# One problem about complemented subspace

Question: For every Banach space $X$ and its subspace $Y$, is there a complemented subspace $Z$ in $X$ such that $Y \subset Z \subset X$ and $\operatorname{card}(Y)=\operatorname{card}(Z)$ i.e., $Y$ and $Z$ have the same cardinality?

The answer is yes if $\operatorname{card}(X)=\mathfrak c$ (i.e., Continuum), the proof is easy by taking a hyperplane $H$ such that $Y \subset H \subset X$, then $\operatorname{card}(Y)=\operatorname{card}(H)=\mathfrak c$.

But I don't know whether it holds for general spaces.

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I assume by "subspace" you mean "closed subspace", right? – Theo Sep 12 '12 at 13:31
Yes, I mean "closed", and I has known the answer is no. See the reference : P.~Koszmider, \emph{A space $C(K)$ where all nontrivial complemented subspaces have big densities.} Studia Math. {\bf 168} (2005) 109--127 – Duanxu Dai Sep 13 '12 at 7:00
For those with institutional access: journals.impan.gov.pl/sm/Inf/168-2-2.html @Duanxu: would you be so kind as to post this comment into an answer box below, so that this question can be considered settled? – t.b. Sep 13 '12 at 7:35

## 1 Answer

The above paper says that there is a Banach space (of continuous functions on a totally disconnected compact Hausdorff space) of density κ bigger than the continuum which has no infinite-dimensional complemented subspaces of density continuum or smaller. In particular no separable infinite-dimensional subspace has a complemented superspace of density continuum or smaller. Hence, for the space C(K) constructed by P.~Koszmider, for each separable subspace Y of C(K), there exist no complemented subspace Z containing Y such that card(Z)=card(Y), in fact, such space Z has card lager than c.

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