Relationship between $c_0$ and $C(K)$

Question

Suppose that $K$ is an infinite compact metric space. Define $c_0=\{ (x_n)_{n \in \mathbb{N}}| \lim_n\| x_n \|=0 \}$ and $C(K)$ the set of real-valued continuous functions on $K$.

Question: Is $C(K) \subseteq c_0$? In this paper, in the proof of Theorem $5$, there is this sentence:

Since the spaces $C(K)$ are absolute Lipschitz retracts, there is a Lipschitz retraction from $c_0$ onto $C(K)$ which maps $0$ to $0$.

From what I understand from the statement above, it seems that $C(K)$ is a subset of $c_0$. Otherwise there won't be any retraction from $C(K)$ onto $c_0$.

Remark: A Banach space is called absolute Lipschitz retract if for every metric space $Y$ containing $X$, there exists a retraction $r:X \rightarrow X$ such that $r$ is also Lipschitz.

Answer 1

The answer to your question is included in the third line of the proof of the theorem you are interested in!

By a classical result of Aharoni (see Theorem 7.11, p. 176 in 1), we know that there is a 3-Lipschitz-homeomorphism between $C(K)$ and some subset of $c_0$.

This proof can be summarized in four steps as follows:

1. By Sobczyk's theorem, any (linear) copy of $c_0$ in $C(K)$ is 2-complemented. (See my answer in this thread.)

2. By Aharoni's theorem $C(K)$ embeds into $c_0$ by a 3-Lipschitz homeomorphism (this time this map cannot be expected to be linear as there are obstacles concerning the Szlenk index) and we get actually a retraction from $c_0$ onto the range of this embedding.

3. We lift these morphisms to linear maps between the free spaces which give linear embeddings with complemented ranges.

4. We apply the Pełczyński decomposition method to get an isomorphism between free spaces.

Answer 2

That proof starts with the sentence that $c_0$ is $2$-complemented in $C(K)$. And indeed $c_0$ is isometrically embedded into $C(K)$. This holds as $K$ is an infinite compact metric space, so has a convergent sequence (with all different terms) $x_n$ with limit $x$ (unequal to all $x_n$). Call $S = \{x_n: n \in \mathbb{N}\} \cup \{x\}$. Then $C(S)$ embeds into $C(K)$ (this is due to Dugundji, I think) and there is a projection of $C(K)$ onto $C(S)$ (restriction), and $\{f \in C(S): f(x) = 0\}$ is isometric to $c_0$.