Question

find :sup, inf bd, int, compact on a set

Answer 1

i just want to know if im right or not here are my answers. thanks you will be a big help

Answer 2

sup = lim((2n-1)/n) = 2

Answer 3

isolated points : (2n-1)/n, n>=2

Answer 4

omg thank you, what about my infimum?

Answer 5

no infimum

Answer 6

why is infimum not 3/2?

Answer 7

because there are numbers less than 3/2 I bet

Answer 8

because there are elements less than 3/2 in the set S

Answer 9

but since unbounded below then inf is none

Answer 10

yep!

Answer 11

GOT IT

Answer 12

what are the definitions of int, bd and S' ?

Answer 13

in isolated point you say n>=2 why. i thought this sequence caps off at 2 and you cant have anything bigger than 2

Answer 14

interior points are point in the set which dont include boundary points

Answer 15

bd points are when if you take a tiny interval at that point you get something in the set and something outside of the set.

Answer 16

S' are the accumulation points

Answer 17

interior points must have some open nbhd, right ?

Answer 18

right which also contain a point in S right

Answer 19

gotcha

Answer 20

your interior points look good to me then

Answer 21

what about compactness

Answer 22

one sec
whenever you use (2n-1)/n, you must specify n>=2

Answer 23

for example, look at ur answer for iv :

when n=1, we have (2*1-1)/1 = 1, but 1 is not an isolated point.

Answer 24

^ i just noticed this. because n=2 produces the first number in the set. Why isnt 2 a boundary point? because 2 is not in S???

Answer 25

il need to review analysis a bit
@ikram002p

Answer 26

or @eliassaab

Answer 27

what about covers and subcovers. i dont understand the whole cover thing

Answer 28

vi requires more justification, you need to show an opencover that doesn't have a finite subcover

Answer 29

i thought any thing with infinity on an interval doesnt have a finite subcover.

Answer 30

right, you're supposed to provide a proper opencover and show that it doesn't have a finite subcover

Answer 31

i think simply saying "anything with infinity.. " wont work in analysis hmm

Answer 32

finite cant match infinity is my logic

Answer 33

because it isnt big enough. LOL

Answer 34

intuition is fine

so in part vi, are you saying (-infty, 1] is an opencover of S ?

Answer 35

that doesn't look like an opencover to me

Answer 36

yes but i think its more like an infinite union as a cover. like (-n,1] U n=1 to n=infinity

Answer 37

sure i'll be free within one hour what else left to solve ?@rational

Answer 38

@rational

Answer 39

just wanted to know if 2 was also a boundary point or not since its not in the set S

Answer 40

http://mathworld.wolfram.com/BoundaryPoint.html

yes 2 is a boundary point

Answer 41

thanks!