Question

is this set compact ?

Answer 1

\[\large \left(3, 5\right)\]

Answer 2

on the real line compact means closed and bounded

Answer 3

yes exactly! i see it is not compact because it is not closed
how do i prove it is not compact if it is not closed ?

Answer 4

compact set : there exists a finite sub cover for every open cover

Answer 5

i think i need to find an open cover without a finite subcover

Answer 6

Consider the open cover \(\{A_n\} = \left\{\left(3-\dfrac{1}{n},5-\dfrac{1}{n}\right)\right\}\). Then \(\displaystyle \bigcup_{n=1}^{\infty}A_n = (3,5)\) but there is no finite subcover that covers (3,5).

Does this make sense?

Answer 7

that looks like nested sets

Answer 8

i can remove all the interior sets for subcover right ?

Answer 9

Excuse me, I meant to say that that \(\displaystyle\bigcup_{n=1}^{\infty}A_n = (2,5)\) (which still covers (3,5)).

Answer 10

that makes sense thanks! on side 2 i can remove everything after n=1, but removing on side 5 is not possible since there are infinitely many sets

Answer 11

That union should be \([2,5)\)

Answer 12

yes side 5 is open