Union of Boundaries Formula The Next CEO of Stack OverflowThe intersection of a connected subspace with the boundary of another subsetRelation between compact ,open and closureConnectedness of a cartesian productApplying the rules of logic.union of two locally connected sets is still locally connectedProofs involving setsProve that $A-(Bcup C) = (A-B) cap (A-C)$Is this proof of the closure of a set being closed correct?$overlineY$ is closedBoundary of union equal union of boundaries
Does the Idaho Potato Commission associate potato skins with healthy eating?
Expressing the idea of having a very busy time
Is there a way to save my career from absolute disaster?
free fall ellipse or parabola?
Help/tips for a first time writer?
Traveling with my 5 year old daughter (as the father) without the mother from Germany to Mexico
What day is it again?
Is it convenient to ask the journal's editor for two additional days to complete a review?
Getting Stale Gas Out of a Gas Tank w/out Dropping the Tank
How to use ReplaceAll on an expression that contains a rule
Does Germany produce more waste than the US?
Does higher Oxidation/ reduction potential translate to higher energy storage in battery?
Why did early computer designers eschew integers?
"Eavesdropping" vs "Listen in on"
Small nick on power cord from an electric alarm clock, and copper wiring exposed but intact
what's the use of '% to gdp' type of variables?
Purpose of level-shifter with same in and out voltages
Yu-Gi-Oh cards in Python 3
Help understanding this unsettling image of Titan, Epimetheus, and Saturn's rings?
Which one is the true statement?
Is dried pee considered dirt?
vector calculus integration identity problem
My ex-girlfriend uses my Apple ID to login to her iPad, do I have to give her my Apple ID password to reset it?
Do I need to write [sic] when including a quotation with a number less than 10 that isn't written out?
Union of Boundaries Formula
The Next CEO of Stack OverflowThe intersection of a connected subspace with the boundary of another subsetRelation between compact ,open and closureConnectedness of a cartesian productApplying the rules of logic.union of two locally connected sets is still locally connectedProofs involving setsProve that $A-(Bcup C) = (A-B) cap (A-C)$Is this proof of the closure of a set being closed correct?$overlineY$ is closedBoundary of union equal union of boundaries
$begingroup$
Say we have a topological space $X$.
I'm trying to show that one can express the union of the boundaries of two sets $A$ and $B$ as
$$partialA cup partialB=partial(A cup B) cup partial(A cap B) cup (partialA cap partialB)$$
It wasn't too difficult to prove the inclusion
$$partial(A cup B) cup partial(A cap B) cup (partialA cap partialB) subseteq partialA cup partialB$$
But I'm not sure how to prove it in the other direction. Here's what I have so far:
Let $x in partialA$ and for the sake of contradiction assume that $x notin partial(A cup B) cup partial(A cap B) cup (partialA cap partialB)$. Then
$$x notin partial(A cup B) land x notin partial(A cap B) land x notin (partialA cap partialB)$$
From the first proposition, it follows that there exists an open neighborhood $U_x$ of $x$ such that
$$U_x cap (A cup B)=varnothing lor U_x cap complement_X(A cup B)=varnothing$$
$$ Rightarrow (U_x cap A) cup (U_x cap B)=varnothing lor U_x cap complement_X(A) cap complement_X(B)=varnothing$$
From $x notin (partial A cap partial B)$, it follows that $x notin partial B$. Thus there's some open neighborhood $V_x$ of $x$ such that
$$V_x cap B=varnothing lor V_x cap complement_X(B)=varnothing$$
But as $x in partial A$, we must have $U_x cap A neq varnothing$ and $V_x cap A neq varnothing$, hence $(U_x cap A) cup (U_x cap B)$ cannot be empty. Therefore $U_x cap complement_X(A) cap complement_X(B)=varnothing$. We also know that there exists an open neighborhood $W_x$ of $x$ such that
$$W_x cap A cap B=varnothing lor W_x cap complement_X(A cap B)=varnothing$$
$$Rightarrow W_x cap A cap B=varnothing lor (W_x cap complement_X(A)) cup (W_x cap complement_X(B))=varnothing$$
But since $x in partial A$, the set $W_x cap complement_X(A)$ must be non-empty. Therefore the above union must also be non-empty, and we conclude that $W_x cap A cap B=varnothing$.
This is where I get stuck, and I'm not sure how to proceed from there. Any hints would be appreciated.
general-topology elementary-set-theory
asked Mar 26 at 21:22
Quantum ChillQuantum Chill
7310
$endgroup$
add a comment |
$begingroup$
Say we have a topological space $X$.
