# On the denseness of sequence space $\ell^1$ in $\ell^2$

$$\ell^1$$ denotes absolutely summable sequences, and $$\ell^2$$ square summable sequences. It is well-known that $$\ell^1$$ is dense in $$\ell^2$$, with the standard argument that for any sequence $$(a_1, a_2, \ldots) \in \ell^2$$ one can take the finite sequence $$(a_1, \ldots, a_N, 0, 0, \ldots) \in \ell^1$$ and approximate the original sequence arbitrarily well as $$N \to \infty$$.

However, consider any $$(a_n) \in \ell^2$$, where I will assume that $$a_n>0$$ for simplicity, take any sequence $$(b_n) \in \ell^2 \setminus \ell^1$$ such as $$b_n = \frac{1}{n}$$, and define the sequence $$(c_n)$$ by $$c_n = a_n + \epsilon b_n$$. Now, $$(c_n) \in \ell^2$$ by Cauchy-Schwarz, but clearly $$(c_n) \notin \ell^1$$ because we're adding a positive sequence and a sequence with a divergent series. But since $$(c_n) \to (a_n)$$ as $$\epsilon \to 0$$, doesn't this mean that $$\ell^2 \setminus \ell^1$$ is dense in $$\ell^2$$? How to reconcile this with the fact that $$\ell^1$$ is dense in $$\ell^2$$?

What is wrong if a set and its complement are both dense. Isn't the set of all rational numbers and its complement both dense in $$\mathbb R$$?