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Let $a,b > 0$, and consider the interval $(a,b)$. Let $a < x_0 < b$. Prove that (a,b) is a neighborhood of $x_0$.

Proof: Assume $a,b>0$ and $a < x_0 < b$. Let $\epsilon = \frac{b-a}{2}$. Then consider the interval $(x_0-\frac{b-a}{2}, x_0 + \frac{b-a}{2})$. Then $(x_0-\frac{b-a}{2}, x_0 + \frac{b-a}{2}) \subset (a,b)$. Thus $(a,b)$ is a neighborhood of $x_0$.

Please let me know if I am going about this correctly. Thanks in advance.

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    It's the right general idea, but I think you've chosen the wrong expression for $\epsilon$. With the $\epsilon$ that you chose, there's no guarantee that $(x_0-\epsilon, x_0+\epsilon)$ is entirely within the interval $(a,b)$.2012-10-02

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What if $a=1$, $x_0=2$, and $b=5$? then $\epsilon=2$, but $(2-2,2+2)=(0,4)\nsubseteq(1,5)$. You need to take $\epsilon$ small enough to make sure that $x_0-\epsilon\ge a$ and $x_0+\epsilon\le b$. Your idea of splitting a distance in half is good, but the distance that you should be splitting is the smaller of $x_0-a$ and $b-x_0$.

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    @tkrm: If you get stuck, or just want me to check what you’ve done, give me another ping.2012-10-02