Show that if X is a normed linear space, then any finite-dimensional subspace M of X must be closed.

Question

It suffices to show that any proper subspace M of X is closed, since if M is not proper the result is trivial.

I am unsure how to approach this proof. Contradiction seems a little messy, as supposing M is not closed implies M is i.) open or ii.) neither closed nor open. Any help would be much appreciated.

Answer 1

Hint: The easiest approach here is sequence continuity. That is, select an arbitrary Cauchy sequence $\{x_n\}_{n=1}^\infty \subset M$ and show that its limit lies in $M$.

Now, what does "finite dimensional" mean, and what does that mean for our sequence $\{x_n\}$?

Answer 2

Let $\dim M = n$. Then, identifying $M$ with $\mathbf{R}^n$, the norm of $X$ restricted to $M$ is equivalent to any of the standard norms on $\mathbf{R}^{n}$. (By equivalent norms I mean ones such that each is bounded by a constant multiple of the other.) Thus $M$ is a complete metric space.

It follows from this that $M$ is a closed subset of $X$. (For let $a_n$ be any sequence of elements in $M$ that converges to $a \in X$. It is a Cauchy sequence, and therefore has a limit in $M$. Thus $a \in M$.)