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I need to proof that if a deleted epsilon neighborhood contains at least one element, it must contain an infinite number of elements. So, this is my proof. Am I doing things wrong? I will try to reach a contradiction.

Consider an deleted epsilon neighborhood of $s_0$: $N_{\epsilon}^-(s_0)$ and define the set S to be the set containing all points of this neighborhood. Suppose, S contains not an infinite number of elements. So S would contain a finite number of elements. Define a new set: $D=\{|s-s_0| | s \in S\}$. Since S was finite, this set is finite and therefore we could pick the smallest element of it. So, there exists an $s^* \in S$ such that $|s^*-s_0|$ is the smallest element of D. Since we know the neighborhood contains at least one element, call it $s_1$, we now that for any $\epsilon$ holds that $$|s_0-s_1|<\epsilon.$$ Now, pick $\epsilon = \frac{1}{2}|s^*-s_0|$, then $$|s_0-s_1|\ge|s^*-s_0|>\frac{1}{2}|s^*-s_0|=\epsilon$$ which contradicts the fact that $|s_0-s_1|<\epsilon$. So the deleted epsilon neighborhood of $s_0$ must contain an infinite number of elements.

Regards, Kevin

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    Neighborhood of what? What is your space? It would be simpler to show that the deleted nhood of radius ${1\over2}|s^*-s_0|$ is empty.2011-11-17
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    Just a deleted neighborhood, $N_{\epsilon}^-(s_0)=\{s|0<|s_0-s|<\epsilon\}$ for any $\epsilon > 0$.2011-11-17
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    I now see you did what I suggested above, but it's badly phrased. Once you find $s^*$, set $\epsilon'={1\over2}|s^*-s_0|$. Then argue that $N_{\epsilon'}^-(s_0)=\emptyset$.2011-11-17
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    Ah ok, thanks. But why use $\epsilon'$? Is that for showing that you pick that certain epsilon? And are the main thoughts in the proof ok?2011-11-17
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    You already used $\epsilon$ for another purpose in the proof; its value is fixed. The main thoughts seem ok. But I still would like to know what set you are working with. Are you talking about a subset $V$ of $\Bbb R$ for which every deleted nhood contains at least one element of $V$ (I assume so)?2011-11-17
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    Ok, I thought because it was fixed, you can use that same epsilon, but that will indeed lead to confusion and you can't see that epsilon was substituted for another value. No, the neighborhood itself contains at least one element. And all sets (/neighborhoods) are in $R$.2011-11-17

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