Fdtxjr

Why does Kotter return in Welcome Back Kotter?

How to write a macro that is braces sensitive?

Is it important to consider tone, melody, and musical form while writing a song?

How do I create uniquely male characters?

What are the differences between the usage of 'it' and 'they'?

Smoothness of finite-dimensional functional calculus

Do VLANs within a subnet need to have their own subnet for router on a stick?

Test whether all array elements are factors of a number

Has the BBC provided arguments for saying Brexit being cancelled is unlikely?

Dragon forelimb placement

Is it tax fraud for an individual to declare non-taxable revenue as taxable income? (US tax laws)

Which models of the Boeing 737 are still in production?

Did Shadowfax go to Valinor?

Show that if two triangles built on parallel lines, with equal bases have the same perimeter only if they are congruent.

What's the output of a record cartridge playing an out-of-speed record

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

Approximately how much travel time was saved by the opening of the Suez Canal in 1869?

In Japanese, what’s the difference between “Tonari ni” (となりに) and “Tsugi” (つぎ)? When would you use one over the other?

Why do falling prices hurt debtors?

How to say job offer in Mandarin/Cantonese?

The use of multiple foreign keys on same column in SQL Server

Font hinting is lost in Chrome-like browsers (for some languages )

How does strength of boric acid solution increase in presence of salicylic acid?

Why did the Germans forbid the possession of pet pigeons in Rostov-on-Don in 1941?

The boundary of union is the union of boundaries when the sets have disjoint closures

Help proving $partial (A cup B) = partial Acuppartial B$?Boundary of union of two sets equals the union of their boundariesProve that if Cl(A) ∩ Cl(B) = ∅, then Bd(A ∪ B) = Bd(A) ∪ Bd(B).If $U,,Vsubseteq mathbbR^2$, $Ucap V = emptyset$ then $partial (V cup U)=(partial V) cup (partial U)$?When boundary of union is equal to union of boundariesIf $Asubset X$, and $partial A$ and $X$ are connected, then $Cl(A)$ is connected.Boundary of union equal union of boundariesCorrectness of topological reasoning (interior, closure and boundary of sets)Prove that $E$ is disconnected iff there exists two open disjoint sets $A$,$B$ in $X$Boundary of the boundary of a closed set?Continuous function crosses the boundary?Boundary of union of disjoint open sets.In a normed set the boundary of a subset is contained in the boundary of the closure of the set.Open sets intersecting on boundaryWhat is the boundary of $mathbbQ times mathbbQ$ in $mathbbR times mathbbQ$?The interior of union of two boundary open sets is emptyBoundary of union equal union of boundaries

8

3

$begingroup$

Assume $bar Acapbar B=emptyset$. Is $partial (A cup B)=partial Acuppartial B$, where $partial A$ and $bar A$ mean the boundary set and closure of set $A$?

I can prove that $partial (A cup B)subset partial Acuppartial B$ but for proving $partial Acuppartial Bsubset partial (A cup B)$ it seems not trivial. I tried to show that for $xin partial Acuppartial B$ WLOG, $xin partial A$ so $B(x)cap A$ and $B(x)cap A^c$ not equal to $emptyset$ but it seems not enough to show the result.

general-topology

share|cite|improve this question

edited Nov 1 '15 at 18:11

user147263

asked Oct 22 '12 at 16:21

MathematicsMathematics

2,20322145

$endgroup$

add a comment | 

8

3

$begingroup$

Assume $bar Acapbar B=emptyset$. Is $partial (A cup B)=partial Acuppartial B$, where $partial A$ and $bar A$ mean the boundary set and closure of set $A$?

I can prove that $partial (A cup B)subset partial Acuppartial B$ but for proving $partial Acuppartial Bsubset partial (A cup B)$ it seems not trivial. I tried to show that for $xin partial Acuppartial B$ WLOG, $xin partial A$ so $B(x)cap A$ and $B(x)cap A^c$ not equal to $emptyset$ but it seems not enough to show the result.

general-topology

share|cite|improve this question

edited Nov 1 '15 at 18:11

user147263

asked Oct 22 '12 at 16:21

MathematicsMathematics

2,20322145

$endgroup$

add a comment | 

$begingroup$

Assume $bar Acapbar B=emptyset$. Is $partial (A cup B)=partial Acuppartial B$, where $partial A$ and $bar A$ mean the boundary set and closure of set $A$?

I can prove that $partial (A cup B)subset partial Acuppartial B$ but for proving $partial Acuppartial Bsubset partial (A cup B)$ it seems not trivial. I tried to show that for $xin partial Acuppartial B$ WLOG, $xin partial A$ so $B(x)cap A$ and $B(x)cap A^c$ not equal to $emptyset$ but it seems not enough to show the result.

general-topology

share|cite|improve this question

edited Nov 1 '15 at 18:11

user147263

asked Oct 22 '12 at 16:21

MathematicsMathematics

2,20322145

$endgroup$

share|cite|improve this question

edited Nov 1 '15 at 18:11

user147263

asked Oct 22 '12 at 16:21

MathematicsMathematics

2,20322145

share|cite|improve this question

edited Nov 1 '15 at 18:11

user147263

asked Oct 22 '12 at 16:21

MathematicsMathematics

2,20322145

asked Oct 22 '12 at 16:21

MathematicsMathematics

2,20322145

asked Oct 22 '12 at 16:21

MathematicsMathematics

2,20322145

1 Answer
1

active

oldest

votes

10

$begingroup$

You can actually get by with a little less: If $overlineA cap B = emptyset = A cap overlineB$, the result holds.

