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$M_i \cap M_j = \{0\}$ and $M_i + M_j = Y+Z$ dense in $X$.

This corresponds to the pair $M_i M_j$. It is easy to see that $M_i + M_j = Y+Z$ which is dense in $X$.

Therefore Remark 3 yields that independently to what is the initial quasi-complemented pair $Y,Z$( complemented or not) the $(M_i)_i$ resulting from Remark 3, since $\operatorname{dist} (B_{M_i}, B_{M_j}) = 0$, are pairwise quasi-complemented but not complemented. Hence Remark 3 answers completely the posed question.

$M_i \cap M_j = \{0\}$ and $M_i + M_j$ dense in $X$.

This corresponds to the pair $M_i M_j$. It is easy to see that $ (y_k)_k \cup (z_k)_k \subset M_i + M_j $ hence $M_i + M_j$ is dense in $X$.

Therefore Remark 3 yields that independently to what is the initial quasi-complemented pair $Y,Z$( complemented or not) the $(M_i)_i$ resulting from Remark 3, since $\operatorname{dist} (B_{M_i}, B_{M_j}) = 0$, are pairwise quasi-complemented but not complemented. Hence Remark 3 answers completely the posed question for separable and WCG spaces.

Therefore Remark 3 yields that independently to what is the initial quasi-complemented pair $Y,Z$( complemented or not) the $(M_i)_i$ resulting from Remark 3, since $\operatorname{dist} (B_{M_i}, B_{M_j}) = 0$, are pairwise quasi-complemented but not complemented.

Therefore Remark 3 yields that independently to what is the initial quasi-complemented pair $Y,Z$( complemented or not) the $(M_i)_i$ resulting from Remark 3, since $\operatorname{dist} (B_{M_i}, B_{M_j}) = 0$, are pairwise quasi-complemented but not complemented. Hence Remark 3 answers completely the posed question.

It is well known that every subspace $Y$ of a separable Banach space $X$ is quasi-complemented.We denote its quasi-complement by $Z$. This is a classical result due to F. J. Murray and G. Mackey.The proof is based on the existence of $\{(y_n, f_n)_{n} \cup (z_n, g_n)_n \}$ which is a biorthogonal system of $X$ with $(y_n)_n, (z_n)_n $ generate the subspaces $Y,Z$ respectively and $(\cap_n Kerf_n) \cap(\cap_n Kerg_n) = \{ 0 \}, $

$<\{ b_jf_k-a_jg_k \}_k \cup \{ b_if_k-a_ig_k \}_k > = <\{f_k\}_k \cup \{g_k\}_k > $

$(\cap_k Ker b_jf_k- a_jg_k) \cap(\cap_k Ker b_if_k-a_ig_k) = \{ 0 \}$

Indeed let $X$ be a WCG space. From Amir Lindenstrauss theorem there exists a family $(P_{\alpha})_{\alpha\in A}$ mutually orthogonal bounded projections such that for each $\alpha\in A$ we have that $P_{\alpha}(X)$ is separable and $<\cup_{\alpha}P_{\alpha}(X)>$ is dense in $X$. For a fixed $ n\in N$ and every $ \alpha\in A$ choose a family $(M_i^{\alpha})_{i=1}^n$ of closed subspaces of $P_{\alpha}(X)$ which are pairwise quasi-complemented in $P_{\alpha}(X)$.

For $ i= 1\dots n$ set $M_i = \overline{<\cup_{\alpha}M_i^{\alpha}>}$. The family $(M_i)_{i=1}^n$ has the desided property.Indeed it is clear that for $i \neq j$ $ M_i + M_j $ is dense in $X$. To see that $M_i \cap M_j =\{ 0 \}$ assume on the contrary that there exists $x \in M_i \cap M_j$ with $x\neq 0$.Then there exists $ \alpha\in A$ such that $P_{\alpha}(x) \neq 0$.Since $P_{\alpha}(M_q) = M_q^{\alpha}$ for all $q= 1\dots n$ we derive a contradiction.

$M_i = \overline { <(a_i^ky_k + b_i^kz_k)_k>}$

Remark 4 : There is something which I could not understand. If $M_i, M_j$ are spaces as in Remark 3 then $dist (B_{M_i}, B_{M_j}) = 0$. Hence $M_i, M_j$ are not a complemented pair. But if the initial subspaces $Y, Z$ are a complemented pair then $M_i + M_j = Y+Z= X$ hence $M_i, M_j$ is a complemented pair. Any help?

