How to check if two inequalities express the same thing? Suppose:
$$
\delta_1>0 \\
\delta_2>0 \\
\delta=\text{min}(\delta_1,\delta_2)
$$
If we know that the following is true:
$$
a-\delta_2<a-\delta<x<a \hspace{2cm} \text{ or } \hspace{2cm}  a<x<a+\delta<a+\delta_1
$$
Can we conclude that the followng is true too?
$$
a-\delta_2\leq a-\delta < x < a+\delta \leq a+\delta_1
$$
I don't know how to formally show this.
Thank you in advance for any help provided.
 A: Yes. if you chose $|x-a|<\delta $.
maybe i dont get this question but  your inequality is true for this specific $a$.
A: Ok, let us say you know
$a-\delta_2 < a-\delta <x <a $
Then,
obviously
$a-\delta+2 < a -\delta <x$
Now, to complete it, we keep in mind the transitive property of comparison.
So
\begin{align}
a-\delta+2 < a -\delta &<x\\
&<a\\
&<a+\delta &\text{since $\delta>0$}\\
\end{align}
Now, $\delta = \min(\delta_1, \delta_2) \leq \delta_1$.
So
\begin{align}
a+\delta \leq a+\delta_1
\end{align}
If you know the second case, just do it in reverse.
A: Note that $0<\delta\le \delta_{i}$ for $i=1,2$. So $0>\delta\ge -\delta_{i}$ for $i=1,2$. Thus 
$a-\delta_{2}\le a-\delta$ and $a+\delta\le a+\delta_{1}$. This holds in general by the definitions.
If we know $a-\delta<x<a$ then $a-\delta_{2}\le a-\delta<x<x+\delta<a+\delta\le a+\delta_{1}$.
If we know $a<x<a+\delta$ then $a-\delta_{2}\le a-\delta<a<x<a+\delta\le a+\delta_{1}$.
A: Yes. Note that $0<\delta\leq\delta_1$ and $0<\delta\leq\delta_2$.
Case 1: Suppose that $a-\delta_2<a-\delta<x<a$. Note that:


*

*$a-\delta_2<a-\delta \implies a-\delta_2\leq a-\delta $

*$x<a$ and $0<\delta \implies x<a+\delta$

*$\delta\leq\delta_1 \implies a+\delta\leq a+\delta_1$


Hence, we obtain $a-\delta_2\leq a-\delta < x < a+\delta \leq a+\delta_1$.

Case 2: Suppose that $a<x<a+\delta<a+\delta_1$. Note that:


*

*$a+\delta<a+\delta_1 \implies a+\delta\leq a+\delta_1 $

*$a<x$ and $0<\delta \implies a<x+\delta \implies a-\delta<x$

*$\delta\leq\delta_2 \implies -\delta_2\leq-\delta \implies a-\delta_2\leq a-\delta$


Hence, we obtain $a-\delta_2\leq a-\delta < x < a+\delta \leq a+\delta_1$.
