Lemma 8 Let $X$ and $Y$ be topological spaces. Then there is a finite list of formulas $\Phi$ and a countable set $S$ such that whenever $M\prec (\phi ;S)$, then $(X\times Y)_M = X_M\times Y_M$.

Proof. By [Reference Correa, Cúth and Somaglia6, Lemma 7], there is a finite list of formulas $\Phi$ and a countable set $S$ such that whenever $M\prec (\phi ;S)$, then $(x,\,y)\in (X\times Y)\cap M$ if and only if $(x,\,y)\in (X\cap M)\times (Y\cap M)$. From this the fact easily follows.

Proof. First, assume (i) holds. By [Reference Correa, Cúth and Somaglia7, Lemma 15 and Proposition 16], there is a finite list of formulas $\Phi$ and a countable set $S$ such that whenever $M\prec (\phi ;S)$, then $X = X_M + (D\cap M)_\perp$. Using Fact 8, we obtain a rectangular exceedingly large family $\mathcal {S}\subset \mathcal {P}(X\times D)$ in the sense of suitable models such that whenever $V\times W\in \mathcal {S}$, then $V + W_\perp = X$. Using theorem 7, we obtain (ii).

Assume (ii) holds. By theorem 7 together with Fact 8 and [Reference Correa, Cúth and Somaglia7, Lemma 15], there is a finite list of formulas $\Phi$ and a countable set $S$ such that whenever $M\prec (\phi ;S)$, then $(X\times D)_M = X_M\times D_M$ and there exists a projection $P:X\to X$ with $\operatorname {Rng} P = X_M$ and $\operatorname {Ker} P = (D\cap M)_\perp$. Thus, there is a a rectangular exceedingly large family $\mathcal {S}\subset \mathcal {P}(X\times D)$ in the sense of suitable models such that for every $V\times W\in \mathcal {S}$ there exists a projection $P:X\to X$ satisfying $\operatorname {Rng} P = V$ and $\operatorname {Ker} P = W_\perp$. By theorem 7, this implies (iii).

Finally, assuming (iii) holds, we easily check that for $\Gamma = \{(V,\,W)\in \mathcal {S}\colon V\times W\text { is separable}\}$ ordered by inclusion, the projections $(P_{V\times W})_{V\times W\in \Gamma }$ satisfying $\operatorname {Rng} P_{V\times W} = V$ and $\operatorname {Ker} P_{V\times W} = W_\perp$ form a projectional skeleton (see e.g. [Reference Correa, Cúth and Somaglia7, Definition 1]) and the set induced by the skeleton is in this case equal to $\bigcup _{V\times W\in \Gamma } (P_{V\times W})^*[X^*] = \bigcup _{V\times W\in \Gamma } (\operatorname {Ker} P_{V\times W})^\perp = \bigcup _{V\times W\in \Gamma } \overline {W}^{w^*}\supset D$. Thus, (i) holds.

Proof. Let us denote by $P_U$ the projection $P_U:U\oplus V\to U$. Further, since all the norms on a finite-dimensional space are equivalent, pick $C>0$ such that $\|\sum _{i=1}^{\dim U} a_ie_i\|\geq C\|a\|_1$ for any $a\in \mathbb {R}^{\dim U}$.

Pick any $a\in S_{\ell _1^{\dim U}}$ and $x\in nB_V$. Let $b\in A$ and $y\in N$ be such that $\|a-b\|_1\leq \varepsilon$ and $\|x-y\|\leq \varepsilon$. Then we have

(3.1)\begin{equation} \begin{aligned} & \Bigg| \Bigg\| x + \sum_{i=1}^{\dim U} a_i e_i \Bigg\| - \Bigg\| x + \sum_{i=1}^{\dim U} a_i f_i \Bigg\| \Bigg| \\ & \quad \leq \| x - y \| + \Bigg\| \sum_{i=1}^{\dim U} a_i e_i - \sum_{i=1}^{\dim U} b_i e_i \Bigg\| + \|x-y\| + \Bigg\| \sum_{i=1}^{\dim U} a_i f_i - \sum_{i=1}^{\dim U} b_i f_i \Bigg\| \\ & \quad + \Bigg| \Bigg\| y + \sum_{i=1}^{\dim U} b_i e_i \Bigg\| - \Bigg\| y + \sum_{i=1}^{\dim U} b_i f_i \Bigg\| \Bigg| \\ & \quad< \varepsilon + \varepsilon m + \varepsilon + \varepsilon m + \varepsilon = \varepsilon(3 + 2m). \end{aligned} \end{equation}

Moreover, for any $z\in V$ and $c\in \mathbb {R}^{\dim U}$ we have

\[ \Bigg\|z + \sum_{i=1}^{\dim U}c_i e_i\Bigg\| \geq \frac{1}{\|P_U\|}\Bigg\|\sum_{i=1}^{\dim U}c_i e_i\Bigg\|\geq \frac{C\|c\|_1}{\|P_U\|}, \]

so from (3.1) we obtain

\[ \Bigg| \Bigg\| x + \sum_{i=1}^{\dim U} a_i e_i \Bigg\| - \Bigg\| x + \sum_{i=1}^{\dim U} a_i f_i \Bigg\| \Bigg|\leq \frac{\varepsilon(3+2m)\|P_U\|}{C}\Bigg\| x + \sum_{i=1}^{\dim U} a_i e_i \Bigg\|, \]

which implies that for $f(\varepsilon ):=\min \{\frac {\varepsilon (3+2m)\|P_U\|}{C}, {1}/{2}\}$ we have

