# Prove that a subset of an infinite dimensional complete space is uncountable

Let $X$ be a Banach space. A subset $S ⊂ X$ is called a Hamel basis of $X$ if $S$ is linearly independent and every element of X is a finite linear combination of elements of $S$.

(i) Prove that if $X$ is infinite dimensional, then $S$ is uncountable. (Use the fact that every finite dimensional linear subspace of $X$ is closed.)

(ii) Give an example of an infinite dimensional normed linear space which has a countable Hamel basis.

My approach for part (i): I'm trying to prove that $S$ is complete, and $S$ has no isolated points, but failed at any of these. I was trying to use the fact that let dim $(S)=k$, then exist a unique representation of L.I. elements $x_1, x_2,...,x_n \in X$ such that for every $x$ in $X$, $x = a_1x_1+...+a_nx_n + y$ where $y$ in $S$. By setting $x=0$, I can show that the space $X$ has dimension $n+k$, which is finite, so the statement is not true for finite case:P. But I don't understand why when $X$ infinite, this statement can be true. Hope someone can help me with this problem.

For part (ii), completely stacked at these kinds of questions:P

(i) Assume that $X$ has a countable Hamel basis $\{x_n\}_{n\in\mathbb N}$, and set $$X_n=\mathrm{span}\,\{x_1,\ldots,x_n\}.$$ Then $X_n$ is closed and nowhere dense. (If it contained an open set $U$, then would contain a open neighborhood of zero, as we can suitably translate $U$ to contain zero, and thus it would contain the whole $X$, as it closed to scalar multiplication.) Then $$X=\bigcup_{n\in\mathbb N}X_n$$ which contradicts Baire's Theorem - A complete space can not be written as a countable union of nowhere dense subsets.
(ii) A simple example is the set of eventually vanishing sequences, with the supremum norm, i.e., $\{a_n\}_{n\in\mathbb N}$, for which there exists an $n_0$, such that $a_n=$, for $n\ge n_0$. Basis are the sequences $$\{\delta_{jn}\}_{n\in\mathbb N}, \quad j\in\mathbb N,$$ as $$\{a_n\}_{n\in\mathbb N}=\sum_{j\in\mathbb N}a_j\{\delta_{jn}\}_{n\in\mathbb N},$$ and clearly the above sum is finite.
• How do you know that $X_n$ is closed? And how can we always translate $U$ to contain zero? And why containing zero implies containing the whole X? – user177196 Oct 2 '14 at 2:22
• In the OP it is said that 'finite dimensional subspaces are closed'. For the second question, if $U\subset X_n$ and $x\in U$, then $-x\in X$, and thus $0\in -x+U\subset X_n$. – Yiorgos S. Smyrlis Oct 2 '14 at 6:19