# For which infinite dimensional real normed linear spaces $X$ , can we say that every infinite dimensional subspace of it is closed in $X$ ?

For which infinite dimensional real normed linear spaces $X$ , can we say that every infinite dimensional subspace of it is closed in $X$ ? Or , does every infinite dimensional normed linear space has an infinite dimensional proper subspace which is not closed ?

Every infinite-dimensional normed space has a non-closed subspace.

Let $X$ be an infinite-dimensional normed space, let $a$ be a nonzero vector. Assume by induction that we have found vectors $x_1, x_2, \dots, x_{n-1}$ for which $|x_i - a| < 1/i$ and $a \not\in V_{n-1} = \Sigma_{i=1}^{n-1} \mathbf{R}x_i$. We will extend this sequence by finding an $n$th term $x_n$. This will be sufficient, since the space $\bigcup_{i=1}^{+\infty} V_i$ will not contain $a$, but will have $a$ in its closure.

Clearly, because $X$ is infinite-dimensional, the ball $B$ of radius $1/n$ around $a$ cannot be contained in any finite-dimensional subspace. It is therefore possible to pick some $x_n \in B$ that does not belong to $V_{n-1} \oplus \mathbf{R}a$. This $x_n$ works.

Hard to beat the simple solution of @Keith:

We'll give a proof for $X$ infinite dimensional Banach space (extra condition).

First, show that there exists a sequence $x_n$ in $X$ such that $||x_n|| =1$ and $d(x_n, \langle x_1, \ldots x_{n-1}\rangle ) \ge \frac{1}{2}$. One constructs the sequence inductively. Once $x_1$, $\ldots x_{n-1}$ are obtained, take $y \not \in \langle x_1, \ldots x_{n-1}\rangle$ , $x = y - \sum_{i=1}^{n-1} c_i x_i$ so that $||y|| \le 2 d (x, \langle x_1, \ldots x_{n-1}\rangle )$, and $x_n = \frac{x}{||x||}$.

Now that we have the sequence $(x_n)$, consider sums $s=\sum \alpha_n x_n$, with $\sum |\alpha_n| < \infty$. If the sequence $\alpha_n$ decreases fast enough all these sums will not be in the space $\langle x_n \rangle_{n\ge 0}$. Indeed: $$d(s,\langle x_1, \ldots x_{n-1}\rangle) = d(\alpha_n x_n + \cdots, \langle x_1, \ldots x_{n-1}\rangle)\ge \\ \ge \alpha_n d ( x_n, \langle x_1, \ldots x_{n-1}\rangle) - || \alpha_{n+1} x_{n+1} + \cdots ||\ge \frac{\alpha_n}{2} - \sum_{k \ge n+1} \alpha_k$$

For instance, we can take $\alpha_n = q^n$ with $|q| < \frac{1}{3}$.

In fact, this shows that any countable dimensional subspace of a Banach space is not closed.