# finite subcover

#### dwsmith

##### Well-known member
The collection $F$ of open intervals of the form $(1/n,2/n)$ where $n = 2,3,\ldots$ is an open covering of the open interval $(0,1)$.
Prove that no finite subcollection of $F$ covers $(0,1)$ without using the fact it isn't compact.

Suppose there exists a finite subcollection of $F$ that covers $(0,1)$. By Heine-Borel, there exist a countable collection of $F$ such that
$$(0,1) = \bigcup_{n = 1}^mI_n$$
How should I continue on?

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#### Ackbach

##### Indicium Physicus
Staff member

1. The finite collection you're allowed to exhibit (by hypothesis) may not be in order. I would recommend using a finite index set for your union.

2. I would pick the largest $n$ in your finite index set, and show that there is a number in $(0,1)$ smaller than $1/n$.

#### dwsmith

##### Well-known member

2. I would pick the largest $n$ in your finite index set, and show that there is a number in $(0,1)$ smaller than $1/n$.
I don't understand this.

#### Ackbach

##### Indicium Physicus
Staff member
I would go like this: suppose there is a finite subcovering $S\subset F$ of $(0,1)$. Then there exists a finite index set $J\subset\mathbb{Z}$ such that
$$S=\{A_{j}|j\in J \land A_{j}\in F\}.$$
By the definition of subcovering, it must be that
$$(0,1)=\bigcup_{j\in J}A_{j}=\bigcup_{j\in J}\left(\frac{1}{j},\frac{2}{j}\right).$$
Let
$$M=\max_{j\in J}j.$$
Then
$$A_{M}=\left(\frac{1}{M},\frac{2}{M}\right),$$
and $1/M$ is as far left as the finite subcovering goes. Argue that this is not far left enough.