Proof. By the Hurewicz theorem (see, e.g. [Reference Kechris34, Theorem 21.18]), there is a set $C\subseteq X$ homeomorphic to the Cantor space, such that $C\cap B$ is countable dense in C. Then $C\cap B$ is an $F_\sigma $ -set but not a $G_\delta $ -set in the zero-dimensional Polish space C, and so it is $F_\sigma $ -complete in C by Wadge’s theorem (see, e.g. [Reference Kechris34, Theorem 22.10]). So for any zero-dimensional Polish space Y and any $F_\sigma $ -subset A of Y, there is a Wadge reduction of $A\subseteq Y$ to $C\cap B\subseteq C$ . But any such reduction is also a reduction of $A\subseteq Y$ to $B\subseteq X$ , and so B is $F_\sigma $ -hard.

The argument with the roles of $F_\sigma $ and $G_\delta $ interchanged is similar.

Notation 1.6. Let Z be a separable Banach space, and let $\mathcal {I}$ be a subset of $\mathcal {P}$ . We put

$$\begin{align*}{\langle {Z}\rangle_{\equiv}^{\mathcal{I}}}:=\{\mu\in\mathcal{I}:\; X_\mu\equiv Z\} \quad \text{and} \quad {\langle {Z}\rangle_{\simeq}^{\mathcal{I}}}:=\{\mu\in\mathcal{I}:\; X_\mu\simeq Z\}.\end{align*}$$

If $\mathcal {I}$ is clear from the context, we write ${\langle Z\rangle _{\equiv }}$ and $\langle Z\rangle _{\simeq }$ instead of ${\langle Z\rangle _{\equiv }^{\mathcal {I}}}$ and ${\langle Z\rangle _{\simeq }^{\mathcal {I}}}$ , respectively.

Proof. Hilbert spaces are characterized among Banach spaces as those Banach spaces whose norm satisfies the parallelogram law, that is $\|x+y\|^2+\|x-y\|^2=2(\|x\|^2+\|y\|^2)$ for any pair of elements $x,y$ . It is clear that a norm satisfies the parallelogram law if and only if it satisfies it on a dense set of vectors, therefore, every norm, respectively, pseudonorm, from $\mathcal {B}$ , respectively, $\mathcal {P}_\infty $ , satisfying the parallelogram law on V defines a Hilbert space. Since norms, respectively, pseudonorms, from $\mathcal {B}$ , respectively, $\mathcal {P}_\infty $ , define only infinite-dimensional spaces, they define spaces isometric to $\ell _2(\mathbb {N})$ . Since the parallelogram law is clearly a closed condition, we are done.

Proof. We recall the fact that a separable infinite-dimensional Banach space is isometric to a subspace of $L_p[0,1]$ if and only if it is finitely representable in $\ell _p$ (see, e.g. [Reference Albiac and Kalton1, Theorem 12.1.9]). The rest follows from Proposition 1.8. We refer the reader to Section 3 for a definition of the class $\mathcal {L}_{p,1+}$ .

Proof. Fix $p\in \left [1,2\right )\cup \left (2,\infty \right )$ . Pick $\mu \in QSL_p$ , such that $X_\mu $ does not isometrically embed as a subspace into $L_p[0,1]$ (such a space exists, see the examples above). Let U be now a basic open neighbourhood of some $\nu \in QSL_p$ . Since the class of $QSL_p$ -spaces is clearly closed under taking $\ell _p$ -sums (see, e.g. [Reference Runde49]), $X_\nu \oplus _p X_\mu $ is still a $QSL_p$ -space. It is easy to define $\nu '\in U$ , so that $X_{\nu '}$ is isometric to $X_\nu \oplus _p X_\mu $ . Now, since $X_\mu $ does not isometrically embed as a subspace into $L_p[0,1]$ , neither $X_{\nu '}$ does. By [Reference Albiac and Kalton1, Theorem 12.1.9], $X_{\nu '}$ is not finitely representable in $\ell _p$ , so also not in $L_p[0,1]$ (by [Reference Albiac and Kalton1, Proposition 12.1.8]). It follows from Proposition 1.8 that there exists a basic open neighbourhood $U'$ of $\nu '$ avoiding $M_p$ . Now, $U\cap U'$ is a nonempty open subset of U avoiding $M_p$ , and we are done.

Proof. Let X be a separable infinite-dimensional Banach space. We show that ${\langle X\rangle _{\simeq }}$ is dense (we show the argument only for $\mathcal {P}_\infty $ , the other cases are analogous). Let F be a finite-dimensional Banach space. It is well-known that every finite-dimensional space is complemented in any infinite-dimensional Banach space, so we have $X\simeq F\oplus _1 Y$ for some Banach space Y. Since F was arbitrary, it follows that every separable Banach space is finitely representable in ${\langle X\rangle _{\simeq }}$ , so by Proposition 1.8, $\overline {{\langle X\rangle _{\simeq }}}=\mathcal {P}_\infty $ , hence, ${\langle X\rangle _{\simeq }}$ is dense.

It follows that ${\langle X\rangle _{\simeq }}$ cannot be closed for any X because it is dense and there are obviously two nonisomorphic spaces. Moreover, if X is not isomorphic to the Gurariı̆ space, then ${\langle X\rangle _{\simeq }}$ cannot be a $G_\delta $ -set since by Theorem 1.10, the isometry class of the Gurariı̆ space is a dense $G_\delta $ -set, so it would have nonempty intersection with ${\langle X\rangle _{\simeq }}$ otherwise.

Proof of Theorem 2.10.

First, we show that isomorphism class is $F_\sigma $ . This was proved for an admissible topology on $SB_{\infty} $ in [Reference Godefroy and Saint-Raymond28, Theorem 4.3]. The same proof, which we briefly sketch, works also for $\mathcal {P}_\infty $ and $\mathcal {B}$ . By Kwapień’s theorem (see, e.g. [Reference Albiac and Kalton1, Theorem 7.4.1]), a separable infinite-dimensional Banach space is isomorphic to $\ell _2$ if and only if it is of type $2$ and of cotype $2$ . It is clear from the definition of type and cotype (see, e.g. [Reference Albiac and Kalton1, Definition 6.2.10]) that these properties are $F_\sigma $ -conditions. So to show that ${\langle \ell _2\rangle _{\simeq }}$ is even $F_\sigma $ -complete, by Lemma 1.1, it suffices to show that ${\langle \ell _2\rangle _{\simeq }}$ is not a $G_\delta $ -set, which we have already proved. An alternative proof showing that the isomorphism class ${\langle \ell _2\rangle _{\simeq }}$ is an $F_\sigma $ -set follows from [Reference Kwapien37, Theorem 2’] (see also Remark 4 therein), which provides a formula defining spaces isomorphic to $\ell _2$ and which obviously defines an $F_\sigma $ -set (in $\mathcal {P}_\infty $ and $\mathcal {B}$ ).

Next, we show that if a separable infinite-dimensional Banach space X is not isomorphic to $\ell _2$ , then ${\langle X\rangle _{\simeq }}$ is not $F_\sigma $ in $\mathcal {B}$ . In what follows, we denote by $\mathcal {T}$ the set of finite tuples (including empty) of natural numbers without repetition. The length of $\gamma \in \mathcal {T}$ is denoted by $|\gamma |$ , and its range by $\operatorname {rng}(\gamma )$ . Moreover, for every $\gamma \in \mathcal {T}$ and every $\mu \in \mathcal {B}$ , we put

$$\begin{align*}\mathbb{M}^{\gamma}_\mu:=\Big\{\nu\in\mathcal{B}\colon \text{for every } (a_i)_{i=1}^{|\gamma|}\in\mathbb{Q}^{|\gamma|} \text{ we have } \nu\Big(\sum_{i=1}^{|\gamma|} a_i e_i\Big) = \mu\Big(\sum_{i=1}^{|\gamma|} a_i e_{\gamma(i)}\Big)\Big\}. \end{align*}$$

In order to get a contradiction, assume that $(F_n)_{n=1}^\infty $ are closed sets in $\mathcal {B}$ such that ${\langle X\rangle _{\simeq }} = \bigcup _{n=1}^\infty F_n$ .

