G } $ is open in $mathbb{C} (Y,Z) $ with the open-compact topology.
I don't know how to prove this.
general-topology
edited Dec 4 '18 at 0:41
Bernard
119k740113
asked Dec 4 '18 at 0:35
Brau Morales MtzBrau Morales Mtz
61
$endgroup$
Let $Y,Z$ Hausdorff spaces, $Y$ compact, let $Fsubseteq Z$ closed, $G subseteq Y$ open.
Then ${ f | f^{-1} (F)subseteq G } $ is open in $mathbb{C} (Y,Z) $ with the open-compact topology.
I don't know how to prove this.
general-topology
general-topology
edited Dec 4 '18 at 0:41
Bernard
119k740113
asked Dec 4 '18 at 0:35
Brau Morales MtzBrau Morales Mtz
61
edited Dec 4 '18 at 0:41
Bernard
119k740113
asked Dec 4 '18 at 0:35
Brau Morales MtzBrau Morales Mtz
61
edited Dec 4 '18 at 0:41
Bernard
119k740113
edited Dec 4 '18 at 0:41
Bernard
119k740113
edited Dec 4 '18 at 0:41
Bernard
119k740113
119k740113
asked Dec 4 '18 at 0:35
Brau Morales MtzBrau Morales Mtz
61
asked Dec 4 '18 at 0:35
Brau Morales MtzBrau Morales Mtz
61
asked Dec 4 '18 at 0:35
Brau Morales MtzBrau Morales Mtz
61
61
add a comment |
add a comment |
2 Answers
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${ f | f^{-1} (F)subseteq G }={ f | f(Ysetminus G)subseteq Zsetminus F}. $
answered Dec 4 '18 at 6:47
Alex RavskyAlex Ravsky
39.9k32282
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The basic idea is Alex' identity:
$${ f mid f^{-1}[F]subseteq G }={ f mid f[Ysetminus G]subseteq Zsetminus F}$$
Proof: if $f$ is in the left hand set, take $y in Y setminus G$ arbitrarily. Then $f(y) notin F$, or otherwise (by definition) $y in f^{-1}[F]$ and thus $f(y) in G$ by $f$ being in the left hand set, contradiction, so $f[Y setminus G] subseteq Z setminus F$, hence $f$ in the right hand set.
If $f$ is in the right hand set, let $y in f^{-1}[F]$ be arbitrary. If $y notin G$ we'd have $y in Ysetminus G$ and so $f(y) in f[Ysetminus G] subseteq Z setminus F$ and so $f(y) notin F$, contradicting $y in f^{-1}[F]$. So $y in G$ and so $f$ is in the left hand set.
As $Ysetminus G$ is closed in $Y$ and $Y$ is compact, so is $Ysetminus G$. Moreover $Z setminus F$ is open in $Z$, as $F$ is closed. So the right hand set is by definition open in the compact-open topology on $C(Y,Z)$ and hence so is the set we are interested in.
answered Dec 5 '18 at 16:03
Henno BrandsmaHenno Brandsma
106k347114
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2 Answers
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2 Answers
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$begingroup$
${ f | f^{-1} (F)subseteq G }={ f | f(Ysetminus G)subseteq Zsetminus F}. $
answered Dec 4 '18 at 6:47
Alex RavskyAlex Ravsky
39.9k32282
$endgroup$
add a comment |
$begingroup$
${ f | f^{-1} (F)subseteq G }={ f | f(Ysetminus G)subseteq Zsetminus F}. $
answered Dec 4 '18 at 6:47
Alex RavskyAlex Ravsky
39.9k32282
$endgroup$
add a comment |
$begingroup$
${ f | f^{-1} (F)subseteq G }={ f | f(Ysetminus G)subseteq Zsetminus F}. $
answered Dec 4 '18 at 6:47
Alex RavskyAlex Ravsky
39.9k32282
$endgroup$
${ f | f^{-1} (F)subseteq G }={ f | f(Ysetminus G)subseteq Zsetminus F}. $
answered Dec 4 '18 at 6:47
Alex RavskyAlex Ravsky
39.9k32282
answered Dec 4 '18 at 6:47
Alex RavskyAlex Ravsky
39.9k32282
answered Dec 4 '18 at 6:47
Alex RavskyAlex Ravsky
39.9k32282
answered Dec 4 '18 at 6:47
Alex RavskyAlex Ravsky
39.9k32282
39.9k32282
add a comment |
add a comment |
$begingroup$
The basic idea is Alex' identity:
$${ f mid f^{-1}[F]subseteq G }={ f mid f[Ysetminus G]subseteq Zsetminus F}$$
Proof: if $f$ is in the left hand set, take $y in Y setminus G$ arbitrarily. Then $f(y) notin F$, or otherwise (by definition) $y in f^{-1}[F]$ and thus $f(y) in G$ by $f$ being in the left hand set, contradiction, so $f[Y setminus G] subseteq Z setminus F$, hence $f$ in the right hand set.
