Open interval, prove that you can derive new numbers 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.
 A: 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.
A: 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...

A: $$I\in(a,b)\iff a<x<b\iff x-a>0\land b-x>0.$$
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.
$$[x-d,x+d]\subset(a,b).$$
A: 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 $d=1$.