Proof of the claim.

Suppose the statement is not true. In particular, it does not hold for $\gamma = \emptyset $ and $m=1$ . That is, there is some $\gamma ^{\prime }_1\in \mathcal {T}$ so that $\mathbb {M}^{\gamma _1^{\prime }}_\mu \cap F_1=\emptyset $ . If $1\in \operatorname {rng}(\gamma ^{\prime }_1)$ , we set $\gamma _1=\gamma ^{\prime }_1$ . Otherwise, we set $\gamma _1={\gamma ^{\prime }_1}^{\frown } (1)$ .

In the next step, we use that the statement is not true for $\gamma _1$ and $m=2$ to obtain $\gamma ^{\prime }_2\in \mathcal {T}$ , $\gamma ^{\prime }_2\supseteq \gamma _1$ so that $\mathbb {M}^{\gamma ^{\prime }_2}_\mu \cap F_2=\emptyset $ . If $2\in \operatorname {rng}(\gamma ^{\prime }_2)$ , we set $\gamma _2=\gamma ^{\prime }_2$ . Otherwise, we set $\gamma _2={\gamma ^{\prime }_2}^\frown {} (2)$ .

We continue analogously. At the end of the recursion, we obtain a bijection $\pi :\mathbb {N}\to \mathbb {N}$ , such that $\pi \supseteq \gamma _n$ for every $n\in \mathbb {N}$ and $\mathbb {M}^{\gamma _n}_\mu \cap F_n=\emptyset $ for every $n\in \mathbb {N}$ . Consider $\mu _0\in \mathcal {B}$ given as

$$\begin{align*}\mu_0\Big(\sum_{i=1}^k a_i e_i\Big):=\mu\Big(\sum_{i=1}^k a_i e_{\pi(i)}\Big),\quad k\in\mathbb{N},\; (a_i)_{i=1}^k\in \mathbb{Q}^k. \end{align*}$$

Then the linear mapping given by $e_i\mapsto e_{\pi (i)}$ , $i\in \mathbb {N}$ , witnesses that $X_{\mu _0}\equiv X_\mu $ and $\mu _0\in \mathbb {M}^{\gamma _n}_\mu $ for every $n\in \mathbb {N}$ . Thus, $\mu _0\notin \bigcup _{n=1}^\infty F_n$ , which is in contradiction with $\mu _0\in {\langle X_\mu \rangle _{\equiv }}\subseteq \bigcup _{n=1}^\infty F_n$ .

Since $X\not \simeq \ell _2$ , by the celebrated solution to the homogeneous subspace problem following from the results of Komorowski and Tomczak-Jaegermann ([Reference Komorowski and Tomczak-Jaegermann35]) and of Gowers ([Reference Gowers29]), it must contain an infinite-dimensional closed subspace $Y\subseteq X$ that is not isomorphic to X. Let $I\subseteq \mathbb {N}$ be an infinite subset and $\{x_n\}_{n\in \mathbb {N}}$ a sequence of linearly independent vectors in X so that

1. (1) $\overline {\operatorname {span}}\{x_n\}_{n\in \mathbb {N}}=X$ , $\overline {\operatorname {span}}\{x_n\colon n\in I\}=Y$ , and

2. (2) for every finite set $F\subseteq \mathbb {N}$ , we have

   $$\begin{align*}\overline{\operatorname{span}}\{x_n\}_{n\in I \cup F} = \overline{\operatorname{span}}\{x_n\}_{n\in F\cap I}\oplus \overline{\operatorname{span}}\{x_n\}_{n\in F\setminus I}\oplus \overline{\operatorname{span}}\{x_n\}_{n\in I \setminus F}.\end{align*}$$

Such a choice is possible, for example, by finding a Markushevich basis $\{x_n\}_{n\in I}$ on Y (see [Reference Hájek, Montesinos Santalucía, Vanderwerff and Zizler30, Theorem 1.22]), extending it to a Markushevich basis $\{x_n\}_{n\in \mathbb {N}}$ on X (see [Reference Hájek, Montesinos Santalucía, Vanderwerff and Zizler30, Theorem 1.45]) and then observing that for any $J\subseteq \mathbb {N}$ , we have that $\{x_n\}_{n\in J}$ is a fundamental biorthogonal system of $\overline {\operatorname {span}}\{x_n\}_{n\in J}$ and so (2) holds (see [Reference Hájek, Montesinos Santalucía, Vanderwerff and Zizler30, Fact 1.5]).

We define $\mu \in \mathcal {B}$ as

$$\begin{align*}\mu\Big(\sum_{i=1}^k a_i e_i\Big):=\Big\|\sum_{i=1}^k a_i x_i\Big\|_X,\quad k\in\mathbb{N},\;(a_i)_{i=1}^k\in\mathbb{Q}^k. \end{align*}$$

Then ${\langle X_\mu \rangle _{\equiv }}={\langle X\rangle _{\equiv }}\subseteq \bigcup _{n=1}^\infty F_n$ and so, by the claim above, there exist $\gamma \in \mathcal {T}$ and $m\in \mathbb {N}$ with $\mathbb {M}^{\gamma '}_\mu \cap F_m\neq \emptyset $ for every $\gamma '\in \mathcal {T}$ with $\gamma '\supseteq \gamma $ . Consider now the space $Z:=(\overline {\operatorname {span}}\{e_i\colon i\in I\cup \operatorname {rng}(\gamma )\},\mu )\subseteq X_\mu $ . Fix some $\widetilde I\subseteq I$ such that $|I\setminus \widetilde I|=|\operatorname {rng}(\gamma )\setminus I|$ and such that $(I\setminus \widetilde I)\cap \operatorname {rng}(\gamma )=\emptyset $ . Define $\widetilde Z:=(\overline {\operatorname {span}}\{e_i\colon i\in \widetilde I\cup \operatorname {rng}(\gamma )\},\mu )$ . Then, using (2), we have

$$\begin{align*}\widetilde{Z} \simeq \overline{\operatorname{span}}\{x_n\}_{n\in \widetilde{I}}\oplus \mathbb{R}^{| \operatorname{rng}(\gamma)\setminus\widetilde{I}|}\simeq \overline{\operatorname{span}}\{x_n\}_{n\in \widetilde{I}}\oplus \mathbb{R}^{| I\setminus\widetilde{I}|}\simeq Y, \end{align*}$$

and so $\widetilde Z\not \simeq X$ . Let $\varphi :\mathbb {N}\to \operatorname {rng}(\gamma )\cup \widetilde I$ be a bijection with $\varphi \supseteq \gamma $ . We define $\nu \in \mathcal {B}$ by

$$\begin{align*}\nu\Big(\sum_{i=1}^k a_i e_i\Big):=\mu\Big(\sum_{i=1}^k a_i e_{\varphi(i)}\Big),\quad k\in\mathbb{N},\;(a_i)_{i=1}^k\in\mathbb{Q}^k. \end{align*}$$

Clearly, $X_\nu \equiv \widetilde Z\not \simeq X$ .