If $f$ is in the right hand set, let $y in f^{-1}[F]$ be arbitrary. If $y notin G$ we'd have $y in Ysetminus G$ and so $f(y) in f[Ysetminus G] subseteq Z setminus F$ and so $f(y) notin F$, contradicting $y in f^{-1}[F]$. So $y in G$ and so $f$ is in the left hand set.
As $Ysetminus G$ is closed in $Y$ and $Y$ is compact, so is $Ysetminus G$. Moreover $Z setminus F$ is open in $Z$, as $F$ is closed. So the right hand set is by definition open in the compact-open topology on $C(Y,Z)$ and hence so is the set we are interested in.
answered Dec 5 '18 at 16:03
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
The basic idea is Alex' identity:
$${ f mid f^{-1}[F]subseteq G }={ f mid f[Ysetminus G]subseteq Zsetminus F}$$
Proof: if $f$ is in the left hand set, take $y in Y setminus G$ arbitrarily. Then $f(y) notin F$, or otherwise (by definition) $y in f^{-1}[F]$ and thus $f(y) in G$ by $f$ being in the left hand set, contradiction, so $f[Y setminus G] subseteq Z setminus F$, hence $f$ in the right hand set.
If $f$ is in the right hand set, let $y in f^{-1}[F]$ be arbitrary. If $y notin G$ we'd have $y in Ysetminus G$ and so $f(y) in f[Ysetminus G] subseteq Z setminus F$ and so $f(y) notin F$, contradicting $y in f^{-1}[F]$. So $y in G$ and so $f$ is in the left hand set.
As $Ysetminus G$ is closed in $Y$ and $Y$ is compact, so is $Ysetminus G$. Moreover $Z setminus F$ is open in $Z$, as $F$ is closed. So the right hand set is by definition open in the compact-open topology on $C(Y,Z)$ and hence so is the set we are interested in.
answered Dec 5 '18 at 16:03
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
The basic idea is Alex' identity:
$${ f mid f^{-1}[F]subseteq G }={ f mid f[Ysetminus G]subseteq Zsetminus F}$$
Proof: if $f$ is in the left hand set, take $y in Y setminus G$ arbitrarily. Then $f(y) notin F$, or otherwise (by definition) $y in f^{-1}[F]$ and thus $f(y) in G$ by $f$ being in the left hand set, contradiction, so $f[Y setminus G] subseteq Z setminus F$, hence $f$ in the right hand set.
If $f$ is in the right hand set, let $y in f^{-1}[F]$ be arbitrary. If $y notin G$ we'd have $y in Ysetminus G$ and so $f(y) in f[Ysetminus G] subseteq Z setminus F$ and so $f(y) notin F$, contradicting $y in f^{-1}[F]$. So $y in G$ and so $f$ is in the left hand set.
As $Ysetminus G$ is closed in $Y$ and $Y$ is compact, so is $Ysetminus G$. Moreover $Z setminus F$ is open in $Z$, as $F$ is closed. So the right hand set is by definition open in the compact-open topology on $C(Y,Z)$ and hence so is the set we are interested in.
answered Dec 5 '18 at 16:03
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
The basic idea is Alex' identity:
$${ f mid f^{-1}[F]subseteq G }={ f mid f[Ysetminus G]subseteq Zsetminus F}$$
Proof: if $f$ is in the left hand set, take $y in Y setminus G$ arbitrarily. Then $f(y) notin F$, or otherwise (by definition) $y in f^{-1}[F]$ and thus $f(y) in G$ by $f$ being in the left hand set, contradiction, so $f[Y setminus G] subseteq Z setminus F$, hence $f$ in the right hand set.
If $f$ is in the right hand set, let $y in f^{-1}[F]$ be arbitrary. If $y notin G$ we'd have $y in Ysetminus G$ and so $f(y) in f[Ysetminus G] subseteq Z setminus F$ and so $f(y) notin F$, contradicting $y in f^{-1}[F]$. So $y in G$ and so $f$ is in the left hand set.
As $Ysetminus G$ is closed in $Y$ and $Y$ is compact, so is $Ysetminus G$. Moreover $Z setminus F$ is open in $Z$, as $F$ is closed. So the right hand set is by definition open in the compact-open topology on $C(Y,Z)$ and hence so is the set we are interested in.
answered Dec 5 '18 at 16:03
Henno BrandsmaHenno Brandsma
106k347114
answered Dec 5 '18 at 16:03
Henno BrandsmaHenno Brandsma
106k347114
answered Dec 5 '18 at 16:03
Henno BrandsmaHenno Brandsma
106k347114
answered Dec 5 '18 at 16:03
Henno BrandsmaHenno Brandsma
106k347114
106k347114
add a comment |
add a comment |
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