Rrerku

What are the advantages and disadvantages of running one shots compared to campaigns?

I’m planning on buying a laser printer but concerned about the life cycle of toner in the machine

Could a US political party gain complete control over the government by removing checks & balances?

Why do we use polarized capacitors?

What is GPS' 19 year rollover and does it present a cybersecurity issue?

If a centaur druid Wild Shapes into a Giant Elk, do their Charge features stack?

Is domain driven design an anti-SQL pattern?

Copycat chess is back

How to deal with fear of taking dependencies

Unbreakable Formation vs. Cry of the Carnarium

Where else does the Shulchan Aruch quote an authority by name?

Does bootstrapped regression allow for inference?

Is there a familial term for apples and pears?

"listening to me about as much as you're listening to this pole here"

Prime joint compound before latex paint?

Can I legally use front facing blue light in the UK?

Does a dangling wire really electrocute me if I'm standing in water?

Email Account under attack (really) - anything I can do?

LWC and complex parameters

Domain expired, GoDaddy holds it and is asking more money

What do you call something that goes against the spirit of the law, but is legal when interpreting the law to the letter?

Is ipsum/ipsa/ipse a third person pronoun, or can it serve other functions?

Is every set a filtered colimit of finite sets?

Add an angle to a sphere

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.

4

$begingroup$

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

share|cite|improve this question

asked yesterday

BlondCaféBlondCafé

364

$endgroup$

add a comment | 

4

$begingroup$

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

share|cite|improve this question

asked yesterday

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 $Asubseteq X land x_0inoverlineA implies f(x_0)inoverlinef(A)$.

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

share|cite|improve this question

asked yesterday

BlondCaféBlondCafé

364

$endgroup$

share|cite|improve this question

asked yesterday

BlondCaféBlondCafé

364

share|cite|improve this question

asked yesterday

BlondCaféBlondCafé

364

asked yesterday

BlondCaféBlondCafé

364

asked yesterday

BlondCaféBlondCafé

364

2 Answers
2

active

oldest

votes

3

$begingroup$

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 $x_0$.

share|cite|improve this answer

answered yesterday

guchiheguchihe

21918

$endgroup$

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

answered yesterday

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
);



);

draft saved

draft discarded

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

Name

Email

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

Name

Email

Required, but never shown

3

$begingroup$

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 $x_0$.

share|cite|improve this answer

answered yesterday

guchiheguchihe

21918

$endgroup$

add a comment | 

3

$begingroup$

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 $x_0$.

share|cite|improve this answer

answered yesterday

guchiheguchihe

21918

$endgroup$

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 $x_0$.

share|cite|improve this answer

answered yesterday

guchiheguchihe

21918

$endgroup$

share|cite|improve this answer

answered yesterday

guchiheguchihe

21918

answered yesterday

guchiheguchihe

21918

answered yesterday

guchiheguchihe

21918

2

$begingroup$

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.

share|cite|improve this answer

answered yesterday

Henno BrandsmaHenno Brandsma

115k349125

$endgroup$

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

answered yesterday

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.

share|cite|improve this answer

answered yesterday

Henno BrandsmaHenno Brandsma

115k349125

$endgroup$

share|cite|improve this answer

answered yesterday

Henno BrandsmaHenno Brandsma

115k349125

answered yesterday

Henno BrandsmaHenno Brandsma

115k349125

answered yesterday

Henno BrandsmaHenno Brandsma

115k349125