I'm trying to show that one can express the union of the boundaries of two sets $A$ and $B$ as
$$partialA cup partialB=partial(A cup B) cup partial(A cap B) cup (partialA cap partialB)$$
It wasn't too difficult to prove the inclusion
$$partial(A cup B) cup partial(A cap B) cup (partialA cap partialB) subseteq partialA cup partialB$$
But I'm not sure how to prove it in the other direction. Here's what I have so far:
Let $x in partialA$ and for the sake of contradiction assume that $x notin partial(A cup B) cup partial(A cap B) cup (partialA cap partialB)$. Then
$$x notin partial(A cup B) land x notin partial(A cap B) land x notin (partialA cap partialB)$$
From the first proposition, it follows that there exists an open neighborhood $U_x$ of $x$ such that
$$U_x cap (A cup B)=varnothing lor U_x cap complement_X(A cup B)=varnothing$$
$$ Rightarrow (U_x cap A) cup (U_x cap B)=varnothing lor U_x cap complement_X(A) cap complement_X(B)=varnothing$$
From $x notin (partial A cap partial B)$, it follows that $x notin partial B$. Thus there's some open neighborhood $V_x$ of $x$ such that
$$V_x cap B=varnothing lor V_x cap complement_X(B)=varnothing$$
But as $x in partial A$, we must have $U_x cap A neq varnothing$ and $V_x cap A neq varnothing$, hence $(U_x cap A) cup (U_x cap B)$ cannot be empty. Therefore $U_x cap complement_X(A) cap complement_X(B)=varnothing$. We also know that there exists an open neighborhood $W_x$ of $x$ such that
$$W_x cap A cap B=varnothing lor W_x cap complement_X(A cap B)=varnothing$$
$$Rightarrow W_x cap A cap B=varnothing lor (W_x cap complement_X(A)) cup (W_x cap complement_X(B))=varnothing$$
But since $x in partial A$, the set $W_x cap complement_X(A)$ must be non-empty. Therefore the above union must also be non-empty, and we conclude that $W_x cap A cap B=varnothing$.
This is where I get stuck, and I'm not sure how to proceed from there. Any hints would be appreciated.
general-topology elementary-set-theory
asked Mar 26 at 21:22
Quantum ChillQuantum Chill
7310
$endgroup$
$begingroup$
Is your $subset$ symbol in the statement that "wasn't too difficult to prove" pointed in the direction you meant? Because the other direction reads (from right to left) as $$Acup B subset Y cup (Acup B)$$ which is trivial, being one of the set theory axioms. The direction you have written is the hard direction.
$endgroup$
– Mark Fischler
Mar 26 at 21:30
$begingroup$
@MarkFischler w...what? Are you implying that $partialA cup partialB=partial(A cup B)$? That only holds true for separated sets. The boundary of $A$ refers to the set $A^- setminus A^circ$. There's nothing purely set-theoretic about this.
$endgroup$
– Quantum Chill
Mar 26 at 21:38
$begingroup$
$x notin partial(A cap B)$ does not imply $x notin partial B$. Simple counterexamples exist for that. (that's in the sentence before you introduce $V_x$).
$endgroup$
– Henno Brandsma
Mar 26 at 21:46
$begingroup$
@HennoBrandsma That's a simple typo on my part, I meant $x notin (partial A cap partial B)$ . Thanks for catching that!
$endgroup$
– Quantum Chill
Mar 26 at 21:49
add a comment |
$begingroup$
Say we have a topological space $X$.