We claim that $\nu \in F_m$ . This will be in contradiction with the fact that $F_m\subseteq {\langle X\rangle _{\simeq }}$ . Since $F_m$ is closed, it suffices to check that each basic open neighbourhood of $\nu $ intersects $F_m$ . Pick $v_1,\ldots , v_l\in V$ and $\varepsilon>0$ . We need to find $\mu '\in F_m$ so that $|\mu '(v_j)-\nu (v_j)|<\varepsilon $ for every $j\leq l$ .

Let $L\in \mathbb {N}$ , $L\geq |\gamma |$ , be such that $v_1,\ldots ,v_l\in \operatorname {span}\{e_i\colon i\leq L\}$ . Since $\varphi |_{\{1,\ldots ,L\}}\supseteq \gamma $ , we may pick $\mu '\in \mathbb {M}^{\varphi |_{\{1,\ldots ,L\}}}_\mu \cap F_m$ . Then

$$\begin{align*}\mu'\Big(\sum_{i=1}^L a_i e_i\Big) = \mu\Big(\sum_{i=1}^L a_i e_{\varphi(i)}\Big) = \nu\Big(\sum_{i=1}^L a_i e_i\Big),\quad (a_i)_{i=1}^L\in\mathbb{Q}^L. \end{align*}$$

In particular, $\mu '(v_j)=\nu (v_j)$ , $j\leq l$ , as desired.

Proof of Theorem 2.13.

For each $E_n$ , $n\in \mathbb {N}$ , we fix its basis (which will be used only in order to know what $Z{\stackrel {K}{\sim } E_n}$ means for a finite-dimensional space Z and $K>1$ ). For each finite tuple $\vec v$ of elements from V, we set $S_{\vec v}$ to be the set of all finite tuples $\vec w$ , $\mathbb {Q}$ -linearly independent in V, such that each element of $\vec v$ (considered as an element of $c_{00}$ ) lies in $\operatorname {span}\vec w$ . For $m\in \mathbb {N}$ and a finite tuple $\vec v$ of elements from V, we set

$$\begin{align*}\begin{aligned} P(\vec v,m):=\big\{\mu\in\mathcal{P}\colon & \text{if } \mu \text{ restricted to }\operatorname{span}\{v_1,\ldots,v_{|\vec v|}\}\subseteq c_{00}\text{ is a norm, then }\\ & \text{there exist } \vec w\in S_{\vec v} \text{ and } n\in\mathbb{N}, \text{ such that }\\ &\mu\text{ restricted to }\operatorname{span}\{w_1,\ldots,w_{|\vec w|}\}\subseteq c_{00}\text{ is a norm and }\\ & (((\operatorname{span}\{w_1,\ldots,w_{|\vec w|}\},\mu),\vec w){\stackrel{\sqrt{\lambda + \frac{1}{m}}}{\sim} E_{n}})\big\}. \end{aligned} \end{align*}$$

Then, using the observation that $\{\mu \in \mathcal {P}\colon \mu \text { restricted to }\operatorname {span}\{v_1,\ldots ,v_{|\vec v|}\}\subseteq c_{00}\text { is a norm}\}$ is open due to Lemma 1.5, $P(\vec v,m)$ is the union of a closed and an open set, so it is a $G_\delta $ -set.

Denote by $LI$ the set of all finite tuples $\vec v = (v_1,\ldots ,v_{|\vec v|})$ of elements from V, which are linearly independent in $c_{00}$ . We now set

$$\begin{align*}\mathcal{G}:=\bigcap_{m\in\mathbb{N}} \bigcap_{\vec v\in LI} P(\vec v,m), \end{align*}$$

which is clearly a $G_{\delta }$ -set. We shall prove that $\mathcal {G}=\mathcal {Z}$ .

If $\mu \in \mathcal {G}$ , it is clear that for every m, we can recursively build an increasing sequence $\{F_k\}_{k=1}^\infty $ of finite-dimensional subspaces whose union is dense in $X_\mu $ , such that we have $d_{BM}(F_k,E_{n_k})\leq \lambda + \frac {1}{m}$ for every $k\in \mathbb {N}$ . It follows that $\mu \in \mathcal {Z}$ .

On the other hand, pick $\mu \in \mathcal {Z}$ . In what follows for $x\in c_{00}$ , we denote by $[x]\in X_\mu $ the equivalence class corresponding to x. Pick some $m\in \mathbb {N}$ and an n-tuple $\vec v\in LI$ , such that $\mu $ restricted to $\operatorname {span}\{v_1,\ldots ,v_n\}\subseteq c_{00}$ is a norm, so $\{v_1,\ldots ,v_n\}$ is a basis of $\operatorname {span}\{v_1,\ldots ,v_n\}$ . Pick $\lambda '\in (\lambda ,\lambda + \tfrac {1}{m})$ and $\delta>1$ with $\delta \lambda '<\lambda + \frac {1}{m}$ . Since $\mu \in \mathcal {Z}$ , there is an increasing sequence $\{F_k\}_{k=1}^\infty $ of finite-dimensional subspaces whose union is dense in $X_\mu $ , such that $\sup _{k\in \mathbb {N}}d_{BM}(F_k,E_{n_k})\leq \lambda '$ . By [Reference Lacey38, Section 17, Theorem 6], we can find a finite dimensional subspace $\operatorname {span}\{[v_1],\ldots ,[v_n]\}\subseteq Y\subseteq X_\mu $ and $k\in \mathbb {N}$ , such that $d_{BM}(Y,F_k)\leq \delta $ so $d_{BM}(Y,E_{n_k})\leq \delta \lambda '$ . Select $y_{n+1},\ldots , y_{\dim Y}\in Y$ , such that $\mathfrak {b} = \{[v_1],\ldots , [v_n], y_{n+1},\ldots , y_{\dim Y}\}$ is a basis of Y. Let $\phi _2^{\mathfrak {b}}$ be the function from Lemma 2.14, and let $\eta>0$ be such that $\delta \lambda '(1+\phi _2^{\mathfrak {b}}(\eta ))^2<\lambda +\frac {1}{m}$ . Further, for every $n+1\leq i\leq \dim Y$ , pick $v_i\in V$ with $\|[v_i]-y_i\|_{X_\mu }<\eta $ . Then $(\mu ,[v_1],\ldots ,[v_{\dim Y}]){\stackrel {1+\phi _2^{\mathfrak {b}}(\eta )}{\sim } Y}$ so $\mu $ restricted to $\operatorname {span}\{v_1,\ldots ,v_{\dim Y}\}\subseteq c_{00}$ is a norm and $\operatorname {span}\{[v_1],\ldots ,[v_{\dim Y}]\}\subseteq X_\mu $ is isometric to $(\operatorname {span}\{v_1,\ldots ,v_{\dim Y}\},\mu )$ . Since $d_{BM}((\operatorname {span}\{v_1,\ldots ,v_{\dim Y}\},\mu ),E_{n_k})<\delta \lambda '\cdot (1+\phi _2^{\mathfrak {b}}(\eta ))^2<\lambda + \frac {1}{m}$ , there exists a surjective isomorphism $T:E_{n_k}\to (\operatorname {span}\{v_1,\ldots ,v_{\dim Y}\},\mu )$ with $\max \{\|T\|,\|T^{-1}\|\}<\sqrt {\lambda + \frac {1}{m}}$ . By Lemma 2.14, we may without loss of generality assume that $w_i:=T(e_i)\in V$ for every $i\leq \dim Y$ . Then $\mu $ restricted to $\operatorname {span}\{w_1,\ldots ,w_{\dim Y}\}\subseteq c_{00}$ is a norm, $\vec w\in S_{\vec v}$ and $((\mu ,\vec w){\stackrel {\sqrt {\lambda + \frac {1}{m}}}{\sim } E_{n_k}})$ .