(3.2)\begin{equation} (1-f(\varepsilon))\Bigg\| x + \sum_{i=1}^{\dim U} a_i e_i \Bigg\|\leq \Bigg\| x + \sum_{i=1}^{\dim U} a_i f_i \Bigg\|\leq (1+f(\varepsilon))\Bigg\| x + \sum_{i=1}^{\dim U} a_i e_i \Bigg\|\end{equation}

and for $h(\varepsilon ) = {f(\varepsilon )}/{1-f(\varepsilon )}$ we have $(1+h(\varepsilon ))^{-1} = (1-f(\varepsilon ))$ and therefore

(3.3)\begin{equation} \frac{1}{1+h(\varepsilon)}\Bigg\| x + \sum_{i=1}^{\dim U} a_i e_i \Bigg\|\leq \Bigg\| x + \sum_{i=1}^{\dim U} a_i f_i \Bigg\|\leq (1+h(\varepsilon))\Bigg\| x + \sum_{i=1}^{\dim U} a_i e_i \Bigg\| \end{equation}

whenever $x\in nB_V$ and $a\in S_{\ell _1^{\dim U}}$.

Pick arbitrary $a \in S_{\ell _1^{\dim U}}$ and $x \in X \setminus mB_U$. We observe

\[ \|x\| - m \leq \| x \| - \left\| \sum_{i=1}^{\dim U} a_i e_i \right\| \leq \left\| x + \sum_{i=1}^{\dim U} a_i e_i \right\| \leq \|x\| + m. \]

The same argument yields

\[ \|x\| -m \leq \left\| x + \sum_{i=1}^{\dim U} a_i f_i \right\| \leq \|x\| + m. \]

This means

(3.4)\begin{equation} \frac{\left\| x + \sum_{i=1}^{\dim U} a_i f_i \right\|}{\left\| x + \sum_{i=1}^{\dim U} a_i e_i \right\|} \leq \frac{\|x\| + m}{\|x\| - m} \text{ and } \frac{\left\| x + \sum_{i=1}^{\dim U} a_i e_i \right\|}{\left\| x + \sum_{i=1}^{\dim U} a_i f_i \right\|} \leq \frac{\|x\| + m}{\|x\| - m}. \end{equation}

We consider the real function $g: z \mapsto {z+m}/{z-m} - 1$. Because $\| x \| > n\geq {1}/{\varepsilon }$ and the function $g$ is decreasing to $0$, we have ${\|x\| + m}/{\|x\| - m} = g(\|x\|) +1< g({1}/{\varepsilon }) + 1$. From this and (3.4) we obtain

(3.5)\begin{equation} \frac{1}{1 + g(\tfrac{1}{\varepsilon})} \Bigg\| x + \sum_{i=1}^{\dim U} a_i e_i \Bigg\|\leq \Bigg\| x + \sum_{i=1}^{\dim U} a_i f_i \Bigg\| \leq (1 + g(\tfrac{1}{\varepsilon})) \Bigg\| x + \sum_{i=1}^{\dim U} a_i e_i \Bigg\|. \end{equation}

Thus, for $\zeta (\varepsilon ):=2\max \{h(\varepsilon ),\,g({1}/{\varepsilon }),\, {\varepsilon }/{C} \|P_U\|\}$ from (3.3) and (3.5) we obtain that

\[ \frac{1}{1 + \zeta(\varepsilon)} \Bigg\| x + \sum_{i=1}^{\dim U} a_i e_i \Bigg\| < \Bigg\| T\Bigg(x + \sum_{i=1}^{\dim U} a_i e_i\Bigg) \Bigg\| < (1 + \zeta(\varepsilon)) \Bigg\| x + \sum_{i=1}^{\dim U} a_i e_i \Bigg\| \]

for every $a\in S_{\ell _1^{\dim U}}$ and every $x\in V$, which easily using a homogeneous argument implies that $T$ is $(1+\zeta (\varepsilon ))$-isomorphism.

In order to prove the ‘Moreover’ part, pick $x^*\in X^*$ with $|x^*(e_i-f_i)|<\varepsilon$ and $y\in B_{U\oplus V}$. Let us denote $y = x + \sum _{i=1}^{\dim U} a_i e_i$. Then we have

\begin{align*} \left| x^*(Ty) - x^*(y) \right| & = \left| x^*\left(x + \sum_{i=1}^{\dim U} a_i f_i\right) - x^*\left(x + \sum_{i=1}^{\dim U} a_i e_i\right) \right| \\ & = \left| \sum_{i=1}^{\dim U} a_i x^*(e_i - f_i) \right| < \|x^*\|\sum_{i=1}^{\dim U} |a_i| \varepsilon = \|x^*\|\varepsilon \|a\|_{\ell_1^{\dim U}} \\ & \leq \frac{\varepsilon\|x^*\|}{C} \left\| \sum_{i=1}^{\dim U} a_i e_i \right\| = \frac{\varepsilon\|x^*\|}{C} \| P_U(y) \| \leq \frac{\varepsilon\|x^*\|}{C} \|P_U\| \|y\|. \end{align*}

Thus, we have that $\|(x^*\circ T - x^*)|_{U\oplus V}\| \leq {\varepsilon }/{C} \|P_U\|\|x^*\| < \zeta (\varepsilon )\|x^*\|$.

