# Context:

I'm trying to algebraically prove that an open interval is an open set. If I sketch it, as suggested by @rschwieb in this answer, 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.

# Attempt:

For definiteness, let us consider: $$$$(c,d) = \{ x \in \mathbb{R} \mid c < x < d \}$$$$ Furthermore, let $$a \in (c,d)$$, and recall that the $$\epsilon$$-neighborhood of $$a$$ is the set: $$$$V_\epsilon (a) = \{ x \in \mathbb{R} \mid |x-a| < \epsilon \}$$$$ Now, if we take $$\epsilon = \mathrm{min} \{ a-c,d-a \}$$, then $$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)$$?

• 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. Aug 22, 2014 at 15:14

Choose $\epsilon \lt \mathrm{min} \{ a-c,d-a \}$.

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,

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

• 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 $a$. Aug 22, 2014 at 15:42
• @Morgan O: Correct. Edited. Aug 22, 2014 at 15:43
• @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. Aug 22, 2014 at 16:08
• Please correct the first line. As Morgan O said, $\epsilon = \mathrm{min} \{ a-c,d-a \}$ works. This question and your answer has been read by hundreds of people by now, so it's important to keep the content correct.
– user147263
Jan 25, 2015 at 17:00

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

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

My attempt of a complete answer:

Let $$( c,d ) = \left\{ x \in \mathbb{R} : c < x < d \right\} \subseteq \mathbb{R}$$ be an open interval and let $$a \in ( c,d )$$ be any point from it. Notice that $$a-c>0$$ and $$d-a>0$$ are distances from $$a$$ to the interval's endpoints $$c \notin ( c,d )$$ and $$d \notin ( c,d )$$. Let $$\epsilon = \min( a-c, d-a )$$. Then, $$a - \epsilon \geq c$$ and $$a + \epsilon \leq d$$. A point with distance $$\epsilon$$ from $$a$$ isn't contained in the interval (it's the interval's boundary point). Yet, if we create a neighborhood (open interval) $$V_{\epsilon}( a )$$ containing points for which $$|x - a| < \epsilon \iff a - \epsilon < x < a + \epsilon$$, we only require the distance from $$a$$ to be smaller than $$\epsilon$$, which means that the endpoints $$c, d$$ are never included in the neighboorhod and hence $$V_{\epsilon}( a )$$ is contained in $$( c,d )$$:

$$\forall a \in ( c,d ) \; {and} \; \epsilon=\min( a-c,d-a ),$$ $$x \in V_{\epsilon}( a ) \iff c \leq a - \epsilon < x < a + \epsilon \leq d$$ $$x \in V_{\epsilon}( a ) \iff c < x < d$$ $$x \in V_{\epsilon}( a ) \iff x \in ( c,d )$$ Hence $$\forall x \in V_{\epsilon}( a ), x \in ( c,d ) \implies V_{\epsilon}( a ) \subseteq ( c,d ) .$$

Remark: To me it seems that the main reason for an open interval to be an open set is the strict inequality in the definition of $$V_{\epsilon}( a )$$, which requires that there must be an $$\epsilon>0$$ distance to the neighbor. The proof would fail if a point $$a$$ could be a neighboring point for itself.

Yes, an open interval in $$\mathbb{R}$$ is an open set in the topology generated by the standard metric.

For any point $$x$$ in an open interval $$(a,b)$$, let $$p=|x-a|$$ and $$q = |b-x|,$$ then $$x \in (a+\frac{p}{2}, b-\frac{q}{2}).$$ Since $$(a+\frac{p}{2}, b-\frac{q}{2})$$ is entirely contained in $$(a, b)$$, then $$(a,b)$$ is an open set