## 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 $z_2$.

Then $(a_n+b_n)$ is a sequence in $Y$ that converges to $z_1+z_2$. $(\alpha a_n)$ is a sequence in $Y$ that converges to $\alpha z_1$.