Proof of Theorem 2.12.

Pick a paving $\{E_n\}_{n=1}^\infty $ of X. It is easy to see that for every $\mu \in \mathcal {P}_\infty $ , the Banach space $X_\mu $ is isomorphic to X if and only if $\mu $ belongs to the set $\mathcal {Z}$ from Theorem 2.13 for some $\lambda \geq 1$ . The ‘In particular’ part follows, since $\mathcal {P}_\infty $ is a $G_\delta $ -set in $\mathcal {P}$ . For admissible topologies, the result follows by applying [Reference Cúth, Doležal, Doucha and Kurka14, Theorem 3.3].

Proof. This is an immediate consequence of Theorem 2.13, Remark 3.5.

Proof. Note that any finite-dimensional space is an $\mathcal {L}_p$ -space, so the class of separable $\mathcal {L}_p$ -spaces may be written as the union of $\mathcal {P}\setminus \mathcal {P}_\infty $ (which is an $F_\sigma $ -set by [Reference Cúth, Doležal, Doucha and Kurka14, Corollary 2.5]) and $\{\mu \in \mathcal {P}_\infty \colon X_\mu \text { is an } \mathcal {L}_p\text {-space}\}$ (which is $G_{\delta \sigma }$ -set by Theorem 3.6 above).

Proof. By [Reference Lindenstrauss and Pełczyński41, Section 7, Corollaries 4 and 5], a separable Banach space is an $\mathcal {L}_{p,1+}$ -space if and only if it is isometric to an $L_p(\mu )$ space for some measure $\mu $ . Finally, note that every separable infinite-dimensional $L_p(\mu )$ space is isometric to one of the spaces mentioned above (see, e.g. [Reference Albiac and Kalton1, pages 137–138]).

Proof. $(i)\implies (ii)$ : Pick $f\in S_{L_p[0,1]}$ and $N\in \mathbb {N}$ . Then, using the continuity of the mapping $[0,1]\ni x\mapsto \int _0^x |f|$ , we find $0=x_0<x_1<\ldots <x_N=1$ , such that $\int _{x_{i-1}}^{x_i} |f|^p = \frac {1}{N} \int _0^1 |f|^p$ for every $i=1,\ldots ,N$ . We put $f_i:=N^{1/p}\cdot f\cdot \chi _{[x_{i-1},x_i]}$ , $i=1,\ldots ,N$ . Then, since the supports of $f_i$ are disjoint and since $f_i$ are normalized, we have $(f_i)_{i=1}^N \equiv \ell _p^N$ . Further, we obviously have $N^{1/p}\cdot f = \sum _{i=1}^N f_i$ .

Obviously, we have $(ii)\implies (iii)$ and $(iii)\implies (iv)$ .

$(iii)\implies (i)$ : In order to get a contradiction, let us assume that X is not isometric to $L_p[0,1]$ , which, by Theorem 3.8, implies that X is isometric to $L_p(\mu )$ , where $(\Omega ,S,\mu )$ is a measure space for which there is $\omega \in \Omega $ with $\mu (\{\omega \}) = 1$ . Fix $\varepsilon> 0$ small enough (to be specified later). Suppose, to the contrary, that there are $f,g\in L_p(\mu )$ , such that $(f,g){\stackrel {1+\varepsilon }{\sim } \ell _p^2}$ and $2^{\frac 1p}\cdot \delta _{\omega }=f+g$ , where $\delta _{\omega }$ is the Dirac function supported by the point $\omega $ . For $\mu $ -a.e. $x\in \Omega \setminus \{\omega \}$ , we have $f(x)+g(x)=0$ , so we assume this holds for all $x\in \Omega \setminus \{\omega \}$ . We without loss of generality assume that $f(\omega )\geq g(\omega )$ .

We claim that both $f(\omega )$ and $g(\omega )$ are positive and $|f(\omega ) - g(\omega )|^p<\frac {1}{2}$ if $\varepsilon>0$ is chosen sufficiently small. Indeed, we have

$$\begin{align*}(1+\varepsilon)^p - \frac{1}{(1+\varepsilon)^p}\geq \Big|\|f\|_p^p - \|g\|_p^p\Big| = \Big||f(\omega)|^p - |g(\omega)|^p\Big|,\end{align*}$$

which implies $||f(\omega )| - |g(\omega )||<2^{-1/p}$ for sufficiently small $\varepsilon>0$ . The claim follows, since if both $f(\omega )$ and $g(\omega )$ were not positive, we would have $2^{1/p}> 2^{-1/p} > ||f(\omega )| - |g(\omega )|| = |(f+g)(\omega )| = 2^{1/p}$ , a contradiction.

First, let us handle the case when $1\leq p<2$ . We have

$$\begin{align*}\begin{aligned} \|2f\|_p^p & = \int_{\Omega\setminus\{\omega\}} |2f|^p \, \mathrm{d} \mu + (2f(\omega))^p = \int_{\Omega\setminus\{\omega\}} |f-g|^p \, \mathrm{d} \mu + (2f(\omega))^p\\ & = \|f-g\|_p^p + (2f(\omega))^p - ((f-g)(\omega))^p\\ & \geq \|f-g\|_p^p + ((f+g)(\omega))^p = \|f-g\|_p^p + \|f+g\|_p^p, \end{aligned} \end{align*}$$

where in the inequality we used superadditivity of the function $[0,\infty )\ni t\mapsto t^p$ . Thus, $(f,g){\stackrel {1+\varepsilon }{\sim } \ell _2^p}$ implies

$$\begin{align*}(1+\varepsilon)^p\geq \|f\|_p^p\geq \frac{\|f-g\|_p^p + \|f+g\|_p^p}{2^p}\geq \frac{4}{2^p(1+\varepsilon)^p}; \end{align*}$$

hence, if $1\leq p<2$ , we get a contradiction for sufficiently small $\varepsilon>0$ .

Finally, let us handle the case when $p>2$ . Note that since $f(\omega )\geq g(\omega )\geq 0$ and $f(\omega ) + g(\omega ) = 2^{1/p}$ , we have $g(\omega )\leq 2^{1/p-1}$ . Further, we have

$$\begin{align*}\begin{aligned} \|2g\|_p^p & = \int_{\Omega\setminus\{\omega\}} |f-g|^p \, \mathrm{d} \mu + (2g(\omega))^p \leq \|f-g\|_p^p + 2. \end{aligned}\end{align*}$$

Thus, $(f,g){\stackrel {1+\varepsilon }{\sim } \ell _2^p}$ implies

$$\begin{align*}\frac{1}{(1+\varepsilon)^p}\leq \|g\|_p^p\leq \frac{\|f-g\|_p^p + 2}{2^p}\leq \frac{2(1+\varepsilon)^p + 2}{2^p}; \end{align*}$$

hence, if $p>2$ , we get a contradiction for sufficiently small $\varepsilon>0$ .

$(iv)\implies (iii)$ : Fix $x\in S_X$ and $\varepsilon> 0$ . Pick $\delta>0$ small enough (to be specified later). Applying Condition $(iv)$ , we obtain $x_1^{\prime },x_2^{\prime }\in X$ , such that $(x_1^{\prime},x_2^{\prime}){\stackrel{1 + \tfrac{\varepsilon}{2}}{\sim}} \ell_p^2$ and $\|2^{1/p}\cdot x - x_1^{\prime } - x_2^{\prime }\| < \delta $ . Now, set $x_1=x^{\prime }_1+(2^{1/p}\cdot x-(x^{\prime }_1+x^{\prime }_2))/2$ and $x_2=x^{\prime }_2+(2^{1/p}\cdot x-(x^{\prime }_1+x^{\prime }_2))/2$ . If $\delta $ was chosen sufficiently small, we have $(x_1,x_2){\stackrel {1+\varepsilon }{\sim } \ell _p^2}$ , and clearly, $2^{1/p}\cdot x=x_1+x_2$ .

