## cinar Group Title 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

1. helder_edwin Group Title

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?

2. helder_edwin Group Title

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?

3. cinar Group Title

yes

4. cinar Group Title

sorry I was working on it, didn't see your comment..

5. cinar Group Title

we know A is closed because of definition, right..

6. helder_edwin Group Title

A is closed when A=cl(A).

7. helder_edwin Group Title

in (a) u r supposed to prove that cl(A) is closed.

8. cinar Group Title

I meant cl(A) is closed, sory

9. helder_edwin Group Title

that is what u r supposed to prove.

10. helder_edwin Group Title

listen. i have to go. i'll try to log-in later. good luck with this.

11. cinar Group Title

thanks, but due is tomorrow, hurry up (:

12. cinar Group Title

quick question

13. cinar Group Title

I know how to prove in R, is it the same in M

14. cinar Group Title

or if the theorem valid for R, can we use for M ( metric space)

15. cinar Group Title

@TuringTest any idea?

16. TuringTest Group Title

all new to me, sorry sounds cool though!

17. cinar Group Title

it is (:

18. helder_edwin Group Title

(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 $$x_0\in cl(A)$$.

19. helder_edwin Group Title

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 $$cl(cl(A))=cl(A)$$.

20. helder_edwin Group Title

(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 $$cl(A\cup B)=cl(A)\cup cl(B)$$.

21. cinar Group Title

wow man nice job, thanks a lot..

22. helder_edwin Group Title

u r welcome. it was fun remembering all this.

23. TuringTest Group Title

24. cinar Group Title

It is mixed of topology and real analysis, in particularly, closed set, neighborhood

25. cinar Group Title

@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..

26. cinar Group Title

$\large Let x0∈cl(A∩B) and\, ε>0\, then \\ \large∅≠B(x0,ε)∩(A∩B)=[B(x0,ε)∩A]∩B=[B(x0,ε)∩B]∩A\\ \large \text{this means that either} \,B(x0,ε)∩A \,and\, B(x0,ε)∩B \, is\, non-empty\\ \large That\, is\, x0∈cl(A)\, and\, x0∈cl(B), hence\, x0∈cl(A)∩cl(B)\\ \large Therefore\, cl(A∩B)⊂cl(A)∩cl(B)$

27. cinar Group Title

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)

28. cinar Group Title

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)

29. helder_edwin Group Title

(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, $$cl(A\cap B)\subseteq cl(A)\cap cl(B)$$.