As you have noticed, we have that $partial ( A cup B ) subseteq partial A cup partial B$. Suppose that $x notin partial ( A cup B )$. There are two cases:

1. $x notin overline A cup B = overlineA cup overlineB$. In this case it can easily be shown that $x notin partial A cup partial B$.


2. $x notin overline X setminus ( A cup B ) $. Then $x in X setminus overline X setminus ( A cup B ) = mathrmInt ( A cup B )$. Without loss of generality assume that $x in A$, and since $Acap overline B=emptyset$, then $xin Xsetminusoverline B$, which implies that $xnotin partial (B)$. Furthermore, it can be shown that $U = mathrmInt ( A cup B ) setminus overlineB$ is a neighbourhood of $x$ that is contained in $A$, and so $x notin partial A$. Thus $x notin partial A cup partial B$.

share|cite|improve this answer

edited Apr 2 '17 at 18:53

Community♦

1

answered Oct 22 '12 at 17:16

user642796user642796

44.9k565119

$endgroup$

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%2f218805%2fthe-boundary-of-union-is-the-union-of-boundaries-when-the-sets-have-disjoint-clo%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

10

$begingroup$

You can actually get by with a little less: If $overlineA cap B = emptyset = A cap overlineB$, the result holds.

As you have noticed, we have that $partial ( A cup B ) subseteq partial A cup partial B$. Suppose that $x notin partial ( A cup B )$. There are two cases:

1. $x notin overline A cup B = overlineA cup overlineB$. In this case it can easily be shown that $x notin partial A cup partial B$.


2. $x notin overline X setminus ( A cup B ) $. Then $x in X setminus overline X setminus ( A cup B ) = mathrmInt ( A cup B )$. Without loss of generality assume that $x in A$, and since $Acap overline B=emptyset$, then $xin Xsetminusoverline B$, which implies that $xnotin partial (B)$. Furthermore, it can be shown that $U = mathrmInt ( A cup B ) setminus overlineB$ is a neighbourhood of $x$ that is contained in $A$, and so $x notin partial A$. Thus $x notin partial A cup partial B$.

share|cite|improve this answer

edited Apr 2 '17 at 18:53

Community♦

1

answered Oct 22 '12 at 17:16

user642796user642796

44.9k565119

$endgroup$

add a comment | 

10

$begingroup$

You can actually get by with a little less: If $overlineA cap B = emptyset = A cap overlineB$, the result holds.

As you have noticed, we have that $partial ( A cup B ) subseteq partial A cup partial B$. Suppose that $x notin partial ( A cup B )$. There are two cases:

1. $x notin overline A cup B = overlineA cup overlineB$. In this case it can easily be shown that $x notin partial A cup partial B$.


2. $x notin overline X setminus ( A cup B ) $. Then $x in X setminus overline X setminus ( A cup B ) = mathrmInt ( A cup B )$. Without loss of generality assume that $x in A$, and since $Acap overline B=emptyset$, then $xin Xsetminusoverline B$, which implies that $xnotin partial (B)$. Furthermore, it can be shown that $U = mathrmInt ( A cup B ) setminus overlineB$ is a neighbourhood of $x$ that is contained in $A$, and so $x notin partial A$. Thus $x notin partial A cup partial B$.

share|cite|improve this answer

edited Apr 2 '17 at 18:53

Community♦

1

answered Oct 22 '12 at 17:16

user642796user642796

44.9k565119

$endgroup$

add a comment | 

$begingroup$

You can actually get by with a little less: If $overlineA cap B = emptyset = A cap overlineB$, the result holds.

As you have noticed, we have that $partial ( A cup B ) subseteq partial A cup partial B$. Suppose that $x notin partial ( A cup B )$. There are two cases:

1. $x notin overline A cup B = overlineA cup overlineB$. In this case it can easily be shown that $x notin partial A cup partial B$.


2. $x notin overline X setminus ( A cup B ) $. Then $x in X setminus overline X setminus ( A cup B ) = mathrmInt ( A cup B )$. Without loss of generality assume that $x in A$, and since $Acap overline B=emptyset$, then $xin Xsetminusoverline B$, which implies that $xnotin partial (B)$. Furthermore, it can be shown that $U = mathrmInt ( A cup B ) setminus overlineB$ is a neighbourhood of $x$ that is contained in $A$, and so $x notin partial A$. Thus $x notin partial A cup partial B$.

share|cite|improve this answer

edited Apr 2 '17 at 18:53

Community♦

1

answered Oct 22 '12 at 17:16

user642796user642796

44.9k565119

$endgroup$

share|cite|improve this answer

edited Apr 2 '17 at 18:53

Community♦

1

answered Oct 22 '12 at 17:16

user642796user642796

44.9k565119

edited Apr 2 '17 at 18:53

Community♦

1

edited Apr 2 '17 at 18:53

Community♦

1

answered Oct 22 '12 at 17:16

user642796user642796

44.9k565119

answered Oct 22 '12 at 17:16

user642796user642796

44.9k565119