Let $C_0(Q)=\{f:\Bbb{N}→\Bbb{Q}:(\forall ε>0)(\exists N∈\Bbb{N})(\forall m>N)(|f(m)|<ε)\}$. If $\{a_n\},\{b_n\}∈C_0(Q)$ then $\{a_n+b_n\}∈C_0(Q)$. Suppose that we have $C_{0}(Q)=\{f:\mathbb{N}\rightarrow\mathbb{Q}:(\forall\epsilon>0)(\exists N\in\mathbb{N})(\forall m>N)(|f(m)|<\epsilon)\}$ and we want to show that if $\{a_{n}\}$,$\{b_{n}\}\in C_{0}(Q)$ it holds that $\{a_{n}+b_{n}\}\in C_{0}(Q)$.
I have looked at the problem, and I think it can be solved using the Triangle Inequality.
We have that in general $|a_{n}+b_{n}|\leq|a_{n}|+|b_{n}|$. Then, given the first proposition, it must be that $|a_{n}|<\epsilon,$ and it must be that $|b_{n}|<\epsilon,$ and so, it must be that $|a_{n}|+|b_{n}|<2\epsilon.$ By the triangle inequality, $|a_{n}+b_{n}|\leq|a_{n}|+|b_{n}|,$ and we have that $|a_{n}+b_{n}|<2\epsilon$. Now, here is the main question: this manipulation implies that $\frac{1}{2}|a_{n}+b_{n}|<\epsilon$ but not entirely $|a_{n}+b_{n}|<\epsilon,$ and I have yet to think of a manipulation from the former to the latter.
This is the most direct way I can think of proving this. Any thoughts?
 A: Consider two sequences $\{a_n\}$ and $\{b_n\}$ in $C_0(Q).$ Given any $\varepsilon > 0,$ there exists an integer $N(a)$ such that for all integers $m \geq N(a) + 1,$ we have that $|a_m| < \frac \varepsilon 2.$ Likewise, there exists an integer $N(b)$ such that for all integers $m \geq N(b) + 1,$ we have that $|b_m| < \frac \varepsilon 2.$ Consider the integer $N = \max \{N(a), N(b)\}.$ Observe that for all integers $m \geq N + 1,$ we have that $$|a_m + b_m| \leq |a_m| + |b_m| < \frac \varepsilon 2 + \frac \varepsilon 2 = \varepsilon$$ by the Triangle Inequality. We conclude that $\{a_n + b_n \}$ is an element of $C_0(Q),$ as desired.
A: What you try to show about the triangle inequality is not true. Namely, if $\vert a_n\vert, \vert b_n\vert < \varepsilon$, then we do in general not have $\vert a_n + b_n \vert < \varepsilon$. A counterexample would be $a_n = \frac{2\varepsilon}{3} = b_n$. Then we have $\vert a_n \vert = \vert b_n \vert = \frac{2\varepsilon}{3}< \varepsilon$. However, we have $\vert a_n + b_n \vert = \frac{4 \varepsilon}{3} > \varepsilon$.
So this is not the way to show the statement you want. However, you can do it differently.
Hint: If you have $\vert a_n \vert < \varepsilon_1$ and $\vert b_n\vert < \varepsilon_2$, then we have
$$ \vert a_n + b_n \vert \leq \vert a_n \vert + \vert b_n \vert < \varepsilon_1 + \varepsilon_2. $$
So if you want $\vert a_n + b_n \vert < \varepsilon$ how could you choose $\varepsilon_1, \varepsilon_2$ such that things work out?
