Quotient of $\ell^\infty$ has infinite dimension

Question

Let $\ell^\infty$ be the bounded sequence space over the complex numbers and let $c_0$ the subspace of all sequences converging to $0$.

I am attempting to show that $\ell^\infty/c_0$ has infinite dimension.

I have looked at several different ways to show this but I think perhaps the easiest is to show that there exists an infinite linearly independent set in the quotient. It is easy to find an infinite countable set in $\ell^\infty$ (just take the sequences of the form $(0,\dots,0,1,0,\dots)$, but of course this set is identified in the quotient). I also was thinking of doing a proof by contradiction, and show that no matter what finite linearly independent set in $\ell^\infty/c_0$ you give, we can construct a sequence not in the span. But the precise way to construct such a sequence is not evident to me at this point (some sort of diagonal argument, perhaps).

Will a cardinality argument be necessary?

Answer 1

For each $n$ let $v_n$ be the sequence of zeroes and ones which has ones exactly in the positions which are divisible by $n$. Show the set $\{v_n:n\geq1\}$ is linearly independent.

Answer 2

It is possible to prove that the dimension of $\ell^{\infty}/c_0$ is $|\mathbb{R}|$ under Axiom of Choice. We find a subset of $\ell^{\infty}/c_0$ whose cardinality is $|\mathbb{R}|$, and it is linearly independent (any finite subset is linearly independent). Then the dimension is at least $|\mathbb{R}|$. Moreover, as $|\ell^{\infty}/c_0|=|\mathbb{R}|$, the dimension is at most $|\mathbb{R}|$. Hence the dimension is $|\mathbb{R}|$.

Consider a basis $B\subseteq (0,1]$ of the vector space $\mathbb{R}$ over $\mathbb{Q}$. We may require that $1\in B$. Then the following subset $S\subseteq \ell^{\infty}/c_0$ has cardinality $|\mathbb{R}|$ and is linearly independent. $$ S=\{(e^{2\pi i b n})_{n\in\mathbb{N}} | \ b\in B-\{1\}\}. $$ To see this, take any finite subset $\{b_1,\ldots, b_k\}$ in $B-\{1\}$. If $c_1,\ldots, c_k\in \mathbb{C}$ satisfy $$ c_1e^{2\pi i b_1 n}+\cdots+c_ke^{2\pi i b_k n} = 0\in \ell^{\infty}/c_0, $$ then we must have $$ \lim_{n\rightarrow\infty}(c_1e^{2\pi i b_1 n}+\cdots+c_ke^{2\pi i b_k n}) = 0. \ \ \ (*) $$ We apply Kronecker's Approximation Theorem in the following form:

If $1,b_1,\ldots, b_k$ are linearly independent over $\mathbb{Q}$, then the sequence $( (nb_1), \ldots, (nb_k) )_{n\in\mathbb{N}}$ of $k$-tuples of fractional parts of $nb_i$, is dense in $[0,1]^k$.

For each $i\leq k$, we can find a subsequence $n_j$, $j\geq 2$ such that $$((n_j b_1), \ldots, (n_j b_k))\in R_{i,j}$$ where $R_{i,j}$ is a product of closed intervals $[1/4, 1/4+1/j]$ on the $i$-th component, $[0,1/j]$ on all other $k-1$ components. Then $$ \lim_{j\rightarrow\infty}\text{Im} ( c_1e^{2\pi i b_1 n_j}+\cdots+c_ke^{2\pi i b_k n_j}) = \text{Im}(c_i). $$ By (*), the above limit must be $0$. Similar argument shows that $\text{Re}(c_i)=0$. Hence all $c_i$'s are zero.

Alternatively, the construction of linearly independent set of cardinality $|\mathbb{R}|$ can be done through an almost disjoint family. See Can a countable set contain uncountably many infinite subsets such that the intersection of any two such distinct subsets is finite? for example, contains many methods of constructing a set of infinite subsets of $\mathbb{N}$, $\{A_r\}_{r\in \mathbb{R}}$, satisfying $A_r\cap A_s$ is finite if $r\neq s$.

Then the indicators $(1_{A_r}(n))$ form a linearly independent set in $\ell^{\infty}/c_0$. To see this, for any finite subset of $\{ (1_{A_r}(n)) | r\in\mathbb{R}\}$, there is an index $K$ such that for $A_r$'s in this finite set, $A_r\cap [K,\infty)$ become disjoint.