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Math Help - Prove set is countable

  1. #1
    Junior Member utopiaNow's Avatar
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    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
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  2. #2
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    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.
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  3. #3
    Junior Member utopiaNow's Avatar
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    Quote Originally Posted by Jose27 View Post
    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} \}?
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  4. #4
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    Quote Originally Posted by utopiaNow View Post
    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
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