Prove that the intervals $(a,\infty)$ and $(-\infty,b)$ are open 
Show that the intervals $(a,\infty)$ and $(-\infty,b)$ are open in
  $\mathbb{R}$.

I'm having trouble choosing a radius for every $x \in (a,\infty),\ y\in (-\infty,b)$ such that $B_1(x,r_1)\subset (a,\infty)$ and $B_2(y,r_2) \subset (-\infty,b)$ where both $B_1$ and $B_2$ are open balls. 
 A: For every $x\in (a,\infty)$, for any $y >0$ we have $x+y \in (a,\infty)$.
Now, $a<x$, so, exists a $\delta >0$ such that $a<x+\delta\Rightarrow x+\delta \in (a,\infty)$. So, use that $\delta$, and you have the ball centered at $x$ with radius $\delta$ contained in $(a,\infty)$
A: Let's look at $(a, \infty)$.  We want to show that for every $x \in (a,\infty)$ we can find an $r$ (based on $x$) so that $B(x,r) \subset (a, \infty)$.
Note two things:
1)  $B(x,r) = (x-r, x+r)$
2) $(x-r, x+r)\subset (a,\infty) \iff a \le x-r$ and $x +r \le \infty$.
So we want $a \le x-r$ or  $r \le x -a$.
And that's it.  Any $r$ so that $0 < r \le x-a$ will do.
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Proof of 1 (if it isn't imediately obvious)
$B(x,r) = \{y\in \mathbb R: |x - y| < r\}$
$= \{y\in \mathbb R: -r < y - x < r\}$
$= \{y \in \mathbb R: x-r < y < x + r\}$
$= (x-r, x+r)$.
Restatement of 2: $(m,n) \subset (u,v) \iff u \le m$ and $v \ge n$.
Proof of 2 (if it's not immediately obvious)
$(m,n) =\{x \in \mathbb R| m < x < n\}$ so 
$(m,n)\subset (u,v)$ means for all $x| m < x < n$ then $u < x < v$
Assuming neither $(m,n)$ nor $(u,v)$ are empty then there exists an $x_0$ so that $m < x_0 < y$ and $u < x_0 < v$.  If $m < u$ then there is a $y$ so that $m < y < u < x_0 < n$ and $y \in (m,n)$ but $y \not \in (u,v)$ which contradicts $(m,n) \subset (u,v)$.  So $m\ge n$.  Similarly $n \le v$.
A: Another way to see that $(a,\infty)$ and $(-\infty,b)$ are open, is to note that they can be written as a union of open sets, in fact
$$(a,\infty)=\bigcup_{n\in\mathbb{N}}(a,a+n)$$
and
$$(-\infty,b)=\bigcup_{n\in\mathbb{N}}(b-n,b) $$
A: A variant:


*

*For $x\in (a,+\infty)$, consider the interval centred at $x$: $B(x,x-a)=(a,2x-a)$.

*For $x\in (-\infty,b)$, the interval $B(x, b-x)=(2b-x,b)$.
Note that both intervals have bounds $a$ (resp. $b$) and the symmetric of $a$ (resp. $b$) w.r.t. $x$.
