## § Subspaces need not have complement

- Clearly, one can have open subspaces that cannot be complemented. For example, the subspace of polynomials in $C[0, 1]$ is dense, and thus has no complement, as a complemented subspace must be closed.

#### § Closed subspace need not have complement

- Apparently, in $l^\infty$, the subspace $c_0$ of sequences that converge to zero does not have a complement.
- Proof is given in a paper "projecting $m$ onto $c_0$"

#### § Lemma: countable set $I$ has uncountable family of countable subsets $S$ which are almost disjoint

- Let $I$ be countable.
- We must prove that (1) there exists a $S \subset 2^I$ that is uncountable, such that (2) every set $K \in S$ is countable, and (3) every two sets $K, L \in S$ have finite intersection $K \cap L$.
- Proof: let $I$ be rationals in $(0, 1)$. For each irrational $r \in (0, 1)$ create a set $S_r$ to be the set of sequences of rationals that converge to $r$.
- TODO: why is each $S_r$ countable?

#### § Proof of theorem

- Suppose that there is a continuous projection of $l^\infty$ into $c_0$,
- Then we must have $l^\infty = c_0 \oplus R$ for some closed subspace $R$.
- Since $l^\infty / c_0$ is isomorphic to $R$.
- TODO