Proving an arbitrary set is open

Question

I'm currently reading a section of Stephen Abbott's Understanding Analysis and I'm struggling to understand the following concept which I will explain below

Firstly, recall the following definitions:

$\ $1) Given $b$ $\in\mathbb{R}$ and $\epsilon>0$ the $\epsilon$-neighbourhood of $b$ is the set:$ \ $

$$V_{\epsilon}(b)=\{x\in\mathbb{R}:|x-b|<\epsilon\}$$

$\ $In other words, $V_{\epsilon}(b)$ is the open interval $(b-\epsilon,b+\epsilon)$ centered at $b$ with radius $\epsilon.$

$\ $2) A set $O\subseteq\mathbb{R}$ is open if for all points $b\in O$ there exists an $\epsilon$-neighbourhood $V_{\epsilon}(b)\subseteq O.$

Just so there is no ambiguity here (which I'm sure there won't be given the definition), $\subseteq$ is referring to a proper subset.

The author gives the following example of an open interval:

$$(c,d)=\{x\in\mathbb{R}:c<x<d\}$$

With the following explanation:

"To see that$ \ (c,d)\ $is open in the sense just defined [above], let $ \ x\in(c,d) \ $be arbitrary. If we take $ \ \epsilon=min\{x-c,d-x\}, \ $then it follows that $V_{\epsilon}(x)\subseteq (c,d).$ It is important to see where the argument breaks down if the interval includes either one of its endpoints."

My question:

I'm assuming the choice of $ \ \epsilon=min\{x-c,d-x\} \ $was due to the fact that we know:

$$x>c, x<d$$

Therefore $ \ x-c \ $ will be the distance between $ \ x $ and $ c \ $(likewise for $ \ d-x $ being the distance between $ \ x $ and $ d \ $), but why is it necessary to choose the minimum from the set of these distances for $ \epsilon? \ $

Answer 1

If $c<d$ and $x=(c+d)/2$ and $\epsilon= \min(x-c,d-x)$ then $\epsilon =(d-c)/2=x-c=d-x,$ and $(-\epsilon+x,\epsilon+x)=(c,d)$ which is NOT a proper subset of $(c,d).$

The usual convention is that $\subset , \subseteq,$ and $\subseteqq$ are synonymous and that $A\subset A.$ And that $A\subsetneq B$ or $A\subsetneqq B$ is used to denote that $A$ is a proper subset of $B.$ The author is following this convention, as he writes $V_{\epsilon}(x)\subseteq (c,d),$ which would be a mistake if $x= (c+d)/2$ and if $\subseteq$ meant proper subset.

When $c<x<d, $ in order that $(-e+x,e+x)\subset (c,d)$ it is necessary that $(-e+x\geq c\land e+x\leq d),$ that is, $(e\leq x-c\land e\leq d-x),$ that is, $e\leq \min (x-c,d-x).$

If $-e+x<c$ then we have $(-e+x,e+x)\supset (-e+x,x)\supset (-e+x,c)\ne \emptyset$ and we also have $(-e+x,c)\cap (c,d)=\emptyset,$ implying that $(-e+x,e+x)$ contains points not belonging to $(c,d).$ Similarly we cannot have $e+x>d.$

Answer 2

To comply with the definition: "In other words, Vϵ(b) is the open interval (b−ϵ,b+ϵ) centered at b with radius ϵ."

Answer 3

Yes, you are correct, due to transitivity, $c < d$. In addition, we need to choose the minimum distance because we want $V_\epsilon(x)\subseteq(c,d)$. If $c$ is $3$, $d$ is $6$ and let $x$ be $5$, if we don't choose the minimum of the two distances and instead choose the distance $x-c=2$, then the open interval $(3,7)$ obviously fails the criteria. I believe it is trivial to see that choosing the minimum distance works. Hence, choosing the minimum works and not choosing the minimum doesn't and that's why we do choose the minimum.

Answer 4

Let's see what happens if you take $\epsilon=max\{x-c,x-d\}$.

Suppose that $V_\epsilon$ is as you defined it to be, and consider any element x in the open interval $(a, b)$.
Now what happens if we look at the open ball $V_\epsilon$ around $x$? Well you'll notice that unless the distance between $x$ and $a$ is the same as the distance between $x$ and $b$, then you'll have elements in $V_\epsilon$ that aren't in $(a,b)$. So int that case $V_\epsilon\nsubseteq(a,b)$.

This fact is very easy to see if you draw a picture!