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 2 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)
 2 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
 2 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
 2 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?

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

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

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

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

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

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

helder_edwin
 2 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.

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

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

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

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

helder_edwin
 2 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
 2 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
 2 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 \(cl(A\cup B)=cl(A)\cup cl(B)\).

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

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

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

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

cinar
 2 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..

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

cinar
 2 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)

cinar
 2 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
 2 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, \(cl(A\cap B)\subseteq cl(A)\cap cl(B)\).
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