Proof. We start the proof by defining certain sets which we will later use to construct the desired Skolem-like function. Given sets $A\subset X\times X^*$, finite sets $B\subset X$ and $D\subset X^*$, for each $n,\,k \in \mathbb {N}$ and $\delta \in \mathbb {Q}_+$, we pick a finite $\delta$-net $A_{k,\delta }$ of $S_{\ell _1^k}$ and a finite $\delta$-net $N_{n,\delta }(B)\subset \operatorname {span} B$ of $nB_{\operatorname {span} B}$. Now since the mapping $\psi _{n,k,\delta }: (B_X)^{k} \to (\mathbb {R}^{N_{n,\delta }(B) \times A_{k,\delta }} \times \mathbb {R}^{D \times k},\,\|\cdot \|_\infty )$, defined by the formula

\[ \psi_{n,k,\delta}(u) := \left(\left(\left\|x + \sum_{i=1}^{k} a_iu_i \right\|\right)_{x\in N_{n,\delta}(B),a\in A_{k,\delta}},\left(d(u_i)\right)_{d\in D, i\leq k}\right), \]

has separable range, we may pick a countable set $F_{n,k,\delta }(B,\,D)\subset B_X$ such that $\psi _{n,k,\delta }(F_{n,k,\delta }(B,\,D)^k)$ is dense in the range of $\psi _{n,k,\delta }$. Next, if $B(A):=\operatorname {span}_\mathbb {Q} \pi _X(A)\subset X$ and $D(A):=\operatorname {span}_\mathbb {Q} \pi _{X^*}(A)\subset X^*$ for every $A\subset X\times X^*$, we inductively define $\xi (A)\subset X$ for $A\subset X\times X^*$ as follows:

\begin{align*} \xi_0(A)& :=B(A),\\ \xi_{n+1}(A)& :=\operatorname{span}_\mathbb{Q}\Bigg(\xi_n(A)\cup \bigcup \Bigg\{F_{n,k,\delta}(B,D)\colon n,k\in\mathbb{N},\; \delta\in\mathbb{Q}_+,\\ & \quad B\subset \xi_n(A) \text{ and } D\subset D(A) \text{ are finite sets}\Bigg\}\Bigg),\\ \xi(A)& :=\bigcup_{n=0}^\infty \xi_n(A). \end{align*}

Finally, we put $\phi (A):=\xi (A)\times D(A)$ for $A\subset X\times X^*$. We claim that $\phi$ is indeed a rectangular Skolem-like function. It is evident that $\phi$ satisfies conditions (S-a) and (S-b) from definition 3. To check (S-c) we pick an up-directed family $\mathcal {A}\subset P(X\times X^*)$. From (S-b) we have $\phi (\bigcup \mathcal {A})\supset \bigcup _{A\in \mathcal {A}}\phi (A)$. To check the reverse inclusion, we start by proving the following claim: for every $n<\omega$, if $x\in \xi _n(\bigcup \mathcal {A})$, then there is $A\in \mathcal {A}$ such that $x\in \xi _n(A)$.

We proceed by induction on $n$. Since the map $A\mapsto B(A)$ satisfies (S-c), and $\mathcal {A}$ is up-directed, the claim is easily seen to be true for $n=0$. Assume that it holds for $n$, and let $x$ be an element in $\xi _{n+1}(\bigcup \mathcal {A})$. Then

\[ x=q_0y+\sum_{j=1}^mq_jz_j \]

for some $q_0,\,q_1,\,\ldots,\,q_m\in \mathbb {Q}$, $y\in \xi _n(\bigcup \mathcal {A})$ and $z_j\in F_{n_j,k_j,\delta _j}(B_j,\,D_j)$, $j\leq m$ where $n_j,\,k_j\in \mathbb {N}$, $\delta _j\in \mathbb {Q}_+$, and $B_j$ and $D_j$ are finite subsets of $\xi _n(\bigcup \mathcal {A})$ and $D(\bigcup \mathcal {A})$ respectively. By the induction hypothesis, there is $A_0\in \mathcal {A}$ such that $y\in \xi _n(A_0)$. Moreover, since $\mathcal {A}$ is up-directed, by applying the induction hypothesis again, we deduce that there is $A_1\in \mathcal {A}$ such that, $A_0\subset A_1$ and $\bigcup _{j=1}^mB_j\subset \xi _n(A_1)$. Since $A\mapsto D(A)$ satisfies (S-c), we may also assume that $\bigcup _{j=1}^mD_j\subset D(A_1)$. Hence $x\in \xi _{n+1}(A_1)$ and the claim is established by induction.

Now let $(x,\,x^*)\in \phi (\bigcup \mathcal {A})$ be arbitrary. Then $x^*\in D(\bigcup \mathcal {A})$ and there is $n<\omega$ such that $x\in \xi _n(\mathcal {A})$. Since the map $A\mapsto D(A)$ satisfies (S-c), there is $A_0\in \mathcal {A}$ such that $x^*\in D(A_0)$. In accordance with our claim, there exists $A_1\in \mathcal {A}$ such that $x\in \xi _n(A_1)$. Finally, since $\mathcal {A}$ is up-directed, there is $A_2\in \mathcal {A}$ such that $A_0\cup A_1\subset A_2$ and $(x,\,x^*)\in \xi _n(A_1)\times D(A_0)\subset \phi (A_2)$. This establishes that $\phi (\bigcup \mathcal {A})\subset \bigcup _{A\in \mathcal {A}}\phi (A)$. The fact that $\phi$ is rectangular is evident from the definition of $\phi$.