Fact 3.10. Let $v,w\in V$ , $v\neq 0$ and $a,b\in \mathbb {R}$ . Then the set

$$\begin{align*}\left\{\mu\in\mathcal{P}:\; \mu(v)\neq 0\text{ and }\mu(a\cdot\frac{v}{\mu(v)} - w)<b\right\}\end{align*}$$

is open in $\mathcal {P}$ .

Proof of Theorem 3.4.

Let $\mathcal {F}$ be the set of those $\nu \in \mathcal {P}_\infty $ for which $X_\nu $ is an $\mathcal {L}_{p,1+}$ -space. By Theorem 3.6, $\mathcal {F}\subseteq \mathcal {P}_\infty $ is a $G_\delta $ -set. By Theorem 3.9, using the obvious observation that Condition $(iv)$ may be verified on a dense subset, we have

$$\begin{align*}{\langle L_p[0,1]\rangle_{\equiv}^{\mathcal{P}_\infty}} = \mathcal{F}\cap\bigcap_{v\in V}\bigcap_{n,k\in\mathbb{N}} U_{v,n,k},\end{align*}$$

where $U_{v,n,k}$ are open sets (using Fact 3.10 and Lemma 1.5) defined as

$$\begin{align*}U_{v,n,k}:=\Big\{\mu\in\mathcal{P}_\infty:\; \exists v_1,v_2\in V:\; (v_1,v_2) \stackrel{1+\tfrac{1}{n}}{\sim} \ell_p^2 \text{ and }\mu\big(2^{1/p}\cdot \frac{v}{\mu(v)} - v_1 - v_2\big) < \frac{1}{k}\Big\}.\end{align*}$$

Thus, ${\langle L_p[0,1]\rangle _{\equiv }^{\mathcal {P}_\infty }}$ is a $G_\delta $ -set.

On the other hand, since any $L_p(\mu )$ is finitely representable in $\ell _p$ and vice versa (see, e.g. [Reference Albiac and Kalton1, Proposition 12.1.8]), from Lemma 3.1 and Theorem 3.8, we obtain that there is at most one (up to isometry) $\mathcal {L}_{p,1+}$ space X, such that ${\langle X\rangle _{\equiv }}$ is a $G_\delta $ -set in $\mathcal {B}$ and that ${\langle L_p[0,1]\rangle _{\equiv }}$ is a $G_\delta $ -complete set.

Proof. $(i)\implies (ii)$ : Fix $x\in S_{\ell _p}$ and $\delta \in (0,1)$ . Pick $l\in \mathbb {N}$ with $\sum _{k=1}^l |x(k)|^p> \delta ^p$ and $N\in \mathbb {N}$ , such that $\sum _{k=1}^l (|x(k)| - \frac {3}{\sqrt [p]{N}})^p> \delta ^p$ . Fix a sequence $(\varepsilon _m)_{m\in \mathbb {N}}\in (0,1)^{\mathbb {N}}$ with $\varepsilon _m\to 0$ . In order to get a contradiction, for every $m\in \mathbb {N}$ , pick $x_1^{\varepsilon _m},\ldots ,x_N^{\varepsilon _m}\in \ell _p$ , such that $(N^{1/p}\cdot x_i^{\varepsilon _m})_{i=1}^N{\stackrel {1+\varepsilon _m}{\sim } \ell _p^N}$ and $\|x-\sum _{i=1}^N x_i^{\varepsilon _m}\|\leq \delta $ for every $m\in \mathbb {N}$ .

We claim that there is $m\in \mathbb {N}$ , such that $|x_i^{\varepsilon _m}(k)x_j^{\varepsilon _m}(k)|<\eta :=N^{-(2+2/p)}$ for every $i,j\in \{1,\ldots ,N\}$ , $i\neq j$ , and $k\in \{1,\ldots ,l\}$ . Indeed, otherwise, there are $i,j,k$ , such that $|x_i^{\varepsilon _m}(k)x_j^{\varepsilon _m}(k)|\geq \eta $ for infinitely many m’s. By passing to a subsequence, we may assume that this holds for every $m\in \mathbb {N}$ . Since the sequences $(|x_i^{\varepsilon _m}(k)|)_m$ and $(|x_j^{\varepsilon _m}(k)|)_m$ are bounded, by passing to a subsequence, we may assume there are numbers $a,b\in \mathbb {R}$ with $x_i^{\varepsilon _m}(k)\to a$ , $x_j^{\varepsilon _m}(k)\to b$ and $|ab|\geq \eta> 0$ . Since $(N^{1/p}\cdot x_i^{\varepsilon _m},N^{1/p}\cdot x_j^{\varepsilon _m}){\stackrel {1+\varepsilon _m}{\sim } \ell _p^2}$ , using Lemma 4.5, for $p>2$ , we obtain

$$\begin{align*}\begin{aligned} 0 & \leq |a+b|^p + |a-b|^p - 2|a|^p - 2|b|^p\\ & = \lim_m \big(|x_i^{\varepsilon_m}(k) + x_j^{\varepsilon_m}(k)|^p + |x_i^{\varepsilon_m}(k) - x_j^{\varepsilon_m}(k)|^p - 2|x_i^{\varepsilon_m}(k)|^p - 2|x_j^{\varepsilon_m}(k)|^p\big)\\ & \leq \lim_m \big(\|x_i^{\varepsilon_m} + x_j^{\varepsilon_m}\|^p + \|x_i^{\varepsilon_m} - x_j^{\varepsilon_m}\|^p - 2\|x_i^{\varepsilon_m}\|^p - 2\|x_j^{\varepsilon_m}\|^p\big) = 0; \end{aligned} \end{align*}$$

hence, $|a+b|^p + |a-b|^p=2|a|^p+2|b|^p=0$ , which, by Lemma 4.5, is in contradiction with $|ab|>0$ . The case when $p<2$ is similar.

From now on, we write $x_i$ instead of $x_i^{\varepsilon _m}$ , where $m\in \mathbb {N}$ is chosen to satisfy the claim above. Fix $k\leq l$ . By the claim above, there is at most one $i_0\in \{1,\ldots ,N\}$ with $|x_{i_0}(k)| \geq \sqrt {\eta }$ , and for this $i_0$ , we have $|x_{i_0}(k)|\leq \|x_{i_0}\|\leq 2N^{-1/p}$ . Consequently, we have

$$\begin{align*}\sum_{i=1}^N |x_i(k)|\leq \frac{2}{N^{1/p}} + \sum_{i\in \{1,\ldots,N\}\& |x_i(k)|<\sqrt{\eta}} |x_i(k)|\leq \frac{2}{N^{1/p}} + N\cdot\sqrt{\eta} = \frac{3}{N^{1/p}}. \end{align*}$$

Thus, we have

$$\begin{align*}\|x-\sum_{i=1}^N x_i\|^p\geq \sum_{k=1}^l \Big(|x(k)| - \sum_{i=1}^N |x_i(k)|\Big)^p\geq \sum_{k=1}^l \Big(|x(k)| - \frac{3}{N^{1/p}}\Big)^p> \delta^p,\end{align*}$$

which is in contradiction with $\|x-\sum _{i=1}^N x_i\| = \|x-\sum _{i=1}^N x_i^{\varepsilon _m}\|\leq \delta $ .

