Continuity at a point in terms of closureSet Closure Union and IntersectionContinuity of a function through adherence of subsetsConvex set with empty interior is nowhere dense?Is preimage of closure equal to closure of preimage under continuous topological maps?How to show the logical equivalence of the following two definitions of continuity in a topological space?Given $A subseteq X$ in the discrete and the trivial topology, find closure of $A$Show two notions of dense are equivalentEquivalent definitions of continuity at a pointAbout continuity and clousureEquivalent definition of irreducible topological subspace.
Awk syntax, strange variable?
Possibly bubble sort algorithm
How is it possible for user to changed after storage was encrypted? (on OS X, Android)
How do I create uniquely male characters?
The use of multiple foreign keys on same column in SQL Server
Do Phineas and Ferb ever actually get busted in real time?
Japan - Plan around max visa duration
Find original functions from a composite function
What would happen to a modern skyscraper if it rains micro blackholes?
How to re-create Edward Weson's Pepper No. 30?
Do airline pilots ever risk not hearing communication directed to them specifically, from traffic controllers?
How to report a triplet of septets in NMR tabulation?
Are tax years 2016 & 2017 back taxes deductible for tax year 2018?
Download, install and reboot computer at night if needed
Email Account under attack (really) - anything I can do?
A Journey Through Space and Time
"You are your self first supporter", a more proper way to say it
Is there really no realistic way for a skeleton monster to move around without magic?
Representing power series as a function - what to do with the constant after integration?
Book about a traveler who helps planets in need
DOS, create pipe for stdin/stdout of command.com(or 4dos.com) in C or Batch?
How can bays and straits be determined in a procedurally generated map?
How to make payment on the internet without leaving a money trail?
If I cast Expeditious Retreat, can I Dash as a bonus action on the same turn?
Continuity at a point in terms of closure
Set Closure Union and IntersectionContinuity of a function through adherence of subsetsConvex set with empty interior is nowhere dense?Is preimage of closure equal to closure of preimage under continuous topological maps?How to show the logical equivalence of the following two definitions of continuity in a topological space?Given $A subseteq X$ in the discrete and the trivial topology, find closure of $A$Show two notions of dense are equivalentEquivalent definitions of continuity at a pointAbout continuity and clousureEquivalent definition of irreducible topological subspace.
$begingroup$
If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverlineA implies f(x_0)inoverlinef(A)$.
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 $x_0inoverlineA$, 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 $f(x_0)inoverlinef(A)$.
general-topology continuity
asked 4 hours ago
BlondCaféBlondCafé
314
$endgroup$
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 $Asubseteq X land x_0inoverlineA implies f(x_0)inoverlinef(A)$.
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 $x_0inoverlineA$, 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 $f(x_0)inoverlinef(A)$.
general-topology continuity
asked 4 hours ago
BlondCaféBlondCafé
314
$endgroup$
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 $Asubseteq X land x_0inoverlineA implies f(x_0)inoverlinef(A)$.
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 $x_0inoverlineA$, 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 $f(x_0)inoverlinef(A)$.
general-topology continuity
asked 4 hours ago
BlondCaféBlondCafé
314
$endgroup$
If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverlineA implies f(x_0)inoverlinef(A)$.
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 $x_0inoverlineA$, 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 $f(x_0)inoverlinef(A)$.
general-topology continuity
general-topology continuity
asked 4 hours ago
BlondCaféBlondCafé
314
asked 4 hours ago
BlondCaféBlondCafé
314
asked 4 hours ago
BlondCaféBlondCafé
314
asked 4 hours ago
BlondCaféBlondCafé
314
asked 4 hours ago
BlondCaféBlondCafé
314
314
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overlinef^-1(Ysetminus V)$, then $f(x_o)in overlinef(f^-1(Ysetminus V))subset overlineYsetminus V= Ysetminus V$, a contradiction, so $x_0 notin overlinef^-1(Ysetminus V)$. Then, if $U = Xsetminus overlinef^-1(Ysetminus V)$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.