Now, pick $\varepsilon >0$, $A\subset X\times X^*$, $D\subset D(A) = \phi _{X^*}(A)$ finite, $B\subset \phi _X(A) = \xi (A)$ finite and finite-dimensional space $E\subset X$ with $E\supset B$. Since $\operatorname {span} B$ is finite-dimensional, there is $U\subset E$ such that $E = \operatorname {span} B\oplus U$ and we may pick a basis $\{e_1,\,\ldots,\,e_k\}\subset B_U$ of the space $U$. Let $\zeta$ be the function from lemma 11 applied to $\operatorname {span} B\oplus U$ and let $\delta >0$ be such that $\zeta (\delta )<\varepsilon$. Now, pick $n\in \mathbb {N}$ with $n\geq {1}/{\delta }$. By the choice of $F_{n,k,\delta }(B,\,D)$, there are $f_1,\,\ldots,\,f_k\in F_{n,k,\delta }(B,\,D)\subset \xi (A)$ such that $\|\psi _{n,k,\delta }((f_i)) - \psi _{n,k,\delta }((e_i))\|_\infty$. Then, by lemma 11, since $\zeta (\delta )<\varepsilon$, we have that the mapping $T:E = \operatorname {span} B\oplus U\to \overline {\xi (A)} = \overline {\phi _X(A)}$ given by

\[ T\left(x+\sum_{i=1}^{k}a_i e_i\right):=x+\sum_{i=1}^{k}a_i f_i,\quad x\in \operatorname{span} B,\; a\in \mathbb{R}^{k} \]

is $(1+\varepsilon )$-isomorphic embedding and so (Kb) holds. Finally, since for any $d\in D$ we have $|d(e_i)-d(f_i)|<\delta$, $i=1,\,\ldots,\,k$, the moreover part of lemma 11 implies that (Kc) holds as well.

Proof. By theorem C, it suffices to find a family $\mathcal {S}$ large in the sense of Skolem-like functions such that each $F\in \mathcal {S}$ is ai-ideal in $X$.

Pick the Skolem-like function $\phi :\mathcal {P}(X\times X^*)\to \mathcal {P}(X\times X^*)$ from the Key lemma 12 and consider its projection onto the first coordinate given by $\psi (A):=\phi _X(A\times \{0\})$. It is easy to check that $\psi :\mathcal {P}(X)\to \mathcal {P}(X)$ is Skolem-like function as well. Thus, in order to finish the proof it suffices to check that $\overline {\psi (A)}$ is ai-ideal for every $A\subset X$.

Fix $A\subset X$ and put $Y:=\overline {\psi (A)}$. Given a finite dimensional subspace $E\subset X$ and $\epsilon >0$ (without loss of generality, we may assume that $\epsilon <1$), we let $\delta =\min \{{\epsilon }/{2(1+\dim (E))},\, {\epsilon }/{2}({1-\epsilon }/{1+\epsilon })\}$ and pick a finite ${\delta }/{2(\delta +2)}$-net $N$ for $B_{E\cap Y}$. Since $\psi (A)$ is dense subset of $Y$, there exists a function $\varphi :N\to \psi (A)$ such that $\|\varphi (x)-x\|\leq {\delta }/{2(\delta +2)}$ for every $x \in N$. Since $\phi$ was the function from Key lemma 12, for $E^\prime =\operatorname {span}\{E\cup \varphi [N]\}$, $B=\varphi [N]$, $D=\emptyset$ and $\delta >0$ there exists a $(1+\delta )$-isomorphic embedding $T:E\to Y$ such that $Tx = x$ for $x\in B$.

Now, we shall check that then $\|(Id-T)|_{E\cap Y}\|\leq \delta$. Indeed, for $x\in B_{E\cap Y}$, there is $n\in N$ such that $\|x-n\|\leq {\delta }/{2(\delta +2)}$ and hence $\|x-\varphi (n)\|\leq \|x-n\|+\|\varphi (n)-n\|\leq {\delta }/{\delta +2}$. In other words, there exists $b \in B$ such that $\|x-b\|\leq {\delta }/{\delta +2}$. Since $Tb=b$ we have

\[ \|(\mathrm{Id}-T)x\|\leq \|x-b\|+\|Tb-Tx\|\leq (1+\|T\|)\|x-b\|\leq \delta, \]

so we really have $\|(Id-T)|_{E\cap Y}\|\leq \delta$.

