Subspaces isomorphic to $C[0, \omega_1]$

Question

Let $\omega_1$ be smallest uncountable ordinal. I am trying to understand the possible "large" subspaces of $C[0,\omega_1]$, namely those which are isomorphic to the whole space. Therefore I have the following question:

Does every subspace of $C[0,\omega_1]$ isomorphic to $C[0,\omega_1]$ contain a complemented copy isomorphic to itself? The only (complemented) examples that I can "construct by hand", excluding the finite-codimensional ones, are of the form

$\mbox{cl lin}(\mathbf{1}_{[0,\gamma_{\sigma}]}\colon \sigma\leq \omega_1)$

where $(\gamma_\sigma)_{\sigma<\omega_1}$ is increasing long sequence of limit ordinals and $\sigma_{\omega_1} = \omega_1$ (note that the family $(\{\mathbf{1}_{[0,\alpha]}\colon \alpha \leq \omega_1\})$ forms the long Schauder basis for $C[0,\omega_1]$).

Thank you, T.

Answer 1

Have you searched the literature? There are a large number of papers about uncomplemented and complemented embeddings of $C(K)$ into $C(K)$. In particular, you should look at papers of Bessaga and Pelczynski from the 1960s (especially their fourth one on spaces of continuous functions), papers of Dan Amir from the late 1960s and 1970s, and Alspach and Benyamini from the 1970s. For an overview of part of the material, read Rosenthal's article in volume 2 of the Handbook of the Geometry of Banach Spaces. In particular, in Section 3C of that article you will find that there is an isomorphic copy of $C[0,\omega^\omega]$ in itself that is not complemented; from that it is very easy to prove that there is an isomorphic copy of $C[0,\omega_1]$ in itself that is not complemented.

Answer 2

Yes, it can deduced from Lemma 1.2 combined with Proposition 2 of

D. Alspach and Y. Benyamini, Primariness of spaces of continuous functions on ordinals. Israel J. Math. 27 (1977), 64–92.

^{Another proof can be found here (Cor. 1.12).}