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First, the question I have is very similar to this question, but I hope it doesn't get closed as duplicate since I'm stuck nevertheless.

I'm trying to algebraically prove that an open interval is an open set. If I sketch it, as suggested by @rschwieb in the other question, then it seems quite obvious that this is indeed true. But I would like to be able to show it algebraically and after having looked at various sources found on the internet, I have decided to ask it here.

For definiteness, let us consider: \begin{equation} (c,d) = \{ x \in \mathbb{R} \mid c < x < d \} \end{equation} Furthermore, let $a \in (c,d)$, and recall that the $\epsilon$-neighborhood of $a$ is the set: \begin{equation} V_\epsilon (a) = \{ x \in \mathbb{R} \mid |x-a| < \epsilon \} \end{equation} Now, if we take \begin{equation} \epsilon = \mathrm{min} \{ a-c,d-a \} \end{equation} then $a-\epsilon \geq c$ and $a+\epsilon \leq d$. Up until here I understand everything. But then I don't understand how we can conclude that $V_\epsilon (a) \subseteq (c,d)$? I'm sorry if this conclusion is really obvious (which it probably is), but for some reason I can't wrap my brain around it. Any help is much appreciated.

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Your question is much better than the linked one. Once this has an upvoted answer, I would rather close the other as a duplicate of this one. –  Jonas Meyer Aug 22 at 15:14

3 Answers 3

up vote 4 down vote accepted

I don't think choosing $\epsilon = \mathrm{min} \{ a-c,d-a \}$ would work. I think $\epsilon \lt \mathrm{min} \{ a-c,d-a \}$ is a better choice.


Picture it geometrically by drawing a real line. $ |x - a| \lt \epsilon $ represents all points on the line that are $\epsilon$-distant from the point $a$. By picking $\epsilon \lt \mathrm{min} \{ a-c,d-a \}$ what you do is to pick the smallest distance from the point $c$ to the boundaries of the interval. Now if we create a neighbourhood (an open set) around $c$ again using this minimum distance it will clearly be contained in the original interval.

Rigorously,

$x \in V_{\epsilon}(a) \implies |x - a| \lt \epsilon \iff a - \epsilon\lt x \lt a + \epsilon \tag{1}$

Now $\epsilon \le a-c $ and $\epsilon \le d - a$. Use these to approximate $\epsilon$ in $(1)$. That is,

$$ c=a - (a - c) \lt a - \epsilon\lt x \lt a + \epsilon \lt a + (d - a) = d \iff x \in (c,d) \implies V_{\epsilon}(a) \subseteq (c, d)$$

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On the first line after "Rigorously," shouldn't it read "$\dots \iff a-\epsilon<x<a+\epsilon$"? Also, it is fine to choose $\epsilon=\min\{|a-c|,|a-d|\}$, since the points in $V_\epsilon(a)$ are of distance strictly less than $\epsilon$ from $a$. –  Morgan O Aug 22 at 15:42
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@Morgan O: Correct. Edited. –  Ishfaaq Aug 22 at 15:43
    
Thanks, this is really nice. The way you do it (i.e. picking $\epsilon < \mathrm{min} \{ a-c,d-a \}$ is similar to what is done here. But the way I have done it above is the way it done in Understanding Analysis by Stephen Abbott. Do you know for sure that Stephen Abbott is wrong? –  Hunter Aug 22 at 16:06
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@Hunter. No. In fact I got it wrong as Morgan suggests above. Abbot's proof is spotless. In the final line the outer inequalities should be $\le, \le$. But everything else still holds. –  Ishfaaq Aug 22 at 16:08

Note that $x \in V_{\varepsilon}(a)$ if and only if $x > a-\varepsilon \ge c$ and $x < a+\varepsilon \le d$.

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Thanks! This is very useful –  Hunter Aug 22 at 16:07

Here's an alternate way to go about proving that $V_\epsilon(a) \subset (c,d)$ for some $\epsilon >0$: instead of "measuring distance to the edges," you can "measure distance to the center." Let $p$ be the midpoint of $(c,d)$. Then

$$(c,d) = V_\delta(p),$$

where $\delta:= |p-c|=|p-d|.$ Given $a \in (c,d)$, let $\mu = \delta-|a-p|.$ Given $x \in V_\mu(a)$: $$|x-p|=|x-a+a-p| \leq |x-a|+|a-p| \leq \delta-|a-p|+|a-p|=\delta.$$

Hence $V_\mu(a)\subset V_\delta(p)=(c,d)$.

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