Now pick a projection $P:E\to E$ with $P[E]=E\cap Y$ and norm $\|P\|\leq \dim (E)$, and define $S:E\to Y$ by $S=P+T(\mathrm {Id}-P)$. It is clear that $S(x)=x$ for every $x\in E\cap Y$. Furthermore, for every $x\in E$, the following relation holds:

\[ \|Sx-Tx\| = \|(\mathrm{Id}-T)Px\|\leq \delta\|Px\|\leq \delta\dim(E)\|x\|<\frac{\epsilon}{2}\|x\|. \]

Therefore, for every $x\in E$ we have

\[ \|Sx\|\leq \|Sx-Tx\|+\|Tx\|\leq \frac{\epsilon}{2}\|x\|+(1+\delta)\|x\|\leq (1+\epsilon)\|x\|. \]

and

\begin{align*} \|Sx\|& \geq \|Tx\|-\|Sx-Tx\| \geq (1-\delta)\|x\|-\frac{\epsilon}{2}\|x\|\\ & \geq \left(1-\frac{\epsilon}{2}\cdot \frac{1-\epsilon}{1+\epsilon}-\frac{\epsilon}{2}\right)\|x\|=(1+\epsilon)^{{-}1}\|x\|. \end{align*}

which shows that $S$ is $(1+\varepsilon )$-isomorphic embedding and therefore completes the proof.

Proof. By theorem C, it suffices to find a rectangular family $\mathcal {S}$ large in the sense of Skolem-like functions such that for each $V\times W\in \mathcal {S}$ there exists a norm-one operator $R:X\to V^{**}$ satisfying both (Ra) and (Rb).

Let $\phi :\mathcal {P}(X\times X^*)\to \mathcal {P}(X\times X^*)$ be the rectangular Skolem-like function from Key Lemma 12. Pick $A\subset X\times X^*$ and put $V:=\overline {\phi _X(A)}$, $W:=\overline {\phi _{X^*}(A)}$. Let

\begin{align*} \varGamma=\{(E,B,D,\epsilon)\colon & E\subset X \text{ finite-dimensional},\ B\in [E\cap \phi_X(A)]^{<\omega},\\ & D\in [\phi_{X^*}(A)]^{<\omega}, \ \epsilon>0\} \end{align*}

be directed by the relation $\leq$ defined as follows $(E,\,B,\,D,\,\epsilon )\leq (E^\prime,\,B^\prime,\,D^\prime,\,\epsilon ^\prime )$ if and only if $E\subset E^\prime$, $B\subset B^\prime$, $D\subset D^\prime$ and $\epsilon ^\prime \leq \epsilon$.

Since $\phi$ is as in Key lemma 12, for every $I=(E^\prime,\,B^\prime,\,D^\prime,\,\epsilon ^\prime )$ there exists a bounded linear operator $T_I:E\to V$ satisfying conditions (Ka), (Kb) and (Kc) of the lemma.

We define $R_I:X\to V$ by

\[ R_Ix= \left\{ \begin{array}{ll} T_Ix & \text{ if }x\in E;\\ 0 & \text{ otherwise }. \end{array} \right. \]

We note that for every $x\in X$ and $x^*\in X^*$, $\{x^*(R_Ix):I\in \varGamma \}$ is a bounded set of real numbers and so relatively compact. We fix a directed ultrafilter $\mathcal {U}$ in $\varGamma$ (that is, $\mathcal {U}$ is nonprincipal ultrafilter satisfying $\{i\in \Gamma \colon i\geq i_0\}\in \mathcal {U}$ for every $i_0\in \Gamma$) and define a function $R:X\to V^{**}$ by the formula:

\[ Rx(x^*)=\lim_{\mathcal{U}}x^*(R_Ix),\qquad x\in X,\; x^*\in V^*. \]

It is readily seen that $R$ is a linear map. Moreover, for arbitrary $x \in B_X$, $x^*\in B_{V^*}$ and $\epsilon >0$, if $I_0=(\mathrm {span}\{x\},\,\emptyset,\,\emptyset,\,\epsilon )$ then for every $I\geq I_0$ we have $|x^*(T_Ix)|\leq \|T_I\|\leq (1+\epsilon )$ and therefore

\[ |Rx(x^*)|=\lim_{\mathcal{U}}|x^*(R_Ix)|\leq (1+\epsilon) \]

and we deduce that $\|R\|\leq 1$.

To verify (Ra), we let $x \in \phi _X(A)$ be arbitrary and set $I_0 = (\mathrm {span}\{x\},\, \{x\},\, \emptyset,\, 1)$. Then, for any $I \geq I_0$, recalling (Ka), for all $x^* \in V^*$, we have

\[ x^*(x)=x^*(T_Ix)=x^*(R_Ix), \]

which implies

\[ Rx(x^*) = \lim_{\mathcal{U}} x^*(R_Ix)=\lim_{I\geq I_0} x^*(R_Ix)=x^*(x). \]

Thus, we have $Rx=x$ for $x\in \phi _X(A)$ and by the continuity of $R$ we obtain (Ra).

To verify (Rb), we let $x\in X$ and $x^*\in \phi _{X^*}(A)$ be arbitrary. Given $\delta >0$ we set $I_0=(\mathrm {span}\{x\},\,\emptyset,\, \{x^*\},\, \delta /(1+\|x^*\|\|x\|))$. For every $I=(E,\,B,\,D,\,\epsilon )\geq I_0$, according to (Kc), we have

\[ |(x^*|_{V})(T_Ix)-x^*(x)| \leq \|(x^*\circ T_I-x^*)|_E\| \|x\|\leq \epsilon\|x^*\|\|x\|<\delta. \]

This implies

\[ |Rx(x^*|_{V})- x^*(x)|= \lim_{\mathcal{U}}|(x^*|_{V})(T_Ix)-x^*(x)|\leq \delta. \]

As $\delta >0$ is arbitrary, we can conclude that $Rx(x^*|_{X_M})=x^*(x)$. Furthermore, due to the continuity of $R$, we can deduce that $Rx(x^*|_{V})=x^*(x)$ holds for all $x\in X$ and for all $x^{*}\in \overline {\phi _{X^*}(A)} = W$.

