Open cover of [0,1)

Question

I have to give cover by open intervals of [0,1) that don't accept a finite sub cover of it.

My idea was to use the the Sorgenfrey topology because the open sets in $\mathbb{R}$ are [a,b) so if I take the open sets as A$_n$ = [1/n , 1) for n>2 then $\cup$ A$_n$ with n>2 it will be a open cover of [0,1) with no finite sub cover.

But I don't know if it is right because they asked me to use open intervals and I don't know if there is a cover of [0,1) by intervals like (a,b) with no finite sub cover.

Answer 1

You can't use a different topology from the one you are given to solve this. Presumably they want you to use the usual metric topology on $[0,1)$.

There is an open cover with no finite subcover given by sets of the form $[0,1-1/n)$. You can show that the union of any finitely many of these is equal to the largest one of them, which does not contain the whole interval.

The sets $[0,a)$ are open in the subspace topology. If you are finding an open cover in the ambient space you can use $(-1,1-1/n)$, and the intersection of these with $[0,1)$ is the open cover I gave above.

Answer 2

Take the cover $\{A_n\}_n$ to be as follows.

$A_n= [0,1-1/n)$.