Remark 5 : There is no problem concerning the proof of Remark 3. My false was that I consider that $M_i + M_j = Y+Z$. What actually we have is that $< (y_k)_k \cup (z_k)_k > \subset M_i +M_j$ which does not imply that $\overline {< (y_k)_k >} + \overline{ < (z_k)_k >} \subset M_i +M_j$.

Therefore Remark 3 yields that independently to what is the initial quasi-complemented pair $Y,Z$( complemented or not) the $(M_i)_i$ resulting from Remark 3, since $dist (B_{M_i}, B_{M_j}) = 0$, are pairwise quasi-complemented but not complemented.

It is well known that every subspace $Y$ of a separable Banach space $X$ is quasi-complemented.We denote its quasi-complement by $Z$. This is a classical result due to F. J. Murray and G. Mackey.The proof is based on the existence of $\{(y_n, f_n)_{n} \cup (z_n, g_n)_n \}$ which is a biorthogonal system of $X$ with $(y_n)_n, (z_n)_n $ generate the subspaces $Y,Z$ respectively and $(\bigcap_n \operatorname{Ker} f_n) \cap(\bigcap_n \operatorname{Ker} g_n) = \{ 0 \}, $

$\langle \{ b_jf_k-a_jg_k \}_k \cup \{ b_if_k-a_ig_k \}_k \rangle = \langle \{f_k\}_k \cup \{g_k\}_k \rangle $

$(\bigcap_k \operatorname{Ker} b_jf_k- a_jg_k) \cap(\bigcap_k \operatorname{Ker} b_if_k-a_ig_k) = \{ 0 \}$

Indeed let $X$ be a WCG space. From Amir Lindenstrauss theorem there exists a family $(P_{\alpha})_{\alpha\in A}$ mutually orthogonal bounded projections such that for each $\alpha\in A$ we have that $P_{\alpha}(X)$ is separable and $\langle \bigcup_{\alpha}P_{\alpha}(X) \rangle$ is dense in $X$. For a fixed $ n\in N$ and every $ \alpha\in A$ choose a family $(M_i^{\alpha})_{i=1}^n$ of closed subspaces of $P_{\alpha}(X)$ which are pairwise quasi-complemented in $P_{\alpha}(X)$.

For $ i= 1\dots n$ set $M_i = \overline{\langle\bigcup_{\alpha}M_i^{\alpha}\rangle}$. The family $(M_i)_{i=1}^n$ has the desided property.Indeed it is clear that for $i \neq j$ $ M_i + M_j $ is dense in $X$. To see that $M_i \cap M_j =\{ 0 \}$ assume on the contrary that there exists $x \in M_i \cap M_j$ with $x\neq 0$.Then there exists $ \alpha\in A$ such that $P_{\alpha}(x) \neq 0$.Since $P_{\alpha}(M_q) = M_q^{\alpha}$ for all $q= 1\dots n$ we derive a contradiction.

$M_i = \overline { \langle (a_i^ky_k + b_i^kz_k)_k \rangle}$

Remark 4 : There is something which I could not understand. If $M_i, M_j$ are spaces as in Remark 3 then $\operatorname{dist} (B_{M_i}, B_{M_j}) = 0$. Hence $M_i, M_j$ are not a complemented pair. But if the initial subspaces $Y, Z$ are a complemented pair then $M_i + M_j = Y+Z= X$ hence $M_i, M_j$ is a complemented pair. Any help?

Remark 5 : There is no problem concerning the proof of Remark 3. My false was that I consider that $M_i + M_j = Y+Z$. What actually we have is that $\langle (y_k)_k \cup (z_k)_k \rangle \subset M_i +M_j$ which does not imply that $\overline {\langle (y_k)_k \rangle} + \overline{ \langle (z_k)_k \rangle} \subset M_i +M_j$.

Therefore Remark 3 yields that independently to what is the initial quasi-complemented pair $Y,Z$( complemented or not) the $(M_i)_i$ resulting from Remark 3, since $\operatorname{dist} (B_{M_i}, B_{M_j}) = 0$, are pairwise quasi-complemented but not complemented.