Proof. Let $\mathcal {S}$ be as in theorem 14. Then given $V\times W\in \mathcal {S}$ there exists norm-one mapping $R:X\to V^{**}$ such that (Ra) and (Rb) hold. Define $P:X^*\to X^*$ by

(3.6)\begin{equation} Px^*(x):=Rx(x^*|_{V}),\qquad x\in X,\; x^*\in X^*. \end{equation}

Then indeed $P$ is a norm-one operator and using condition (Ra) we obtain $Px^*|_V = x^*|_V$, $x^*\in X^*$ which implies that $P$ is a projection. Condition (Rb) implies that $\operatorname {Rng} P\supset W$. It is immediate from (3.6) that $Px^* = 0$ for $x^*\in V^\perp$, so we have $V^\perp \subset \operatorname {Ker} P$. Finally, given $x^*\in \operatorname {Ker} P$ we have $x^*|_V = Px^*|_V \equiv 0$ and so $\operatorname {Ker} P\subset V^\perp$.

Proof. By [Reference Kubiś23, Proposition 26], $X$ is Asplund if and only if $X$ is a subset of the set induced by a projectional skeleton on $X^*$. Thus, by lemma 9 (applied to $X\times D = X^*\times X$) we obtain that (i) holds if and only if $X$ is Asplund.

It is immediate that (iii) implies (ii).

By [Reference Kubiś23, Corollary 25], $X$ is WLD if and only if $X^*$ is the set induced by a projectional skeleton on $X$.

First, suppose that $X$ is WLD. By lemma 9, there exists a rectangular exceedingly rich family $\mathcal {S}$ of subspaces of $X\times X^*$ such that for every $V\times W\in \mathcal {S}$ there exists a projection $Q:X\to X$ with $Q[X] = V$ and $\operatorname {Ker} Q = W_\perp$. Then $P = Q^*:X^*\to X^*$ is a $w^*$-$w^*$ continuous projection with $\operatorname {Rng} P = \overline {W}^{w^*}$ and $\operatorname {Ker} P = V^\perp$. Thus, if $X$ is WLD then both (ii) and (iii) hold.

Finally, if (ii) holds, then for every $V\times W\in \mathcal {S}$ we have $X = \overline {V + W_\perp }$ (because otherwise there is $x^*\in X^*\setminus \{0\}$ such that $x^*|_{V + W_\perp }\equiv 0$, and therefore $x^*\in V^\perp \cap (W_\perp )^\perp \subset \operatorname {Rng} P\cap \operatorname {Ker} P = \{0\}$, contradiction). Thus, by lemma 9 we obtain that $X^*$ is the set induced by a projectional skeleton on $X$ and so $X$ is WLD.

Proof. The proof is very similar to the proof of Key lemma 12. The additional ingredient is only to use Lipschitz-free spaces. We pick arbitrary $0\in X$, so $X$ is a pointed metric space.

Given sets $A\subset X\times \operatorname {Lip}_0(X)$, finite sets $B\subset X$ and $D\subset \operatorname {Lip}_0(X)$, for each $n,\,k \in \mathbb {N}$ and $\delta \in \mathbb {Q}_+$, we pick a finite $\delta$-net $A_{k,\delta }$ of $S_{\ell _1^k}$ and a finite $\delta$-net $N_{n,\delta }(B)\subset \mathcal {F}(B) := \operatorname {span} \delta (B) (\subset \mathcal {F}(X))$ of $nB_{\mathcal {F}(B)}$. Now since for any $R\in \mathbb {N}$ the mapping $\psi _{n,k,\delta,R}: (B_X(0,\,R))^k \to (\mathbb {R}^{N_{n,\delta }(B) \times A_{k,\delta }} \times \mathbb {R}^{D \times k},\,\|\cdot \|_\infty )$, defined by the formula

\[ \psi_{n,k,\delta,R}(u) := \left(\left(\left\|x + \sum_{i=1}^{k} a_i\delta(u_i) \right\|\right)_{x\in N_{n,\delta}(B),a\in A_{k,\delta}},\left(d(u_i)\right)_{d\in D, i\leq k}\right), \]

has separable range, we may pick a countable set $F_{n,k,\delta,R}(B,\,D)\subset B_X(0,\,R)$ such that $\psi _{n,k,\delta,R}(F_{n,k,\delta,R}(B,\,D)^k)$ is dense in the range of $\psi _{n,k,\delta,R}$. Next, if $D(A):=\operatorname {span}_\mathbb {Q} \pi _{\operatorname {Lip}_0(X)}(A)\subset X^*$ for every $A\subset X\times \operatorname {Lip}_0(X)$, we inductively define $\xi (A)\subset X$ for $A\subset X\times \operatorname {Lip}_0(X)$ as follows:

\begin{align*} \xi_0(A)& :=\pi_X(A),\\ \xi_{n+1}(A)& :=\xi_n(A)\cup \bigcup\Bigg\{F_{n,k,\delta,R}(B,D)\colon n,k,R\in\mathbb{N},\; \delta\in\mathbb{Q}_+,\\ & \quad B\subset \xi_n(A) \text{ and } D\subset D(A) \text{ are finite sets}\Bigg\},\\ \xi(A)& :=\bigcup_{n=0}^\infty \xi_n(A). \end{align*}

Similarly as in the proof of Key lemma 12 we check that indeed the mapping $\phi$ defined by $\phi (A):=\xi (A)\times D(A)$, $A\subset X\times \operatorname {Lip}_0(X)$ is a rectangular Skolem-like function.

