# covering

• Jul 11th 2008, 10:46 AM
particlejohn
covering
Let $\displaystyle S = (0,1)$ and $\displaystyle \mathcal{O} = \{(1/n,1): n \in \bold{N} \}$. Show that $\displaystyle \mathcal{O}$ is an open cover of $\displaystyle S$.

So choose $\displaystyle x \in (0,1)$. Pick $\displaystyle n_0 \in \bold{N}$ such that $\displaystyle 1/n_0 < x$. This implies that $\displaystyle x \in (1/n_0, 1) \subset \bigcup_{n \in \bold{N}} (1/n,1) = \bigcup_{G \in \mathcal{O}} G$. Is this correct? Could I pick $\displaystyle n_0$ in a different way?

What if we had a finite subset $\displaystyle \mathcal{O'} \subset \mathcal{O}$?
• Jul 11th 2008, 11:26 AM
Plato
If you had a finite subset of the collection, then the smallest left-hand endpoint would a number $\displaystyle 0 < t < 1$.
But $\displaystyle \frac{t}{2} \in \left( {0,1} \right)$ and $\displaystyle \frac{t}{2} < t$.
Therefore, the subcollection does not cover $\displaystyle \left( {0,1} \right)$.
• Jul 11th 2008, 01:01 PM
Plato
The statement that set is compact means every open covering has a finite subcover.
Having one open cover which has a finite subcover does not imply that the set is compact.

Consider: $\displaystyle \left( { - 1,.000009} \right) \cup \left[ {\bigcup\limits_{n \in N} {\left( { n^{ - 1} ,1} \right)} } \right]$
That is clearly an open covering of $\displaystyle \left( {0,1} \right)$ and the collection contains a finite subcover.
But $\displaystyle \left( {0,1} \right)$ is still not compact.