I'm trying to show that one can express the union of the boundaries of two sets $A$ and $B$ as
$$partialA cup partialB=partial(A cup B) cup partial(A cap B) cup (partialA cap partialB)$$
It wasn't too difficult to prove the inclusion
$$partial(A cup B) cup partial(A cap B) cup (partialA cap partialB) subseteq partialA cup partialB$$
But I'm not sure how to prove it in the other direction. Here's what I have so far:
Let $x in partialA$ and for the sake of contradiction assume that $x notin partial(A cup B) cup partial(A cap B) cup (partialA cap partialB)$. Then
$$x notin partial(A cup B) land x notin partial(A cap B) land x notin (partialA cap partialB)$$
From the first proposition, it follows that there exists an open neighborhood $U_x$ of $x$ such that
$$U_x cap (A cup B)=varnothing lor U_x cap complement_X(A cup B)=varnothing$$
$$ Rightarrow (U_x cap A) cup (U_x cap B)=varnothing lor U_x cap complement_X(A) cap complement_X(B)=varnothing$$
From $x notin (partial A cap partial B)$, it follows that $x notin partial B$. Thus there's some open neighborhood $V_x$ of $x$ such that
$$V_x cap B=varnothing lor V_x cap complement_X(B)=varnothing$$
But as $x in partial A$, we must have $U_x cap A neq varnothing$ and $V_x cap A neq varnothing$, hence $(U_x cap A) cup (U_x cap B)$ cannot be empty. Therefore $U_x cap complement_X(A) cap complement_X(B)=varnothing$. We also know that there exists an open neighborhood $W_x$ of $x$ such that
$$W_x cap A cap B=varnothing lor W_x cap complement_X(A cap B)=varnothing$$
$$Rightarrow W_x cap A cap B=varnothing lor (W_x cap complement_X(A)) cup (W_x cap complement_X(B))=varnothing$$
But since $x in partial A$, the set $W_x cap complement_X(A)$ must be non-empty. Therefore the above union must also be non-empty, and we conclude that $W_x cap A cap B=varnothing$.
This is where I get stuck, and I'm not sure how to proceed from there. Any hints would be appreciated.
general-topology elementary-set-theory
asked Mar 26 at 21:22
Quantum ChillQuantum Chill
7310
$endgroup$
Say we have a topological space $X$.
I'm trying to show that one can express the union of the boundaries of two sets $A$ and $B$ as
$$partialA cup partialB=partial(A cup B) cup partial(A cap B) cup (partialA cap partialB)$$
It wasn't too difficult to prove the inclusion
$$partial(A cup B) cup partial(A cap B) cup (partialA cap partialB) subseteq partialA cup partialB$$
But I'm not sure how to prove it in the other direction. Here's what I have so far:
Let $x in partialA$ and for the sake of contradiction assume that $x notin partial(A cup B) cup partial(A cap B) cup (partialA cap partialB)$. Then
$$x notin partial(A cup B) land x notin partial(A cap B) land x notin (partialA cap partialB)$$
From the first proposition, it follows that there exists an open neighborhood $U_x$ of $x$ such that
$$U_x cap (A cup B)=varnothing lor U_x cap complement_X(A cup B)=varnothing$$
$$ Rightarrow (U_x cap A) cup (U_x cap B)=varnothing lor U_x cap complement_X(A) cap complement_X(B)=varnothing$$
From $x notin (partial A cap partial B)$, it follows that $x notin partial B$. Thus there's some open neighborhood $V_x$ of $x$ such that
$$V_x cap B=varnothing lor V_x cap complement_X(B)=varnothing$$
But as $x in partial A$, we must have $U_x cap A neq varnothing$ and $V_x cap A neq varnothing$, hence $(U_x cap A) cup (U_x cap B)$ cannot be empty. Therefore $U_x cap complement_X(A) cap complement_X(B)=varnothing$. We also know that there exists an open neighborhood $W_x$ of $x$ such that
$$W_x cap A cap B=varnothing lor W_x cap complement_X(A cap B)=varnothing$$
$$Rightarrow W_x cap A cap B=varnothing lor (W_x cap complement_X(A)) cup (W_x cap complement_X(B))=varnothing$$
But since $x in partial A$, the set $W_x cap complement_X(A)$ must be non-empty. Therefore the above union must also be non-empty, and we conclude that $W_x cap A cap B=varnothing$.
This is where I get stuck, and I'm not sure how to proceed from there. Any hints would be appreciated.
general-topology elementary-set-theory
general-topology elementary-set-theory
asked Mar 26 at 21:22
Quantum ChillQuantum Chill
7310
asked Mar 26 at 21:22
Quantum ChillQuantum Chill
7310
edited Mar 26 at 22:44
Quantum Chill
asked Mar 26 at 21:22
Quantum ChillQuantum Chill
7310
asked Mar 26 at 21:22
Quantum ChillQuantum Chill
7310
asked Mar 26 at 21:22
Quantum ChillQuantum Chill
7310
7310
$begingroup$
Is your $subset$ symbol in the statement that "wasn't too difficult to prove" pointed in the direction you meant? Because the other direction reads (from right to left) as $$Acup B subset Y cup (Acup B)$$ which is trivial, being one of the set theory axioms. The direction you have written is the hard direction.