Now, pick $\varepsilon >0$, $A\subset X\times X^*$, $D\subset D(A) = \phi _{\operatorname {Lip}_0(X)}(A)$ finite, $B\subset \phi _X(A) = \xi (A)$ finite and finite set $E\subset X$ with $E\supset B$. Let $R\in \mathbb {N}$ be such that $E\subset B(0,\,R)$ and enumerate $\{e_1,\,\ldots,\,e_k\} = E\setminus B$. Let $\zeta$ be the function from lemma 11 applied to $m=R$ and $\mathcal {F}(B)\oplus \mathcal {F}(E\setminus B)$ and let $\delta >0$ be such that $\zeta (\delta )<\varepsilon$. Now, pick $n\in \mathbb {N}$ with $n\geq {1}/{\delta }$. By the choice of $F_{n,k,\delta,R}(B,\,D)$, there are $f_1,\,\ldots,\,f_k\in F_{n,k,\delta,R}(B,\,D)\subset \xi (A)$ such that $\|\psi _{n,k,\delta,R}((f_i)) - \psi _{n,k,\delta,R}((e_i))\|_\infty$. Then, by lemma 11, since $\zeta (\delta )<\varepsilon$, we have that the mapping $\widehat {T}:\mathcal {F}(E) = \mathcal {F}(B)\oplus \mathcal {F}(E\setminus B)\to \mathcal {F}(\overline {\xi (A)}) = \mathcal {F}(\overline {\phi _X(A)})$ given by

\[ \widehat{T}(x+\sum_{i=1}^{k}a_i \delta(e_i)):=x+\sum_{i=1}^{k}a_i \delta(f_i),\quad x\in \mathcal{F}(B),\; a\in \mathbb{R}^{k} \]

is $(1+\varepsilon )$-isomorphic embedding. Thus, the mapping $T:E\to \phi _X(A)$ given by $T(x):=\delta ^{-1}(\widehat {T}(\delta (x)))$ satisfies (K'a) and (K'b). Finally, since for any $d\in D$ we have $|d(e_i)-d(f_i)|<\delta$, $i=1,\,\ldots,\,k$, the moreover part of lemma 11 implies that (K'c) holds as well.

Proof. By theorem C, it suffices to find a rectangular family $\mathcal {S}$ large in the sense of Skolem-like functions such that each $F\in \mathcal {S}$ is almost isometric local retract in $X$.

Pick the rectangular Skolem-like function $\phi :\mathcal {P}(X\times \operatorname {Lip}_0(X))\to \mathcal {P}(X\times \operatorname {Lip}_0(X))$ from the statement of lemma 17 and consider its projection onto the first coordinate given by $\psi (A):=\phi _X(A\times \{0\})$. It is easy to check that $\psi :\mathcal {P}(X)\to \mathcal {P}(X)$ is Skolem-function as well. Thus, in order to finish the proof it suffices to check that $\overline {\psi (A)}$ is almost isometric local retract for every $A\subset X$.

Pick $A\subset X$ and denote $V:=\overline {\psi (A)}$. Fix $\varepsilon >0$ and a finite set $E\subset X$. Pick $\delta >0$ small enough (more concretely, $\delta < \varepsilon /2$ and $\delta <\min \{{\varepsilon d(x,\,y)}/{2}\colon x,\,y\in E,\, x\neq y\}$) and a function $\varphi :E\cap V \to \psi (A)$ such that $d(\varphi (x),\,x)<\delta$ for every $x\in E\cap V$. We let $E' = E\cup \operatorname {Rng} \varphi$, $B=\operatorname {Rng} \varphi$, $D= \emptyset$ and use (K'a) and (K'b) to find $(1+\delta )$-biLipschitz embedding $T:E'\to \psi (A)$ satisfying $Tx=x$ for $x\in B$.

We define $S:E\to V$ by $Sx = x$ for $x\in E\cap V$ and $Sx = T(x)$ for $x\in E\setminus V$. It suffices to show that $S$ is $(1+\varepsilon )$-biLipschitz embedding. Since $T$ is $(1+\varepsilon )$-biLipschitz embedding, it suffices to consider $x\in E\cap V$, $y\in E\setminus V$ and estimate then $d(Sx,\,Sy)$. For $x\in E\cap V$, $y\in E\setminus V$ we have

\begin{align*} |d(Sx,Sy) - d(x,y)| & = |d(x,Ty) - d(x,y)|\leq \delta + |d(Ty,\varphi(x)) - d(x,y)|\\ & = \delta + |d(Ty,T\varphi(x)) - d(x,y)|\leq \delta + \delta d(x,y)\leq \varepsilon d(x,y). \end{align*}

Thus, $S$ is $(1+\varepsilon )$-biLipschitz embedding.

