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Prove If $I$ is an open interval, and if $x\in I$ , then there is some $d > 0$ such that $[x-d; x+d ] \in I$

I, for the life of me can't figure this one out. Despite being preceded by an easy exercise, and being seemingly intuitive, I just can't show how you can continuously derive new numbers without thinking of concrete examples. This proof essentially asks to show that if an interval is open, you can choose random integers which approach the endpoints, but are distinct from everything else, ie: Between .99 and 1, is .999, and between .999 and 1 is .999 and so on.

Roberto Singer
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    One way to derive new numbers: $a<(a+b)/2<b$. – Cheerful Parsnip Jul 05 '19 at 19:29
  • But that doesn't prove the theorem – Roberto Singer Jul 05 '19 at 19:49
  • I was hoping you would find it to be a useful hint. You can in fact prove the theorem using this observation. – Cheerful Parsnip Jul 05 '19 at 22:22
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    Between $x$ and $b$ is $x_1 = \frac {x+b}2$. And between $a$ and $x$ is $x_0 = \frac {a + x} 2$. So $[x_0, x_1]$ is a closed interval that works. (Which is what I think Cheerful Parsnip's point was) But $[x_0, x_1]$ isn't centered on $x$. But just let $d$ be any number so that $d < x-a$ and $d < b-x$ then you will have $x - d > x-(x-a)=a$ and $x+d < x+ (b-x)=b$ and so $[x-d,x+d]\subset (a,b)$. But you must have $d < x-a$ and $d < x+b$. Well just take $k*\min(x-a,x+b)$ where $k < 1$ so $k = .9$ or $k = \frac 12$ or ... whatever. – fleablood Jul 05 '19 at 22:42

4 Answers4

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SO the interval is $(a,b)$ with endpoints $a$ and $b$. You have a point $x \in (a,b)$ so $a < x < b$

You need show that you can find a $d$ so that $x + d < b$ and $a < x-d < x$.

How?

Well you need $d < b-x$ and $d < x-a$. So let $d < \min(b-x, x-a)$.

How?

Let $d = \frac {\min(b-x,x-a)}2$.

That's it.

fleablood
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Let $I = (a,b), x \in I$. Consider $2d = \min(b-x,x-a)$. Then $[x-d,x+d] \subset I$:

  • if $d = \frac{b-x}{2} < \frac{x-a}{2}$, then $a < x - \frac{x-a}{2}< \frac{x-b}{2}= x-d < x+d=\frac{x+b}{2} < b$
  • similarly, if $d = \frac{a-x}{2} < \frac{b-x}{2}$, then...
dcolazin
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$$I\in(a,b)\iff a<x<b\iff x-a>0\land b-x>0.$$

If you choose $2d=\min(x-a,b-x)$, which is positive, you have

$$x+2d=\min(2x-a,b)\le b$$

and

$$x-2d=\max(a,2x-b)\ge a.$$

And this implies

$$x+d\le b-d<b,\\x-d\ge a+d>a$$

i.e.

$$[x-d,x+d]\subset(a,b).$$

  • I love this answer, thank you so much! I never thought to use just the minimum value function, I feel like I should've only adhered to strict algebra and simple definitions. This one's a lot more "free". – Roberto Singer Jul 05 '19 at 21:08
  • @RobertoSinger: the min is required because the available room is asymmetric. –  Jul 06 '19 at 08:05
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Suppose that, for every integer $n>0$, the interval $[x-1/n,x+1/n]$ is not contained in $I$. Then there exists $y_n$ such that $y_n\in[x-1/n,x+1/n]$, $y_n\notin I$.

Prove that the sequence $(y_n)$ converges to $x$. Since $I$ is open, $\mathbb{R}\setminus I$ is closed, so $x\in\mathbb{R}\setminus I$.

The other answers are a bit misleading. An open interval could be of the form $(-\infty,b)$ or $(a,\infty)$. However, in both cases you can reduce to an interval of the form $(a,b)$. Take $a=x-1$ for the case $I=(-\infty,b)$ and $b=x+1$ for the case $I=(a,\infty)$. In both cases $x\in(a,b)\subseteq I$. Thus we can assume, without loss of generality, that $I$ is of the form $(a,b)$.

Now consider $d=(x-a)/2$. If $[x-d,x+d]\subseteq (a,b)$, you're done. Otherwise, $x+d\ge b$, that is, $d\ge b-x > (b-x)/2=d'$.

Then $[x-d',x+d']\subseteq(a,b)$, because $a<x-d<x-d'$ and $x+d'<b$.

If also $\mathbb{R}$ is considered to be an open interval, just take $d=1$.

egreg
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