$(ii)\implies (iii)$ is obvious.

$(iii)\implies (i)$ : Suppose that X is not isometric to $\ell _p$ . By Theorem 3.8, X is isometric to $L_p[0,1]\oplus _p Y$ for some (possibly trivial) Banach space Y. By abusing the notation, we may assume that $X=L_p[0,1]\oplus _p Y$ . Let $\mathbf {1}\in L_p[0,1]$ be the constant 1 function, and define $x\in X=L_p[0,1]\oplus _p Y$ by $x=(\mathbf {1},0)$ . Now, fix $N\in \mathbb {N}$ arbitrarily. Define $x_1,\ldots ,x_N\in X$ by $x_i=(\chi _{[\frac {i-1}n,\frac in]},0)$ . Clearly, $(N^{1/p}\cdot x_i)_{i=1}^N\equiv \ell ^n_p$ , and we have $x = \sum _{i=1}^N x_i$ .

$(ii)\implies (iv)$ is obvious, so it only remains to show that $(iv)\implies (ii)$ . For every $x\in X\setminus \{0\}$ , $\delta \in (0,1)$ , $N\in \mathbb {N}$ , $\varepsilon>0$ and $x_1,\ldots ,x_N\in X$ , we denote by $V(x,\delta ,N,\varepsilon ,(x_i)_{i=1}^N)$ the assertion that if $(N^{1/p}\cdot x_i)_{i=1}^N{\stackrel {1+\varepsilon }{\sim } \ell _p^N}$ , then $\|\frac x{\|x\|}-\sum _{i=1}^N x_i\|\ge \delta $ . The desired implication straightforwardly follows by the following two easy observations. First, if $x\in D\setminus \{0\}$ , $\delta $ , N and $\varepsilon $ are given, such that $V(x,\delta ,N,\varepsilon ,(x_i)_{i=1}^N)$ holds for every $x_1,\ldots ,x_N\in D$ , then $V(x,\delta ,N,\varepsilon ,(x_i)_{i=1}^N)$ holds for every $x_1,\ldots ,x_N\in X$ . Second, if for every $x\in D\setminus \{0\}$ and $\delta $ , there are N and $\varepsilon $ , such that $V(x,\frac {1+\delta }2,N,\varepsilon ,(x_i)_{i=1}^N)$ holds for every $x_1,\ldots ,x_N\in X$ , then for every $x\in X\setminus \{0\}$ and $\delta $ , there are N and $\varepsilon $ , such that $V(x,\delta ,N,\varepsilon ,(x_i)_{i=1}^N)$ holds for every $x_1,\ldots ,x_N\in X$ .

Proof of Proposition 4.4.

Let $\mathcal {F}$ be the set of those $\nu \in \mathcal {P}_\infty $ for which $X_\nu $ is an $\mathcal {L}_{p,1+}$ -space. By Theorem 3.6, $\mathcal {F}\subseteq \mathcal {P}_\infty $ is a $G_\delta $ -set. By Proposition 4.6 $(i)\Leftrightarrow (iv)$ , we have

$$\begin{align*}{\langle \ell_p\rangle_{\equiv}^{\mathcal{P}_\infty}} = \mathcal{F}\cap\bigcap_{v\in V\setminus\{0\}}\bigcap_{m\in\mathbb{N}}\bigcup_{n,k\in\mathbb{N}}V_{v,m,n,k},\end{align*}$$

where the closed (see Fact 3.10 and Lemma 1.5) sets $V_{v,m,n,k}$ are given by

$$\begin{align*}\begin{aligned} V_{v,m,n,k}:=\Big\{\mu\in\mathcal{P}_\infty:\; & \mu(v)=0\text{, or for every } (v_i)_{i=1}^n\in V^n, \text{ we have}\\ & \neg\left((\sqrt[p]nv_i)_{i=1}^n{\stackrel{1+\frac 1k}{\sim} \ell_p^n}\right) \text{ or }\mu\big(\frac{v}{\mu(v)} - \sum_{i=1}^nv_i\big)\ge\frac 1m\Big\}. \end{aligned} \end{align*}$$

Thus, ${\langle \ell _p\rangle _{\equiv }^{\mathcal {P}_\infty }}$ is an $F_{\sigma \delta }$ -set.

Proof. That the mapping is $w^*$ - $w^*$ homeomorphism onto its image follows from the fact that $B_{X^*}$ is $w^*$ -compact and the mapping is one-to-one (because $(x_n)$ separate the points of $B_{X^*}$ ) and $w^*$ - $w^*$ continuous (because on $B_{\ell _\infty }$ , the $w^*$ -topology coincides with the topology of pointwise convergence). It is also straightforward to see that the mapping is isometry. Thus, it suffices to prove that

$$\begin{align*}\{(x^*(x_n))_{n=1}^\infty\colon x^*\in B_{X^{*}}\} = \Omega(X). \end{align*}$$

The inclusion $ \subseteq $ is easy, let us prove $ \supseteq $ . Given numbers $ a_{1}, a_{2}, \dots $ satisfying $ | \sum _{n \in M} a_{n} | \leq \Vert \sum _{n \in M} x_{n} \Vert $ for any finite $ M \subseteq \mathbb {N} $ , we need to find $ x^{*} \in B_{X^{*}} $ , such that $ x^{*}(x_{n}) = a_{n} $ for each $ n $ .

Let us realize first that

• • $ |a_{n} - a_{m}| \leq \Vert x_{n} - x_{m} \Vert $ for every $ n, m $ ,

• • $ |a_{n} + a_{m} - a_{l}| \leq \Vert x_{n} + x_{m} - x_{l} \Vert $ for every $ n, m, l $ .

We check the first inequality only, the second inequality can be checked in the same way. Given $ \varepsilon> 0 $ , let $ n' $ different from $ n $ and $ m $ be such that $ \Vert x_{n} + x_{n'} \Vert < \varepsilon $ . We obtain $ |a_{n} - a_{m}| = |(a_{n} + a_{n'}) - (a_{m} + a_{n'})| \leq |a_{n} + a_{n'}| + |a_{m} + a_{n'}| \leq \Vert x_{n} + x_{n'} \Vert + \Vert x_{m} + x_{n'} \Vert \leq 2 \Vert x_{n} + x_{n'} \Vert + \Vert x_{m} - x_{n} \Vert < 2 \varepsilon + \Vert x_{m} - x_{n} \Vert $ . Since $ \varepsilon> 0 $ was chosen arbitrarily, we arrive at $ |a_{n} - a_{m}| \leq \Vert x_{m} - x_{n} \Vert $ . It follows that there is a function $ f : B_{X} \to \mathbb {R} $ with the Lipschitz constant $ 1 $ , such that $ f(x_{n}) = a_{n} $ for each $ n $ . We claim that $ f(u + v) = f(u) + f(v) $ and $f(\alpha u)=\alpha f(u)$ , whenever $ u, \alpha u, v, u + v \in B_{X} $ . Given $ \varepsilon> 0 $ , we pick $ n, m, l $ , such that $ \Vert x_{n} - u \Vert < \varepsilon , \Vert x_{m} - v \Vert < \varepsilon $ and $ \Vert x_{l} - (u + v) \Vert < \varepsilon $ . Then, $ |f(u) + f(v) - f(u + v)| < |a_{n} + a_{m} - a_{l}| + 3\varepsilon \leq \Vert x_{n} + x_{m} - x_{l} \Vert + 3\varepsilon \leq \Vert u + v - (u + v) \Vert + 3\varepsilon + 3\varepsilon = 6\varepsilon $ . Since $ \varepsilon> 0 $ was chosen arbitrarily, we arrive at $ |f(u) + f(v) - f(u + v)| = 0 $ . This also shows that $f(u/2)=f(u)/2$ , therefore, $f(\alpha u)=\alpha f(u)$ , provided that $\alpha $ is a dyadic rational number. For a general $\alpha $ , we use density of dyadic rationals and continuity of f.