Proof. Given $\epsilon > 0$ and a finite dimensional Banach space $E$ we let $T:E\to Y$ be a linear isometric embedding. If $F$ denotes a finite dimensional Banach containing $E$, since $X$ is a Gurariĭ space (an $L_1$-predual space), for $0<\delta <\sqrt {1+\epsilon }-1$, $T$ admits a linear extension $\hat {T}: F\to X$ of $T$ with $(1+\delta )^{-1}\|x\| \leq \|T(x)\|\leq (1 + \delta )\|x\|$ ($\|T(x)\|\leq (1 + \delta )\|x\|$) for all $x\in E$. Since $Y$ is an ai-ideal in $X$, there exists an $(1+\delta )$-isomorphic embedding $S:\hat {T}[F] \to Y$ such that $S(x)=x$ for every $x\in Y\cap \hat {T}[F]$. Hence $S\circ \hat {T}: F \to Y$ is a linear extension of $T$ satisfying $(1 + \epsilon )^{-1}\|x\|\leq (1 + \delta )^{-2}\|x\|\leq \|S\circ T(x)\| \leq (1 + \delta )^2\|x\|\leq (1 + \epsilon )\|x\|$ ($\|S\circ T(x)\| \leq (1 + \delta )^2\|x\|\leq (1 + \epsilon )\|x\|$) for all $x\in F$.

Proof. We start with handling the first equivalence by distiguishing two cases. In the case that $X$ is Gurariĭ space, by theorem 14 there is exceedingly rich family $\mathcal {S}_1$ such that every $Y\in \mathcal {S}_1$ is ai-ideal in $X$ and therefore, by lemma 21, it is Gurariĭ space. Thus, we have that for any $Y\in \mathcal {S}_1$ it holds that $X$ is Gurariĭ space if and only if $Y$ is Gurariĭ space.

In the case that $X$ is not Gurariĭ space, there is $\epsilon >0$, finite dimensional $E$ and $F$, with $E\subset F$, and an isometric linear embedding $T:E\to X$ that cannot be extended to an $(1+\epsilon )$-isomorphic embedding from $F$ into $X$. Then we let

\[ \mathcal{S}_1:=\{V\subset X\colon V\textit{ is a closed subspace of }X \textit{ and contains } E\cup F\cup T(E)\} \]

and easily check that $\mathcal {S}_1$ is exceedingly rich family with $V$ not being Gurariĭ for $V\in \mathcal {S}_1$. Thus, we have that for any $Y\in \mathcal {S}$ it holds that $X$ is not Gurariĭ if and only if $Y$ is not Gurariĭ.

Thus, in any case there exists exceedingly rich family $\mathcal {S}_1$ such that for any $Y\in \mathcal {S}_1$ we have that $X$ is Gurariĭ if and only if $Y$ is Gurariĭ. Similarly, we find exceedingly rich family $\mathcal {S}_2$ such that for any $Y\in \mathcal {S}_2$ we have that $X$ is $L_1$-predual if and only if $Y$ is $L_1$-predual. Finally it suffices to put $\mathcal {S}:=\mathcal {S}_1\cap \mathcal {S}_2$, which is exceedingly rich family by lemma 2.

Proof. Similarly as in the proof of theorem 22, we distiguish two cases. If $X$ has the Daugavet property, then according to theorem 14 there exists an exceedingly rich family $\mathcal {S}$ such that every $Y\in \mathcal {S}$ is ai-ideal in $X$, and, by [Reference Abrahamsen, Lima and Nygaard2, Proposition 3.8], each of these ai-ideals inherits the Daugavet property. Thus, if $X$ has the Daugavet property, $\mathcal {S}$ does the job.

If $X$ does not have Daugavet property, then, based on the characterization mentioned above, there exist $y\in S_X$, $x^*\in S_{X^*}$ and $\epsilon >0$ such that, for every $x\in S_X$, $x^*(x)<1-\epsilon$ or $\|x + y\|<2 - \epsilon$.

We let $D\subset B_X$ be a countable set witnessing that $\|x^*\|=1$, that is, $\sup _{x\in D}\|x^*(x)\|=1$. The family $\mathcal {S}':=\{Y\subset X\colon$Y$\text { is a closed subspace of }X \text{ and} \text{contains } \{y\}\cup D\}$ is readily seen to be an exceedingly rich family of closed subsets of $X$. Moreover, for each $Y\in \mathcal {S}'$ we have $\|x^*|_Y\|=1$, and since $S_Y\subset S_X$, for each $x\in S_Y$ we have $x^*(x)<1-\epsilon$ or $\|x + y\|<2 - \epsilon$, which proves that $Y$ does not have the Daugavet property. Thus, if $X$ does not have the Daugavet property, $\mathcal {S}'$ does the job.

Proof. Similarly as in the proof of theorem 22 by distinguishing two cases. If $X$ is an absolute ai-local retract, then from theorem 19 we may fix a exceedingly rich family $\mathcal {S}$ comprising ai-local retracts of $X$. With an argument similar to lemma 21 we deduce that each $F\in \mathcal {S}$ is absolute ai-local retract. On the other hand, assuming that $X$ is not an absolute ai-local retract, there exists $\epsilon >0$, finite sets $E$, $F\subset X$ and a isometric mapping $f:E:\to X$ that cannot be extended to a $(1+\epsilon )$-biLipschitz embedding. The family $\mathcal {S}=\{Y\subset X:Y\text { contains }E\cup F\cup T(E)\}$ is clearly a exceedingly rich family such that $Y$ is not an absolute ai-local retract for all $Y\in \mathcal {S}$.