A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
I need some help..
A and B are two subsets of the metric space M. Prove following.
cl=closure of the set
a) cl(cl(A))=cl(A)
b) cl(A union B)=cl(A) union cl(B)
anonymous
 3 years ago
I need some help.. A and B are two subsets of the metric space M. Prove following. cl=closure of the set a) cl(cl(A))=cl(A) b) cl(A union B)=cl(A) union cl(B)

This Question is Closed

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.2Let \((M,d)\) be a metric space and \(B(x_0,\varepsilon)=\{x\in M:d(x_0,x)<\varepsilon\}\). u r familiar with all this. right?

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.2Let \(A\subseteq M\). We call \(y_0\in M\) a "limit point" of \(A\) if \[ \large (\forall\varepsilon>0)[B(y_0,\varepsilon)\cap A\neq\emptyset] \] The "closure" of \(A\) is the set \(cl(A)\) of all the limit points of \(A\). again. u r familiar with this. right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry I was working on it, didn't see your comment..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we know A is closed because of definition, right..

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.2A is closed when A=cl(A).

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.2in (a) u r supposed to prove that cl(A) is closed.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I meant cl(A) is closed, sory

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.2that is what u r supposed to prove.

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.2listen. i have to go. i'll try to login later. good luck with this.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks, but due is tomorrow, hurry up (:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I know how to prove in R, is it the same in M

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or if the theorem valid for R, can we use for M ( metric space)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@TuringTest any idea?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0all new to me, sorry sounds cool though!

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.2(a) we have to prove \(cl(cl(A))=cl(A)\). let \(x_0\in cl(cl(A))\), this means that for every \(\varepsilon>0\) \[ \large B(x_0,\varepsilon)\cap cl(A)\neq\emptyset. \] Question: \(B(x_0,\varepsilon)\cap A\neq\emptyset\) ? Let \(x_1\in B(x_0,\varepsilon)\cap cl(A)\) and \(x_1\neq x_0\). Let \(\delta<d(x_0,x_1)/2\), then \[ \large B(x_1,\delta)\subset B(x_0,\varepsilon)\hspace{2.5in} (1) \] but since \(x_1\in cl(A)\) then \(B(x_1,\delta)\cap A\neq\emptyset\), and this and (1) imply that \[ \large B(x_0,\varepsilon)\cap A\neq\emptyset \] that is \(x_0\in cl(A)\).

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.2Conversely, let \(y_0\in cl(A)\) and \(\varepsilon>0\), then \(B(y_0,\varepsilon)\cap A\neq\emptyset\). Question: \(B(y_0,\varepsilon)\cap cl(A)\neq\emptyset\) ? Let \(y_1\in cl(A)\) with \(y_1\neq y_0\) and let \(\gamma>d(y_0,y_1)\) then \[ \large B(y_0,\gamma)\cap cl(A)\neq\emptyset \] because at least \(y_1\in B(y_0,\gamma)\cap cl(A)\). Hence \(y_0\in cl(cl(A))\). Therefore \(cl(cl(A))=cl(A)\).

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.2(b) we have to prove \(cl(A\cup B)=cl(A)\cup cl(B)\). Let \(x_0\in cl(A\cup B)\) and \(\varepsilon>0\) then \[ \large \emptyset\neq B(x_0,\varepsilon)\cap(A\cup B)= [B(x_0,\varepsilon)\cap A]\cup[B(x_0,\varepsilon)\cap B] \] this means that either \(B(x_0,\varepsilon)\cap A\) or \(B(x_0,\varepsilon)\cap B\) is nonempty. That is \(x_0\in cl(A)\) or \(x_0\in cl(B)\), hence \(x_0\in cl(A)\cup cl(B)\). Conversely, if \(\varepsilon>0\) and \(x_0\in cl(A)\cup cl(B)\) then \[ \large x_0\in cl(A)\vee x_0\in cl(B) \] then \[ \large B(x_0,\varepsilon)\cap A\neq\emptyset\qquad\text{or}\qquad B(x_0,\varepsilon)\cap B\neq\emptyset \] from this it follows that \[ \large \emptyset\neq[B(x_0,\varepsilon)\cap A]\cup [B(x_0,\varepsilon)\cap B]=B(x_0,\varepsilon)\cap(A\cup B) \] that is \(x_0\in cl(A\cup B)\). Therefore \(cl(A\cup B)=cl(A)\cup cl(B)\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wow man nice job, thanks a lot..

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.2u r welcome. it was fun remembering all this.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0What math is this please?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It is mixed of topology and real analysis, in particularly, closed set, neighborhood

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@helder_edwin I forgot part c) cl(A intersection B) C cl(A) intersection cl(B) and find an example to show that equality need not to hold.. C= subset can you also take a look this one.. thanks in advance..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\large Let x0∈cl(A∩B) and\, ε>0\, then \\ \large∅≠B(x0,ε)∩(A∩B)=[B(x0,ε)∩A]∩B=[B(x0,ε)∩B]∩A\\ \large \text{this means that either} \,B(x0,ε)∩A \,and\, B(x0,ε)∩B \, is\, nonempty\\ \large That\, is\, x0∈cl(A)\, and\, x0∈cl(B), hence\, x0∈cl(A)∩cl(B)\\ \large Therefore\, cl(A∩B)⊂cl(A)∩cl(B)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Letx0∈cl(A∩B)andε>0then∅≠B(x0,ε)∩(A∩B)=[B(x0,ε)∩A]∩B=[B(x0,ε)∩B]∩Athis means that eitherB(x0,ε)∩AandB(x0,ε)∩Bisnon−emptyThatisx0∈cl(A)andx0∈cl(B),hencex0∈cl(A)∩cl(B)Thereforecl(A∩B)⊂cl(A)∩cl(B)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Letx0∈cl(A∩B)andε>0then∅≠B(x0,ε)∩(A∩B)=[B(x0,ε)∩A]∩B=[B(x0,ε)∩B]∩Athis means that both B(x0,ε)∩AandB(x0,ε)∩Bisnon−emptyThatisx0∈cl(A)andx0∈cl(B), hencex0∈cl(A)∩cl(B)Thereforecl(A∩B)⊂cl(A)∩cl(B)

helder_edwin
 3 years ago
Best ResponseYou've already chosen the best response.2(c) we have to prove that \(cl(A\cap B)\subseteq cl(A)\cap cl(B)\). Let \(\varepsilon>0\) and \(x_0\in cl(A\cap B)\) then \[ \large \emptyset\neq B(x_0,\varepsilon)\cap(A\cap B)= [B(x_0,\varepsilon)\cap A]\cap[B(x_0,\varepsilon)\cap B] \] this means that both \(B(x_0,\varepsilon)\cap A\) and \(B(x_0,\varepsilon)\cap B\) are not empty. That is, \(x_0\in cl(A)\) and \(x_0\in cl(B)\). Therefore, \(cl(A\cap B)\subseteq cl(A)\cap cl(B)\).
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.