Closure is linear subspace

Let $X$ be normed linear space, $Y$ a subspace of $X$ . The closure of $Y$ , $\bar{Y}$ , is a linear subspace of $X$ .

Proof:
We use the “sequential” equivalent definition of closure, rather than the one using open balls: $\bar Y$ is the set of all limits of all convergent sequences of points in $Y$ . Let $z_1,z_2\in \bar{Y}$ , $\alpha\in\mathbb{R}$ . There is a sequence $(a_n)$ in $Y$ such that $a_n\to z_1$ . Similarly there is a sequence $(b_n)$ in $Y$ which converges to $z_2$ .