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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)
 one year ago
 one year 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)
 one year ago
 one year ago

This Question is Closed

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

helder_edwinBest ResponseYou've already chosen the best response.2
Let \(A\subseteq M\). 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?
 one year ago

cinarBest ResponseYou've already chosen the best response.0
sorry I was working on it, didn't see your comment..
 one year ago

cinarBest ResponseYou've already chosen the best response.0
we know A is closed because of definition, right..
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.2
A is closed when A=cl(A).
 one year ago

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

cinarBest ResponseYou've already chosen the best response.0
I meant cl(A) is closed, sory
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.2
that is what u r supposed to prove.
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.2
listen. i have to go. i'll try to login later. good luck with this.
 one year ago

cinarBest ResponseYou've already chosen the best response.0
thanks, but due is tomorrow, hurry up (:
 one year ago

cinarBest ResponseYou've already chosen the best response.0
I know how to prove in R, is it the same in M
 one year ago

cinarBest ResponseYou've already chosen the best response.0
or if the theorem valid for R, can we use for M ( metric space)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
all new to me, sorry sounds cool though!
 one year ago

helder_edwinBest 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 \(x_0\in cl(A)\).
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.2
Conversely, 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 \(cl(cl(A))=cl(A)\).
 one year ago

helder_edwinBest 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 \(cl(A\cup B)=cl(A)\cup cl(B)\).
 one year ago

cinarBest ResponseYou've already chosen the best response.0
wow man nice job, thanks a lot..
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.2
u r welcome. it was fun remembering all this.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
What math is this please?
 one year ago

cinarBest ResponseYou've already chosen the best response.0
It is mixed of topology and real analysis, in particularly, closed set, neighborhood
 one year ago

cinarBest 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..
 one year ago

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

cinarBest ResponseYou've already chosen the best response.0
Letx0∈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)
 one year ago

cinarBest ResponseYou've already chosen the best response.0
Letx0∈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)
 one year ago

helder_edwinBest 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, \(cl(A\cap B)\subseteq cl(A)\cap cl(B)\).
 one year ago
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