Question

Show that if G is an open set and F is a closed set,then G-F is an open set and F-G is a closed set.

Answer 1

who are u pls?

Answer 2

haha.me lah, ain. triple y, help me please.:)

Answer 3

i still thinking.. sorry..

Answer 4

ok. never mine.

Answer 5

ye ke 3y still thinking??

Answer 6

wiah, nak answer please. hukhuk. :)

Answer 7

wiah 2 sapa??

Answer 8

wiah tu, alawiah ke?

Answer 9

ya. alawiyah la tu. hehe.

Answer 10

pandainy saya

Answer 11

admin la sangat. :p

Answer 12

got clue la.. the intersection of an open set is open, and the intersection of a close set is close.. that what i noe.. hehe

Answer 13

my frenz said use this answer
The simplest way to understand this problem is to understand exactly what G\F means in the context of unions, intersections, and complements.

Quite simply, A/B = A int B' (the intersection of A and not-B)

Now: if F is closed, then F' is open by the definition of a closed set. Equivalently, G being open means that G' is closed.

Thus, G\F = G int F'

F\G = F int G'
This is a countable intersection of closed sets, which must always be closed (another theorem).

Answer 14

what does A/B and G/F means?

Answer 15

A/B means the intersection of A and \[B^{c}\]

Answer 16

same goes to G/F

Answer 17

A/B or A\B??

Answer 18

both also same?if write in simbol how? is it A - B?

Answer 19

ya la..

Answer 20

owh.ok.thanx.do u know how to solve this question?
Let q(n) = 1/2 if n is prime, and -1if otherwise,
and let {an\} be defined recursively by \[a _{1} = 1, a _{n+1} = a _{n} + 2^{-n} q(n), n \ge 1\]
Show that {\[a _{n}\]} is a Cauchy sequence.