Question

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)

Answer 1

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?

Answer 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?

Answer 3

yes

Answer 4

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

Answer 5

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

Answer 6

A is closed when A=cl(A).

Answer 7

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

Answer 8

I meant cl(A) is closed, sory

Answer 9

that is what u r supposed to prove.

Answer 10

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

Answer 11

thanks, but due is tomorrow, hurry up (:

Answer 12

quick question

Answer 13

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

Answer 14

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

Answer 15

@TuringTest any idea?

Answer 16

all new to me, sorry
sounds cool though!

Answer 17

it is (:

Answer 18

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

Answer 19

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 \(cl(cl(A))=cl(A)\).

Answer 20

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

Answer 21

wow man nice job, thanks a lot..

Answer 22

u r welcome.
it was fun remembering all this.

Answer 23

What math is this please?

Answer 24

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

Answer 25

@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..

Answer 26

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

Answer 27

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)

Answer 28

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)

Answer 29

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