$endgroup$
– Mark Fischler
Mar 26 at 21:30
$begingroup$
@MarkFischler w...what? Are you implying that $partialA cup partialB=partial(A cup B)$? That only holds true for separated sets. The boundary of $A$ refers to the set $A^- setminus A^circ$. There's nothing purely set-theoretic about this.
$endgroup$
– Quantum Chill
Mar 26 at 21:38
$begingroup$
$x notin partial(A cap B)$ does not imply $x notin partial B$. Simple counterexamples exist for that. (that's in the sentence before you introduce $V_x$).
$endgroup$
– Henno Brandsma
Mar 26 at 21:46
$begingroup$
@HennoBrandsma That's a simple typo on my part, I meant $x notin (partial A cap partial B)$ . Thanks for catching that!
$endgroup$
– Quantum Chill
Mar 26 at 21:49
add a comment |
$begingroup$
Is your $subset$ symbol in the statement that "wasn't too difficult to prove" pointed in the direction you meant? Because the other direction reads (from right to left) as $$Acup B subset Y cup (Acup B)$$ which is trivial, being one of the set theory axioms. The direction you have written is the hard direction.
$endgroup$
– Mark Fischler
Mar 26 at 21:30
$begingroup$
@MarkFischler w...what? Are you implying that $partialA cup partialB=partial(A cup B)$? That only holds true for separated sets. The boundary of $A$ refers to the set $A^- setminus A^circ$. There's nothing purely set-theoretic about this.
$endgroup$
– Quantum Chill
Mar 26 at 21:38
$begingroup$
$x notin partial(A cap B)$ does not imply $x notin partial B$. Simple counterexamples exist for that. (that's in the sentence before you introduce $V_x$).
$endgroup$
– Henno Brandsma
Mar 26 at 21:46
$begingroup$
@HennoBrandsma That's a simple typo on my part, I meant $x notin (partial A cap partial B)$ . Thanks for catching that!
$endgroup$
– Quantum Chill
Mar 26 at 21:49
$begingroup$
Is your $subset$ symbol in the statement that "wasn't too difficult to prove" pointed in the direction you meant? Because the other direction reads (from right to left) as $$Acup B subset Y cup (Acup B)$$ which is trivial, being one of the set theory axioms. The direction you have written is the hard direction.
$endgroup$
– Mark Fischler
Mar 26 at 21:30
$begingroup$
Is your $subset$ symbol in the statement that "wasn't too difficult to prove" pointed in the direction you meant? Because the other direction reads (from right to left) as $$Acup B subset Y cup (Acup B)$$ which is trivial, being one of the set theory axioms. The direction you have written is the hard direction.
$endgroup$
– Mark Fischler
Mar 26 at 21:30
$begingroup$
@MarkFischler w...what? Are you implying that $partialA cup partialB=partial(A cup B)$? That only holds true for separated sets. The boundary of $A$ refers to the set $A^- setminus A^circ$. There's nothing purely set-theoretic about this.
$endgroup$
– Quantum Chill
Mar 26 at 21:38
$begingroup$
@MarkFischler w...what? Are you implying that $partialA cup partialB=partial(A cup B)$? That only holds true for separated sets. The boundary of $A$ refers to the set $A^- setminus A^circ$. There's nothing purely set-theoretic about this.
$endgroup$
– Quantum Chill
Mar 26 at 21:38
$begingroup$
$x notin partial(A cap B)$ does not imply $x notin partial B$. Simple counterexamples exist for that. (that's in the sentence before you introduce $V_x$).
$endgroup$
– Henno Brandsma
Mar 26 at 21:46
$begingroup$
$x notin partial(A cap B)$ does not imply $x notin partial B$. Simple counterexamples exist for that. (that's in the sentence before you introduce $V_x$).
$endgroup$
– Henno Brandsma
Mar 26 at 21:46
$begingroup$
@HennoBrandsma That's a simple typo on my part, I meant $x notin (partial A cap partial B)$ . Thanks for catching that!
$endgroup$
– Quantum Chill
Mar 26 at 21:49
$begingroup$
@HennoBrandsma That's a simple typo on my part, I meant $x notin (partial A cap partial B)$ . Thanks for catching that!
