Let $a,b,c,\epsilon$ be positive real numbers. This $\epsilon$ is that you can see in $(\epsilon, \delta)$-definition of limit, thus $\epsilon$ is arbitrary though $a,b,c$ is constant.
Now this equation is given.
$a < b < c + \epsilon \ \ \ (1)$
In this case, can I say the below equation is right?
$a < c\ \ \ (2)$
I think it is true. Let me explain my solution.
(i) When $a<c$
Trivial.
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(ii) When $a = c$
$(\mathrm{eq.}1) \Rightarrow c < b < c + \epsilon \ \ \ (3)$
Let $d$ be a positive real number and rewrite $b$ as below.
$b = c + d\ \ \ (4)$
Then (eq.3) becomes
$c < c + d < c + \epsilon \ \ \ (5)$
But, since you can lessen $\epsilon$ arbitrarily, if you decide $\epsilon = d - 0$, the relational equation between $a,b,c,\epsilon$ becomes
$c < c + \epsilon < c + d\ \ \ (6)$
$\Leftrightarrow a < c + \epsilon < b\ \ \ (7)$
This contradict (eq.1). Thus $a \neq c$.
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(iii) When $a > c$
(eq.1) $\Rightarrow c < a < b < c + \epsilon$
This case isn't essentially different from the second case of $a = c$.
$ $
Thus $a < c$.
Could anyone tell me whether this solution is right? Thank you in advance.
Clearly taking limit when $\epsilon\to 0$ we get $b\le c$ hence $a\lt c$.
It is good as an exercise for you using trichotomy (by the order relation in real numbers) but it is not necessary.