Prove set is countable

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October 13th 2009, 05:19 PM
utopiaNow

Prove set is countable

I'll first state the question as it appears:
Let $E \in \mathbb{R}^n$ and suppose that for every $x \in E, B(x;1) \cap E$ is countable. Prove that $E$ is countable.

To me: $E \in \mathbb{R}^n$ , 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 $E \subset \mathbb{R}^n$ . 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 $X \subseteq \mathbb{R}^n$ . If $C$ is an open covering of $X$ , then there is a countable subcollection $D \subseteq C$ that is also a cover of $X$ . 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
October 13th 2009, 05:34 PM
Jose27

$B=\{ B(x,1) : x \in \mathbb{Q} \}$ is countable and covers $\mathbb{R} ^n$ so $(\bigcup _{A \in B} A )\cap E$ covers $E$ now use that the union of countably many countable sets is countable.
October 13th 2009, 06:02 PM
utopiaNow

Quote:

Originally Posted by Jose27

$B=\{ B(x,1) : x \in \mathbb{Q} \}$ is countable and covers $\mathbb{R} ^n$ so $(\bigcup _{A \in B} A )\cap E$ covers $E$ 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 $(\bigcup _{A \in B} A )$ is countable, and since $B(x;1) \cap E$ , is countable $\forall x \in E$ , $E$ must be countable? Is that your line of reasoning?

If, so does this mean you used the fact that $E \subseteq \mathbb{R}^n \subseteq B=\{ B(x,1) : x \in \mathbb{Q} \}$ ?
October 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 $(\bigcup _{A \in B} A )$ is countable, and since $B(x;1) \cap E$ , is countable $\forall x \in E$ , $E$ must be countable? Is that your line of reasoning?

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

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