Now, it is easy to see that f uniquely extends to a linear functional on X.

Convention. Whenever we talk about open (closed, $F_\sigma $ , etc.) subsets of $B_{l_\infty }$ , we always mean open (closed, $F_\sigma $ , etc.) subsets in the weak* topology. On the other hand, whenever we talk about the diameter of a subset of $B_{l_\infty }$ , or about the distance of two subsets of $B_{l_\infty }$ , we always mean the diameter, or the distance, with respect to the metric given by the norm of $\ell _\infty $ . Also, we write only $\mathcal K(B_{l_\infty })$ instead of $\mathcal K(B_{l_\infty },w^*)$ .

Proof. The sets of the form

$$\begin{align*}\{F\in\mathcal K(X)\colon F\subseteq W\}\;\;\;\text{ and }\;\;\;\{F\in\mathcal K(X)\colon F\cap W\neq\emptyset\}, \end{align*}$$

where W ranges over all open subsets of X, form a subbasis of the topology of $\mathcal K(X)$ . So we only need to check that $f^{-1}(U)$ is a $\boldsymbol {\Sigma }_{\alpha +1}^0$ -set for every open set U of one of these forms. For the first case, this follows immediately from the assumptions, and for the second case, if $\{W_n\colon n\in \mathbb {N}\}$ is an open basis for the topology of X, we have

$$\begin{align*}f^{-1}(\{F\in\mathcal K(X)\colon F\cap W\neq\emptyset\}) = \bigcup _{n\in\mathbb{N} \text{, such that } \overline{W_n}\subseteq W} P\setminus \{p\in P\colon f(p)\subseteq X\setminus \overline{W_n}\}, \end{align*}$$

which, by the assumptions, is the countable union of sets from $\boldsymbol {\Sigma }_\alpha ^0(P)\cup \boldsymbol {\Pi }_\alpha ^0(P)$ .

Proof. First of all, we describe the choice of the sets $\{x_n^\nu \colon n\in \mathbb N\}$ , $\nu \in \mathcal P$ . Let $g\colon [0,\infty )\rightarrow [1,\infty )$ be given by $g(t)=1$ for $t\le 1$ and $g(t)=t$ for $t>1$ . Let $\{v_n\colon n\in \mathbb N\}$ be an enumeration of all elements of the vector space V (which is naturally embedded into all Banach spaces $X_\nu $ , $\nu \in \mathcal P$ ). Now, for every $\nu \in \mathcal P$ and every $n\in \mathbb N$ , we define $x_n^\nu \in B_{X_\nu }$ by $x_n^\nu =\frac {v_n}{g(\nu (v_n))}$ . Then, for every $\nu \in \mathcal P$ , we have that $\{x_n^\nu \colon n\in \mathbb N\}$ is a dense subset of $B_{X_\nu }$ . Note also that the set

$$\begin{align*}\left\{\left(\nu,(a_n)_{n=1}^\infty\right)\in\mathcal P\times B_{l_\infty}\colon\left|\sum\limits_{n\in M}a_n\right|>\nu\left(\sum\limits_{n\in M}x_n^\nu\right)\right\} \end{align*}$$

is open in $\mathcal P\times (B_{l_\infty },w^*)$ for every $M\subseteq \mathbb {N}$ finite (the proof is easy and is omitted).

Pick an open subset U of $B_{\ell _{\infty }}$ . We have

$$ \begin{align*} &\Omega^{-1}\left(\{F\in\mathcal K(B_{l_\infty})\colon F\subseteq U\}\right)\\ & \quad =\left\{\nu\in\mathcal P\colon\mathop{\forall}\limits_{(a_n)_{n=1}^\infty\in B_{l_\infty}}\left(\left(\mathop{\exists}\limits_{\substack{M\subseteq\mathbb N\\ \text{finite}}}\left|\sum\limits_{n\in M}a_n\right|>\nu\left(\sum\limits_{n\in M}x_n^\nu\right)\right)\text{ or }\left((a_n)_{n=1}^\infty\in U\right) \right)\right\}. \end{align*} $$

The complement of the last set is the projection of a closed subset of $\mathcal P\times (B_{l_\infty },w^*)$ onto the first coordinate. As the space $(B_{l_\infty },w^*)$ is compact, the complement is a closed subset of $\mathcal P$ . It remains to show that the set $\{\nu \in \mathcal {P}\colon \Omega (\nu )\cap U\neq \emptyset \}$ is open. Pick $\nu \in \mathcal {P}$ with $\Omega (\nu )\cap U\neq \emptyset $ . By Lemma 4.7, there exists $x^*\in B_{X_\nu ^*}$ , such that the sequence $(a_n)_{n=1}^\infty $ given by $a_n=x^*(x_n^\nu )$ , $n\in \mathbb {N}$ , satisfies $(a_n)_{n=1}^\infty \in \Omega (\nu )\cap U$ . Let $\varepsilon>0$ and $N\in \mathbb {N}$ be such that $(b_n)_{n=1}^\infty \in \ell _\infty $ is an element of U whenever $|b_n-a_n|<\varepsilon $ for every $1\leq n\leq N$ . Let us consider subspaces of $c_{00}$ , given as $E=\operatorname {span}\{v_1,\ldots ,v_N\}$ and $F=\{x\in E\colon \nu (x)=0\}$ . Let G be such that $F\oplus G = E$ and $\overline {G\cap V}=G$ (it is enough to pick a basis of E consisting of vectors from V, and using the Gauss elimination to determine which vectors from the basis generate the algebraic complement to F). Let $P_F:E\to F$ and $P_G:E\to G$ be linear projections onto F and G, respectively. Pick $\delta < \min \{1,\tfrac {\varepsilon }{3}\}$ , such that $ \delta \cdot | x^{*}(P_{G}v_{n}) | < \varepsilon / 3 $ for every $1\leq n\leq N$ . Finally, put

$$\begin{align*}\begin{aligned} \mathcal{O}:=& \{\nu'\in\mathcal{P}\colon \frac{1}{1-\delta}\nu(x)>\nu'(x)>(1-\delta)\nu(x)\text{ for every } x\in G\setminus \{ 0 \}\}\cap\\ & \qquad \bigcap_{n=1}^N\{\nu'\in\mathcal{P}\colon \nu'(P_{F}v_{n}) < \delta, | \nu'(v_{n}) - \nu(v_{n}) | < \delta\}. \end{aligned}\end{align*}$$

Then, $\mathcal {O}$ is an open neighbourhood of $\nu $ , which easily follows from Lemma 1.5 and the fact that $G\cap V$ is dense in G.

