# Is my proof valid?

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 therefor 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. – David Mitra Nov 17 '11 at 8:40
Just a deleted neighborhood, $N_{\epsilon}^-(s_0)=\{s|0<|s_0-s|<\epsilon\}$ for any $\epsilon > 0$. – Kevin Nov 17 '11 at 8:43
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 $N_{\epsilon'}^-(s_0)=\emptyset$. – David Mitra Nov 17 '11 at 8:55
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? – Kevin Nov 17 '11 at 9:41
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)? – David Mitra Nov 17 '11 at 9:53