anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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helder_edwin
  • helder_edwin
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?
helder_edwin
  • helder_edwin
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?
anonymous
  • anonymous
yes

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anonymous
  • anonymous
sorry I was working on it, didn't see your comment..
anonymous
  • anonymous
we know A is closed because of definition, right..
helder_edwin
  • helder_edwin
A is closed when A=cl(A).
helder_edwin
  • helder_edwin
in (a) u r supposed to prove that cl(A) is closed.
anonymous
  • anonymous
I meant cl(A) is closed, sory
helder_edwin
  • helder_edwin
that is what u r supposed to prove.
helder_edwin
  • helder_edwin
listen. i have to go. i'll try to log-in later. good luck with this.
anonymous
  • anonymous
thanks, but due is tomorrow, hurry up (:
anonymous
  • anonymous
quick question
anonymous
  • anonymous
I know how to prove in R, is it the same in M
anonymous
  • anonymous
or if the theorem valid for R, can we use for M ( metric space)
anonymous
  • anonymous
@TuringTest any idea?
TuringTest
  • TuringTest
all new to me, sorry sounds cool though!
anonymous
  • anonymous
it is (:
helder_edwin
  • helder_edwin
(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
helder_edwin
  • helder_edwin
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 \(cl(cl(A))=cl(A)\).
helder_edwin
  • helder_edwin
(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 non-empty. 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
  • anonymous
wow man nice job, thanks a lot..
helder_edwin
  • helder_edwin
u r welcome. it was fun remembering all this.
TuringTest
  • TuringTest
What math is this please?
anonymous
  • anonymous
It is mixed of topology and real analysis, in particularly, closed set, neighborhood
anonymous
  • anonymous
@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
  • anonymous
\[\large Let x0∈cl(A∩B) and\, ε>0\, then \\ \large∅≠B(x0,ε)∩(A∩B)=[B(x0,ε)∩A]∩B=[B(x0,ε)∩B]∩A\\ \large \text{this means that either} \,B(x0,ε)∩A \,and\, B(x0,ε)∩B \, is\, non-empty\\ \large That\, is\, x0∈cl(A)\, and\, x0∈cl(B), hence\, x0∈cl(A)∩cl(B)\\ \large Therefore\, cl(A∩B)⊂cl(A)∩cl(B)\]
anonymous
  • anonymous
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)
anonymous
  • anonymous
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)
helder_edwin
  • helder_edwin
(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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