Question

\[\overline{A \cup B}=\overline{A} \ \cup \overline{B}\]

Answer 1

\[a \in \overline{A} \ \iff \forall r>0, \ B(a,r)\cap \ A \ \ne \emptyset\]

Answer 2

thats pretty much all I have

Answer 3

looks like a proof of something i am fairly sure is wrong

Answer 4

hmm, I have it in two books as a propasition

Answer 5

proposition*

Answer 6

oooh sorry, i thought they were sets and compliment, excuse me it is late
it is sets and closures

Answer 7

yeah

Answer 8

try to use that $\overline{A}$ is the smallest closed set containing \(A\)

Answer 9

you have to show containment both ways
depending on your definition of closed sets
try here for a more precise explanation than i can write
http://math.stackexchange.com/questions/195311/union-of-closure-of-sets-is-the-closure-of-the-union

Answer 10

@watchmath it must have been a while
$ does not work use \(

Answer 11

for in line \(\overline{A}\) tex

Answer 12

yes I know satellite. But I am just got use to using dollar.

Answer 13

Remember de Morgan's law?

Answer 14

Haha! I remember learning de morgans xD

Answer 15

Since \[
\overline{A\cup B}=\overline A\cap \overline B
\]You must find where: \[
\overline A\cap \overline B\neq \overline A\cup \overline B
\]

Answer 16

yeah

Answer 17

It's not a contradiction, but rather something that will sometimes be false.

Answer 18

since \(\overline{A}\cup \overline{B}\) is closed and containing \(A\cup B\) byt the property of closure then we have \(\overline{A}\cup \overline{B}\supseteq\overline{A\cup B}\). On the other hand \(\overline{A}\subseteq \overline{A\cup B}\) and \(\overline{B}\subseteq \overline{A\cup B}\). Hence \[\overline{A}\cup \overline{B}\subseteq \overline{A\cup B}\]

Answer 19

ahh easy peasy ty. I guess it makes total sense that \(A \subset \overline{A}\)
they just define closure for us and then give me the iff statement so Im thinking I need to used a closed ball. but I guess not.

Answer 20

use*