$endgroup$
– Quantum Chill
Mar 26 at 21:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Notation: $S^c=X$ $S$ for any $S.$
Let $f(A,B)=partial Acup partial B.$
Let $g(A,B)=partial (Acup B)cup partial (Acap B)cup (partial Acap partial B).$
Observations: $f(A,B)=f(B,A)=f(A^c,B^c)$ and $g(A,B)=g(B,A)=g(A^c,B^c).$
To show $g(A,B)supset f(A,B)$ it suffices to show $g(A,B)supset partial A,$ because then $g(A,B)=g(B,A)supset partial B.$
For any $pin partial A:$
(i). If $pin partial B$ then $pin partial Acap partial Bsubset g(A,B).$
(ii). If $pnot in partial B$ then $pin int(B)$ or $pin int (B^c).$
(ii-a). If $pin int(B):$ If $U$ is any nbhd of $p$ then $Ucap int(B)$ is a nbhd of $p, $ and $pin partial A.$ So there exist $p'in Acap Ucap int (B)subset Acap B$ and $p''in A^ccap Ucap int(B)subset A^ccap Bsubset (Acap B)^c.$
So $pin partial (Acap B)subset g(A,B).$
(ii-b). If $p in int (B^c):$ Then $pin partial A^c$ and $pin int(B^c).$
By (ii-a) applied to $A^c$ and $B^c,$ this implies $pin partial (A^ccap B^c)$.... And we have $partial (A^ccap B^c)=partial ((A^ccap B^c)^c)=partial (Acup B)subset g(A,B).$
answered Mar 28 at 2:41
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
add a comment |
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%2f3163750%2funion-of-boundaries-formula%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Notation: $S^c=X$ $S$ for any $S.$
Let $f(A,B)=partial Acup partial B.$
Let $g(A,B)=partial (Acup B)cup partial (Acap B)cup (partial Acap partial B).$
Observations: $f(A,B)=f(B,A)=f(A^c,B^c)$ and $g(A,B)=g(B,A)=g(A^c,B^c).$
To show $g(A,B)supset f(A,B)$ it suffices to show $g(A,B)supset partial A,$ because then $g(A,B)=g(B,A)supset partial B.$
For any $pin partial A:$
(i). If $pin partial B$ then $pin partial Acap partial Bsubset g(A,B).$
(ii). If $pnot in partial B$ then $pin int(B)$ or $pin int (B^c).$
(ii-a). If $pin int(B):$ If $U$ is any nbhd of $p$ then $Ucap int(B)$ is a nbhd of $p, $ and $pin partial A.$ So there exist $p'in Acap Ucap int (B)subset Acap B$ and $p''in A^ccap Ucap int(B)subset A^ccap Bsubset (Acap B)^c.$
So $pin partial (Acap B)subset g(A,B).$
(ii-b). If $p in int (B^c):$ Then $pin partial A^c$ and $pin int(B^c).$
By (ii-a) applied to $A^c$ and $B^c,$ this implies $pin partial (A^ccap B^c)$.... And we have $partial (A^ccap B^c)=partial ((A^ccap B^c)^c)=partial (Acup B)subset g(A,B).$
answered Mar 28 at 2:41
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
add a comment |
$begingroup$
Notation: $S^c=X$ $S$ for any $S.$
Let $f(A,B)=partial Acup partial B.$
Let $g(A,B)=partial (Acup B)cup partial (Acap B)cup (partial Acap partial B).$
Observations: $f(A,B)=f(B,A)=f(A^c,B^c)$ and $g(A,B)=g(B,A)=g(A^c,B^c).$
To show $g(A,B)supset f(A,B)$ it suffices to show $g(A,B)supset partial A,$ because then $g(A,B)=g(B,A)supset partial B.$
For any $pin partial A:$
(i). If $pin partial B$ then $pin partial Acap partial Bsubset g(A,B).$
(ii). If $pnot in partial B$ then $pin int(B)$ or $pin int (B^c).$
(ii-a). If $pin int(B):$ If $U$ is any nbhd of $p$ then $Ucap int(B)$ is a nbhd of $p, $ and $pin partial A.$ So there exist $p'in Acap Ucap int (B)subset Acap B$ and $p''in A^ccap Ucap int(B)subset A^ccap Bsubset (Acap B)^c.$
So $pin partial (Acap B)subset g(A,B).$
(ii-b). If $p in int (B^c):$ Then $pin partial A^c$ and $pin int(B^c).$
By (ii-a) applied to $A^c$ and $B^c,$ this implies $pin partial (A^ccap B^c)$.... And we have $partial (A^ccap B^c)=partial ((A^ccap B^c)^c)=partial (Acup B)subset g(A,B).$
answered Mar 28 at 2:41
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
add a comment |
$begingroup$