answered 4 hours ago
guchiheguchihe
1978
$endgroup$
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 $x_0 in overlinef^-1[Ysetminus V]$ and so the assumption on $f$ would imply that $y=f(x_0) in overlinef[f^-1[Ysetminus V]]$. But $f[f^-1[B]] subseteq B$ for any $B$ so we'd deduce that $y in overlineYsetminus V = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.
answered 4 hours ago
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
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%2f3178435%2fcontinuity-at-a-point-in-terms-of-closure%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
$begingroup$
Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overlinef^-1(Ysetminus V)$, then $f(x_o)in overlinef(f^-1(Ysetminus V))subset overlineYsetminus V= Ysetminus V$, a contradiction, so $x_0 notin overlinef^-1(Ysetminus V)$. Then, if $U = Xsetminus overlinef^-1(Ysetminus V)$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.
answered 4 hours ago
guchiheguchihe
1978
$endgroup$
add a comment |
$begingroup$
Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overlinef^-1(Ysetminus V)$, then $f(x_o)in overlinef(f^-1(Ysetminus V))subset overlineYsetminus V= Ysetminus V$, a contradiction, so $x_0 notin overlinef^-1(Ysetminus V)$. Then, if $U = Xsetminus overlinef^-1(Ysetminus V)$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.
answered 4 hours ago
guchiheguchihe
1978
$endgroup$
add a comment |
$begingroup$
Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overlinef^-1(Ysetminus V)$, then $f(x_o)in overlinef(f^-1(Ysetminus V))subset overlineYsetminus V= Ysetminus V$, a contradiction, so $x_0 notin overlinef^-1(Ysetminus V)$. Then, if $U = Xsetminus overlinef^-1(Ysetminus V)$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.
answered 4 hours ago
guchiheguchihe
1978
$endgroup$
Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overlinef^-1(Ysetminus V)$, then $f(x_o)in overlinef(f^-1(Ysetminus V))subset overlineYsetminus V= Ysetminus V$, a contradiction, so $x_0 notin overlinef^-1(Ysetminus V)$. Then, if $U = Xsetminus overlinef^-1(Ysetminus V)$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.
answered 4 hours ago
guchiheguchihe
1978
answered 4 hours ago
guchiheguchihe
1978
answered 4 hours ago
guchiheguchihe
1978
answered 4 hours ago
guchiheguchihe
1978
1978
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 $x_0 in overlinef^-1[Ysetminus V]$ and so the assumption on $f$ would imply that $y=f(x_0) in overlinef[f^-1[Ysetminus V]]$. But $f[f^-1[B]] subseteq B$ for any $B$ so we'd deduce that $y in overlineYsetminus V = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.
answered 4 hours ago
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
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 $x_0 in overlinef^-1[Ysetminus V]$ and so the assumption on $f$ would imply that $y=f(x_0) in overlinef[f^-1[Ysetminus V]]$. But $f[f^-1[B]] subseteq B$ for any $B$ so we'd deduce that $y in overlineYsetminus V = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.
answered 4 hours ago
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
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 $x_0 in overlinef^-1[Ysetminus V]$ and so the assumption on $f$ would imply that $y=f(x_0) in overlinef[f^-1[Ysetminus V]]$. But $f[f^-1[B]] subseteq B$ for any $B$ so we'd deduce that $y in overlineYsetminus V = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.
answered 4 hours ago
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
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 $x_0 in overlinef^-1[Ysetminus V]$ and so the assumption on $f$ would imply that $y=f(x_0) in overlinef[f^-1[Ysetminus V]]$. But $f[f^-1[B]] subseteq B$ for any $B$ so we'd deduce that $y in overlineYsetminus V = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.
answered 4 hours ago
Henno BrandsmaHenno Brandsma
115k349125
answered 4 hours ago
Henno BrandsmaHenno Brandsma
115k349125
answered 4 hours ago
Henno BrandsmaHenno Brandsma
115k349125
answered 4 hours ago
Henno BrandsmaHenno Brandsma
115k349125
115k349125
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%2f3178435%2fcontinuity-at-a-point-in-terms-of-closure%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
