Finite Dimensional Subspace of Normed Vector Space is Closed

Theorem
Let $V$ be a normed vector space.

Let $W$ be a finite dimensional subspace of $V$.

Then $W$ is closed.

Proof
Suppose that $\dim W = n$.

Let:


 * $\set {e_1, e_2, \ldots, e_n}$

be a basis for $W$.

that $W$ is not a closed set.

Then there exists a convergent sequence $\sequence {w_k}$ in $W$ such that:


 * $w_k \to w$

where $w \in V \setminus W$.

Note that:


 * $\set {e_1, e_2, \ldots, e_n}$

is linearly independent in $W$, and hence $V$.

Note that since $w \not \in W$, $w$ cannot be written as a linear combination of elements of $\set {e_1, e_2, \ldots, e_n}$.

So:


 * $\set {e_1, e_2, \ldots, e_n, w}$

is linearly independent in $V$.

So consider the subspace:


 * $W^* = \span \set {e_1, e_2, \ldots, e_n, w}$.

Using the sequence $\sequence {w_n}$ from before, write:


 * $w_k = \tuple {w_k^{\paren 1}, w_k^{\paren 2}, \ldots, w_k^{\paren n}, 0} \in W^*$

and:


 * $w = \tuple {0, 0, \ldots, 0, 1} \in W^*$

We necessarily have:


 * $w_k^{\paren j} \to 0$

for each $1 \le j \le n$.

However, we would also have:


 * $0 \to 1$

Clearly this is impossible, so we have derived a contradiction.

So $W$ is necessarily closed.