Question

let \[x \rightarrow x \in M , A =\{x\}\cup\{x_n: n \geq 1\}\] prove A is closed via \[A^c\] is open let \[x \rightarrow x \in M , A =\{x\}\cup\{x_n: n \geq 1\}\] prove A is closed via \[A^c\] is open @Mathematics

Answer 1

\[x_n \rightarrow x \in M , A =\{x\}\cup\{x_n: n \geq 1\}\]

Answer 2

use first order logic

Answer 3

i am trying to show that
\[A^c\] is open and it is eluding me

Answer 4

a set is closed if and only if it contains all of its limit points

Answer 5

yeah i got this after i thought about it, but i think it is not as simple as "the set contains its limit points"

Answer 6

i think you show the complement is open and i managed that without much difficulty after i cleaned the cobwebs out of my brain

Answer 7

A theorem from a book I have (Carothers Real Analysis page 54)

\[\text{Given a set } F\text{ in }(M,d),\text{ the following are equivalent:}\]

\[\text{(i) }F\text{ is closed; that is, }F^{c}=M\backslash F\text{ is open}\]
\[\text{(ii ) }\text{If }B_\epsilon(x)\cap F\neq \emptyset\text{ for every }\epsilon>0,\text{ then }x\in F\]
\[\text{(iii) }\text{If a sequence }(x_n)\subset F\text{ converges to some point }x\in M,\text{ then }x\in F\]

Answer 8

actually this is in fact a problem right from carothers

Answer 9

then use theorem 4.9 ;)

Answer 10

page 58 # .36

Answer 11

nice

Answer 12

but i was told by the questioner that her professor said to show that Ac is open

Answer 13

that problem is one I have boxed...so a long time ago it was assigned to me ;)

Answer 14

that's fine show A is closed and this Ac is open as you already stated

Answer 15

which i did. but i am now confused as to why i cannot use that theorem directly

Answer 16

oh wait. i think i see why not. part iii says "If a sequence converges to some point x in F then x is in F"

Answer 17

which is true for your set A

Answer 18

i have another question. while you have your carothers out

Answer 19

68 #34 without using sequences (which is what it says to do)

Answer 20

again she was told not to for some reason, but i though we could say "inverse image of open sets is open" should work for this one

Answer 21

yes...using the topological def of continuity will work.

Answer 22

so i could say, since the metric is clearly continuous on M that means inverse image of open sets is open in M, hence in MxM inverse image of open sets is open

Answer 23

in MxM of course

Answer 24

yes

Answer 25

though I prefer the term preimage :)

Answer 26

preimage fine. thanks!