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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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    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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