Notation: $S^c=X$ $S$ for any $S.$
Let $f(A,B)=partial Acup partial B.$
Let $g(A,B)=partial (Acup B)cup partial (Acap B)cup (partial Acap partial B).$
Observations: $f(A,B)=f(B,A)=f(A^c,B^c)$ and $g(A,B)=g(B,A)=g(A^c,B^c).$
To show $g(A,B)supset f(A,B)$ it suffices to show $g(A,B)supset partial A,$ because then $g(A,B)=g(B,A)supset partial B.$
For any $pin partial A:$
(i). If $pin partial B$ then $pin partial Acap partial Bsubset g(A,B).$
(ii). If $pnot in partial B$ then $pin int(B)$ or $pin int (B^c).$
(ii-a). If $pin int(B):$ If $U$ is any nbhd of $p$ then $Ucap int(B)$ is a nbhd of $p, $ and $pin partial A.$ So there exist $p'in Acap Ucap int (B)subset Acap B$ and $p''in A^ccap Ucap int(B)subset A^ccap Bsubset (Acap B)^c.$
So $pin partial (Acap B)subset g(A,B).$
(ii-b). If $p in int (B^c):$ Then $pin partial A^c$ and $pin int(B^c).$
By (ii-a) applied to $A^c$ and $B^c,$ this implies $pin partial (A^ccap B^c)$.... And we have $partial (A^ccap B^c)=partial ((A^ccap B^c)^c)=partial (Acup B)subset g(A,B).$
answered Mar 28 at 2:41
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
Notation: $S^c=X$ $S$ for any $S.$
Let $f(A,B)=partial Acup partial B.$
Let $g(A,B)=partial (Acup B)cup partial (Acap B)cup (partial Acap partial B).$
Observations: $f(A,B)=f(B,A)=f(A^c,B^c)$ and $g(A,B)=g(B,A)=g(A^c,B^c).$
To show $g(A,B)supset f(A,B)$ it suffices to show $g(A,B)supset partial A,$ because then $g(A,B)=g(B,A)supset partial B.$
For any $pin partial A:$
(i). If $pin partial B$ then $pin partial Acap partial Bsubset g(A,B).$
(ii). If $pnot in partial B$ then $pin int(B)$ or $pin int (B^c).$
(ii-a). If $pin int(B):$ If $U$ is any nbhd of $p$ then $Ucap int(B)$ is a nbhd of $p, $ and $pin partial A.$ So there exist $p'in Acap Ucap int (B)subset Acap B$ and $p''in A^ccap Ucap int(B)subset A^ccap Bsubset (Acap B)^c.$
So $pin partial (Acap B)subset g(A,B).$
(ii-b). If $p in int (B^c):$ Then $pin partial A^c$ and $pin int(B^c).$
By (ii-a) applied to $A^c$ and $B^c,$ this implies $pin partial (A^ccap B^c)$.... And we have $partial (A^ccap B^c)=partial ((A^ccap B^c)^c)=partial (Acup B)subset g(A,B).$
answered Mar 28 at 2:41
DanielWainfleetDanielWainfleet
35.8k31648
edited Mar 28 at 2:47
answered Mar 28 at 2:41
DanielWainfleetDanielWainfleet
35.8k31648
answered Mar 28 at 2:41
DanielWainfleetDanielWainfleet
35.8k31648
answered Mar 28 at 2:41
DanielWainfleetDanielWainfleet
35.8k31648
35.8k31648
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%2f3163750%2funion-of-boundaries-formula%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
$begingroup$
Is your $subset$ symbol in the statement that "wasn't too difficult to prove" pointed in the direction you meant? Because the other direction reads (from right to left) as $$Acup B subset Y cup (Acup B)$$ which is trivial, being one of the set theory axioms. The direction you have written is the hard direction.
$endgroup$
– Mark Fischler
Mar 26 at 21:30
$begingroup$
@MarkFischler w...what? Are you implying that $partialA cup partialB=partial(A cup B)$? That only holds true for separated sets. The boundary of $A$ refers to the set $A^- setminus A^circ$. There's nothing purely set-theoretic about this.
$endgroup$
– Quantum Chill
Mar 26 at 21:38
$begingroup$
$x notin partial(A cap B)$ does not imply $x notin partial B$. Simple counterexamples exist for that. (that's in the sentence before you introduce $V_x$).
$endgroup$
– Henno Brandsma
Mar 26 at 21:46
$begingroup$
@HennoBrandsma That's a simple typo on my part, I meant $x notin (partial A cap partial B)$ . Thanks for catching that!
$endgroup$
– Quantum Chill
Mar 26 at 21:49