covering

**particlejohn** · July 11th 2008, 10:46 AM

Let $S = (0,1)$ and $\mathcal{O} = \{(1/n,1): n \in \bold{N} \}$ . Show that $\mathcal{O}$ is an open cover of $S$ .

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

What if we had a finite subset $\mathcal{O'} \subset \mathcal{O}$ ?

**Plato** · July 11th 2008, 11:26 AM

If you had a finite subset of the collection, then the smallest left-hand endpoint would a number $0 < t < 1$ .
But $\frac{t}{2} \in \left( {0,1} \right)$ and $\frac{t}{2} < t$ .
Therefore, the subcollection does not cover $\left( {0,1} \right)$ .

**Plato** · July 11th 2008, 01:01 PM

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

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