Suppose that a,b exist in $\Bbb R$ with $a < b$. Show for every $x$ that exists in $(a,b)$, there exists $\epsilon\ > 0$ such that there is an $\epsilon$-neighborhood centered at $x$ that is contained in $(a, b)$.
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Where are you stuck? – rschwieb Feb 12 '13 at 17:47
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Cough Density of rationals Cough – Gautam Shenoy Feb 12 '13 at 17:49
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7Density of the rationals has nothing to do with it... – David Mitra Feb 12 '13 at 17:52
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Hint: draw a picture.
As long as $\epsilon$ is shorter than both the distance from $x$ to $a$ and the distance from $x$ to $b$, you have what you want. (See?)
rschwieb
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I drew a picture and I was working off of what @copper.hat had written above because a proof somewhat similar to this was discussed in class. My problem is I'm confused on how to prove this existence of the defined epsilon for every x in the interval. – Student Feb 12 '13 at 18:06
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@Student copper.hat gave away exactly what my hint was getting at. Since the distance between $x$ and either of $a$ or $b$ always exists, this is certainly the way to go, no? – rschwieb Feb 12 '13 at 18:20
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So for my proof: Suppose some a,b in the reals such that a<b. Consider some epsilon>0 and x in the interval (a, b) such that (x-a)<epsilon. Thus, x<epsilon+a. But x is in the interval (a, b), so this cannot be the case. And prove the same for b-x? – Student Feb 12 '13 at 18:56
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@Student No, there is no "let $\epsilon>0$". You need to chose epsilon. After wisely choosing, you point out why you can move above $x$ and below $x$ by epsilon without escaping the interval $(a,b)$. – rschwieb Feb 12 '13 at 19:37
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