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
2
active
oldest
votes
$begingroup$
${ f | f^{-1} (F)subseteq G }={ f | f(Ysetminus G)subseteq Zsetminus F}. $
answered Dec 4 '18 at 6:47
Alex RavskyAlex Ravsky
40.2k32282
$endgroup$
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
107k347114
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024942%2fopen-sets-in-continuous-function-space%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
${ f | f^{-1} (F)subseteq G }={ f | f(Ysetminus G)subseteq Zsetminus F}. $
answered Dec 4 '18 at 6:47
Alex RavskyAlex Ravsky
40.2k32282
$endgroup$
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
40.2k32282
$endgroup$
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
40.2k32282
$endgroup$
${ f | f^{-1} (F)subseteq G }={ f | f(Ysetminus G)subseteq Zsetminus F}. $
answered Dec 4 '18 at 6:47
Alex RavskyAlex Ravsky
40.2k32282
answered Dec 4 '18 at 6:47
Alex RavskyAlex Ravsky
40.2k32282
answered Dec 4 '18 at 6:47
Alex RavskyAlex Ravsky
40.2k32282
answered Dec 4 '18 at 6:47
Alex RavskyAlex Ravsky
40.2k32282
40.2k32282
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
107k347114
$endgroup$
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
107k347114
$endgroup$
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
107k347114
$endgroup$
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
107k347114
answered Dec 5 '18 at 16:03
Henno BrandsmaHenno Brandsma
107k347114
answered Dec 5 '18 at 16:03
Henno BrandsmaHenno Brandsma
107k347114
answered Dec 5 '18 at 16:03
Henno BrandsmaHenno Brandsma
107k347114
107k347114
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024942%2fopen-sets-in-continuous-function-space%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
