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Theorem:

M is dense in X IFF for all $x \in X$ and $\epsilon >0, B_{\epsilon }\left ( x \right ) \cap M$ is non-empty.

Def: A subset M of a metric space X is called dense in X if the closure $\bar{M}$ is equal to X.

Suppose that M is dense. Then, there exists $\bar{M}=X$.

Suppose to the contrary that there exists some $x \in X$ and $\epsilon >0$: $B_{\epsilon }\left ( x \right ) =\varnothing$. The complement $X \setminus B_{\epsilon }\left ( x \right )$ is a proper closed subset of X. It is proper because $X \setminus B_{\epsilon }\left ( x \right )$ does not equal X. Closed because the open ball is an open set and the complement of an open set is closed.

BUT, here is where things goes awry and despite looking at it for hours, I cannot understand.

My notes mentioned that $X \setminus B_{\epsilon }\left ( x \right )$ contains M.

How can this be true? It seems nonsensical to me. $B_{\epsilon }\left ( x \right ) \cap M$ is non-empty. Drawing a simple picture illustrates the fact that 'removing' the open ball removes elements in M. So anything that encloses $X \setminus B_{\epsilon }\left ( x \right )$ is non-empty so cannot contain M.

Any clarification is appreciated.

Thanks in advance.

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You want to show that $M\cap B_{\epsilon}(x)\neq \emptyset$, for every $x\in X$ and $\epsilon>0$. Since you go by contradiction, this means that you assume that the statement above is not true. So there exist $x\in X$ and $\epsilon>0$ such that $M\cap B_{\epsilon}(x)= \emptyset$. ($M$ doesn't appear in your proof and it is a typo). From this you deduce that $M\subseteq B_{\epsilon}(x)^c$.

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  • $\begingroup$ Wait, The theorem as given by the notes is: >M is dense in X IFF for all $x \in X$ and $\epsilon >0, B_{\epsilon }\left ( x \right ) \cap M$ is non-empty. is the correct theorem " M is dense in X IFF for all $x \in X$ and $\epsilon >0, B_{\epsilon }\left ( x \right ) \cap M$ is empty "? $\endgroup$ – Mathematicing Oct 30 '16 at 12:36
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    $\begingroup$ No, the theorem is correct as stated. The typo is in the proof of it. Since you use contradiction, you need to take the negation of the sentence "For every $x\in X$ and every $\epsilon>0$, $B_{\epsilon}(x)\cap M\neq \emptyset$." $\endgroup$ – tree detective Oct 30 '16 at 14:00
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My notes mentioned that $X \setminus B_{\epsilon }\left ( x \right )$ contains M.

That is simply not true in if $M$ is dense in $X$.

What your notes may have meant by that is anyone's guess. Perhaps it's a typo. Perhaps the notes are simply wrong. Perhaps it's part of a proof-by-contradiction, where it's concluded from an assumption that turns out to be false.

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  • $\begingroup$ Is the given theorem outlined in the OP correct then? $\endgroup$ – Mathematicing Oct 30 '16 at 12:49
  • $\begingroup$ @Mathematicing: Yes. $\endgroup$ – Henning Makholm Oct 30 '16 at 14:24

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