Constructed the quotient space in the space of convergent sequences 
Considered the space $\mathcal{c}$ of convergent sequences and let $Y$ be the closed subset of $\mathcal{c}$ compose by the constant sequences, constructed $\mathcal{c} / Y$

Is these correct?
Let $x,y \in \mathcal{c}$ then $x-y \in \mathcal {c}$, those $x_n - y_n=\alpha$, $\forall n \in \mathbb{N}$ and therefore $\lim_{n \to \infty} x_n= \lim_{n \to \infty} y_n + \alpha$ and I conclude that
$$x+ Y=\{ y \in \mathcal{c} : \exists \alpha \in \mathbb{R} \lim_{n \to \infty} x_n= \lim_{n \to \infty} y_n + \alpha \}$$
I think to the quotient space is isomorphic to $\mathbb{R}$
Any hint or suggestion I will very grateful.

Let $c_0$ denote the space of sequences convergent to $0.$ Consider the mapping $T:c\to c_0$ given by
$$T\{x_n\}=\{x_n-x_0\}, \quad x_0=\lim_nx_n$$ The mapping is equal identity on $c_0.$
Observe that  $\ker T=Y.$ Therefore the quotient space $c/Y=c/\ker T$ is isomorphic to $c_0=T(c).$
The norm on the quotient space $c/Y$ does not coincide with the norm on $c_0.$ The norm on the quotient space is by definition equal $$
\|[x]\|=\inf \{\|x+y\|\,:\, y\in Y\}$$ It can be shown that
$$\|[x]\|={1\over 2}\sup_{n,m}|x_n-x_m|$$ The norm inherited from $c_0$ is equal
$$\|[x]\|_1 =\|Tx\|=\sup_n|x_n-x_0|$$
It can be shown that these norms are equivalent, namely
$$\|[x]\|\le \|[x]\|_1,\quad \|[x]\|_1\le 2\,\|[x]\|$$
