# Open cover of [0,1)

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.

• You can't choose the topology; you have to use the one given, or implied. Here, it's evidently the usual topology. Look at sets of the form $[0,1-1/n)$. – David Mitra Dec 8 '19 at 16:20
• Well, but [a,b) is not an open set in the usual topology, so I can't use it for this – user732763 Dec 8 '19 at 16:23
• Be careful. Sets of the form $[1/n, 1)$ do not even cover your space! In which of these sets do we add $0$? As the hints suggest, you have the right idea, but need to move the other way (using intervals $[0, \frac{n-1}{n})$) – HallaSurvivor Dec 8 '19 at 16:23
• If we take a infinite union when n tend to infinite 1/n is 0. – user732763 Dec 8 '19 at 16:24
• No, $1/n$ is never $0$. It's a fact that $1/n\to0$ as $n\to\infty$; that doesn't change the fact that $0$ is not an element of $\bigcup_n[1/n,1)$. – David C. Ullrich Dec 8 '19 at 16:27

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.

• Okey, the point that I was omitting was using the subspace topology because I was using the usual topology of $\mathbb{R}$. – user732763 Dec 8 '19 at 16:27
• @Alvaro There are two notions of an open cover of a subspace. You can use an open cover by subsets in the larger space, which may contain points not in the subspace, or you can use an open cover in the subspace topology. These are basically equivalent because you can take the intersection. – Matt Samuel Dec 8 '19 at 16:29

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

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

• But the problem is that [a,b) is not a open set in the usual topology. – user732763 Dec 8 '19 at 16:25
• It is, if $a = 0$. What are open sets in subspace topology? – Ajay Kumar Nair Dec 8 '19 at 16:26
• Okey, I was thinking only in the usual topology of $\mathbb{R}$ not in the subspace topology. – user732763 Dec 8 '19 at 16:29