# Prove set is countable

• Oct 13th 2009, 05:19 PM
utopiaNow
Prove set is countable
I'll first state the question as it appears:
Let $\displaystyle E \in \mathbb{R}^n$ and suppose that for every $\displaystyle x \in E, B(x;1) \cap E$ is countable. Prove that $\displaystyle E$ is countable.

To me: $\displaystyle E \in \mathbb{R}^n$, doesn't make sense, I'm interpreting this as one specific n-tuple, so I don't understand how an n-tuple can or cannot be countable. So I interpreted the question as $\displaystyle E \subset \mathbb{R}^n$. This may be wrong right away so if someone can explain to me why it should be the other way that would help.

But here's my attempt at a solution based on my interpretation:
I'm trying to use the Lindelof covering theorem which states that: Suppose $\displaystyle X \subseteq \mathbb{R}^n$. If $\displaystyle C$ is an open covering of $\displaystyle X$, then there is a countable subcollection $\displaystyle D \subseteq C$ that is also a cover of $\displaystyle X$. But I'm not able to make much progress with that method. And as of now I can't think of another way of showing it.

Any tips, or hints would be appreciated. Thanks in advance
• Oct 13th 2009, 05:34 PM
Jose27
$\displaystyle B=\{ B(x,1) : x \in \mathbb{Q} \}$ is countable and covers $\displaystyle \mathbb{R} ^n$ so $\displaystyle (\bigcup _{A \in B} A )\cap E$ covers $\displaystyle E$ now use that the union of countably many countable sets is countable.
• Oct 13th 2009, 06:02 PM
utopiaNow
Quote:

Originally Posted by Jose27
$\displaystyle B=\{ B(x,1) : x \in \mathbb{Q} \}$ is countable and covers $\displaystyle \mathbb{R} ^n$ so $\displaystyle (\bigcup _{A \in B} A )\cap E$ covers $\displaystyle E$ now use that the union of countably many countable sets is countable.

So we use that the union of countably many countable sets is countable to see that $\displaystyle (\bigcup _{A \in B} A )$ is countable, and since $\displaystyle B(x;1) \cap E$, is countable $\displaystyle \forall x \in E$, $\displaystyle E$ must be countable? Is that your line of reasoning?

If, so does this mean you used the fact that $\displaystyle E \subseteq \mathbb{R}^n \subseteq B=\{ B(x,1) : x \in \mathbb{Q} \}$?
• Oct 13th 2009, 06:11 PM
Jose27
Quote:

Originally Posted by utopiaNow
So we use that the union of countably many countable sets is countable to see that $\displaystyle (\bigcup _{A \in B} A )$ is countable, and since $\displaystyle B(x;1) \cap E$, is countable $\displaystyle \forall x \in E$, $\displaystyle E$ must be countable? Is that your line of reasoning?

If, so does this mean you used the fact that $\displaystyle E \subseteq \mathbb{R}^n \subseteq B=\{ B(x,1) : x \in \mathbb{Q} \}$?

Sort of, but not exactly: $\displaystyle E \subset \mathbb{R} ^n \subset (\bigcup _{A \in B} A )$. And this last one obviously cant be countable but $\displaystyle E= E \cap \mathbb{R} ^n = (\bigcup _{A \in B} A ) \cap E = \bigcup_{A \in B} (A \cap E)$ where each intersection in this last union is countable by hypothesis so the union must also be countable