Question

Show that the set is not compact by exhibiting an open cover with no finite subcover
\(A=\{x \in R^n | ||x||<1\}\)
Please, help

Answer 1

try this :

\[\bigcup\limits_{n=1}^{\infty} \left(\frac{-1}{n}, ~\frac{1}{n}\right)\]

Answer 2

what is (-1/n , 1/n) in? is it a point in R^2, or an interval in R?

Answer 3

it is an interval in R but you want an opencover that works for vectors is it

Answer 4

you can extend it to R^n, just think of them as radii of openballs

Answer 5

how about

\[U_n=\left\{x\in\mathbb{R}^n|~\|x\|<1-\frac{1}{n}\right\}\]

let \[B=\{U_n\}_{n=1}^{\infty}\]

Answer 6

How can you guys can pick the specific subcover right after seeing the problem like that?

Answer 7

that is not a subcover...it is a cover

Answer 8

that has no finite subcover

Answer 9

you can cookup many infinite opencovers like that

Answer 10

whatever, I struggle with figure out either of them.
like what ganeshie8 suggested, I am not sure whether I can use R^2 or not since the element in A is in R^n

Answer 11

don't think of them as intervals
think of them as openballs in R^n

Answer 12

I am afraid of that if I pick some cover in R^2, I don't have that cover is one of covers in R^n

Answer 13

also there is a \(\LaTeX\) command for doing the norm of a vector

\|x\|

\[\|x\|\]

Answer 14

the term "openball" is not metric space specific

Answer 15

but i see the example i gave you earlier looks more like an interval in R

Answer 16

One more confirm: a cover of a set is just partial cover, right?

Answer 17

I mean that cover is just cover some elements in set, not the thing like this
which one is the right one?