We will show that $\mathcal {O}\subseteq \{\nu '\in \mathcal {P}\colon \Omega (\nu ')\cap U\neq \emptyset \}$ . Pick $\nu '\in \mathcal {O}$ . If we put $y^*(x):= (1-\delta )x^*(x)$ for $x\in G$ , then $|y^*(x)|=(1-\delta )|x^*(x)| \leq (1 - \delta ) \nu (x) \leq \nu '(x) $ for every $ x \in G $ , and so by the Hahn-Banach theorem, we may extend $y^*$ to a functional (denoted again by $y^*$ ) from the dual unit ball of $X_{\nu '}$ . By Lemma 4.7, the sequence $(b_n)_{n=1}^\infty $ given by $b_n:=y^*(x_n^{\nu '})$ , $n\in \mathbb {N}$ , is in $\Omega (\nu ')$ . Moreover, for every $1\leq n\leq N$ , we have

$$\begin{align*}\begin{aligned} |b_n-a_n| & = \big| \tfrac{1}{g(\nu'(v_{n}))} y^{*}(v_{n}) - \tfrac{1}{g(\nu(v_{n}))} x^{*}(v_{n}) \big|\\[3pt] & = \big| \tfrac{1}{g(\nu'(v_{n}))} y^{*}(P_{F}v_{n}) + \tfrac{1}{g(\nu'(v_{n}))} y^{*}(P_{G}v_{n}) - \tfrac{1}{g(\nu(v_{n}))} x^{*}(P_{F}v_{n} + P_{G}v_{n}) \big|\\[3pt] & = \big| \tfrac{1}{g(\nu'(v_{n}))} y^{*}(P_{F}v_{n}) + \tfrac{1}{g(\nu'(v_{n}))} (1-\delta)x^*(P_{G}v_{n}) - \tfrac{1}{g(\nu(v_{n}))} x^{*}(P_{G}v_{n}) \big|\\[3pt] & \leq \tfrac{1}{g(\nu'(v_{n}))} |y^{*}(P_{F}v_{n})| + \big| \tfrac{1}{g(\nu'(v_{n}))} (1 - \delta) - \tfrac{1}{g(\nu(v_{n}))} \big| | x^{*}(P_{G}v_{n}) |\\[3pt] & \leq \delta + \big(|g(\nu(v_{n})) - g(\nu'(v_{n}))| + \delta\big)|x^{*}(P_{G}v_{n})|\\[3pt] & \leq \delta + 2\delta |x^*(P_{G}v_{n})| < \varepsilon, \end{aligned} \end{align*}$$

and so $(b_n)_{n=1}^\infty \in \Omega (\nu ')\cap U$ . Hence, $\mathcal {O}\subseteq \{\nu '\in \mathcal {P}\colon \Omega (\nu ')\cap U\neq \emptyset \}$ , so $\{\nu \in \mathcal {P}\colon \Omega (\nu )\cap U\neq \emptyset \}$ is an open set and $\Omega $ is a continuous mapping.

Proof. First, we claim that the set

$$\begin{align*}\{F\in\mathcal K(B_{l_\infty})\colon\text{diam}(U\cap F)<\varepsilon\} \end{align*}$$

is an $F_\sigma $ -set for every open subset U of $B_{l_\infty }$ . Indeed, the set above equals

$$ \begin{align*} &\bigcup_{k=1}^\infty\{F\in\mathcal K(B_{l_\infty})\colon\text{diam}(U\cap F)\le\varepsilon-\tfrac 1k\}\\ & \quad =\bigcup_{k=1}^\infty\;\bigcap_{\substack{O_1,O_2\text{ open subsets of }U\\ \text{dist}(O_1,O_2)\ge\varepsilon-\tfrac 1k}}\;\{F\in\mathcal K(B_{l_\infty})\colon F\cap O_1=\emptyset\text{ or }F\cap O_2=\emptyset\}, \end{align*} $$

and our claim immediately follows.

Now, let W be an open subset of $B_{l_\infty }$ . Let $\{U_n\colon n\in \mathbb N\}$ be an open basis for the weak* topology of $B_{l_\infty }$ . Then, we have (using a compactness argument in the last equality) that

$$ \begin{align*} &s_\varepsilon^{-1}(\{F\in\mathcal K(B_{l_\infty})\colon F\subseteq W\})\\ & \quad =\{F\in\mathcal K(B_{l_\infty})\colon\mathop{\exists}\limits_{M\subseteq\mathbb N}\big(\big(\mathop{\forall}\limits_{n\in M}\text{diam}(U_n\cap F)<\varepsilon\big)\text{ and }\big(F\subseteq W\cup\bigcup_{n\in M}U_n\big)\big)\}\\ & \quad =\{F\in\mathcal K(B_{l_\infty})\colon\mathop{\exists}\limits_{\substack{M\subseteq\mathbb N\\ \text{finite}}}\big(\big(\mathop{\forall}\limits_{n\in M}\text{diam}(U_n\cap F)<\varepsilon\big)\text{ and }\big(F\subseteq W\cup\bigcup_{n\in M}U_n\big)\big)\}, \end{align*} $$

and our previous claim implies that the last set is an $F_\sigma $ -set.

Thus, by Lemma 4.8, the mapping $s_\varepsilon $ is $\boldsymbol {\Sigma }_3^0$ -measurable.

Proof. First, we show that $ (B_{c_{0}^{*}})^{\prime }_{2\varepsilon } = (1 - \varepsilon ) B_{c_{0}^{*}} $ (this must be known, but we were unable to find any reference). By a standard argument, $ (1 - \varepsilon ) B_{X^{*}} \subseteq (B_{X^{*}})^{\prime }_{2\varepsilon } $ for any infinite-dimensional $ X $ . (Let $ x^{*} \in (1 - \varepsilon ) B_{X^{*}} $ . Any $ w^{*} $ -open set $ U $ containing $ 0 $ contains also both $ y^{*} $ and $ -y^{*} $ for some $ y^{*} \in S_{X^{*}} $ , and so $ \mathrm {diam}(U \cap B_{X^{*}}) = 2 $ . For this reason, any $ w^{*} $ -open set $ V $ containing $ x^{*} $ fulfils $ \mathrm {diam}(V \cap (x^{*} + \varepsilon B_{X^{*}})) = 2\varepsilon $ , in particular, $ \mathrm {diam}(V \cap B_{X^{*}}) \geq 2\varepsilon $ . This proves that $ x^{*} \in (B_{X^{*}})^{\prime }_{2\varepsilon } $ .)

Let us show that the opposite inclusion takes place for $ X = c_{0} $ . Assuming $ 1 - \varepsilon < \Vert x^{*} \Vert \leq 1 $ , we need to check that $ x^{*} \notin (B_{c_{0}^{*}})^{\prime }_{2\varepsilon } $ . Let $ e_{1}, e_{2}, \dots $ be the canonical basis of $ c_{0} $ . Let $ n $ be large enough that $ \sum _{i=1}^{n} |x^{*}(e_{i})|> 1 - \varepsilon $ , and let $ \delta> 0 $ satisfy $ 2 \delta n < [\sum _{i=1}^{n} |x^{*}(e_{i})|] - (1 - \varepsilon ) $ . Let

$$ \begin{align*}U = \{ y^{*} \in c_{0}^{*} : 1 \leq i \leq n \Rightarrow |y^{*}(e_{i}) - x^{*}(e_{i})| < \delta \}.\end{align*} $$

For $ y^{*}, z^{*} \in U \cap B_{c_{0}^{*}} $ , we have

$$ \begin{align*}\sum_{i=n+1}^{\infty} |y^{*}(e_{i})| = \Vert y^{*} \Vert - \sum_{i=1}^{n} |y^{*}(e_{i})| \leq 1 - \sum_{i=1}^{n} |x^{*}(e_{i})| + \delta n,\end{align*} $$

and the same for $ z^{*} $ , thus

$$ \begin{align*}\Vert y^{*} - z^{*} \Vert &\leq \sum_{i=1}^{n} |y^{*}(e_{i}) - z^{*}(e_{i})| + \sum_{i=n+1}^{\infty} |y^{*}(e_{i})| + \sum_{i=n+1}^{\infty} |z^{*}(e_{i})|\\&\leq 2 \delta n + 2 \Big[ 1 - \sum_{i=1}^{n} |x^{*}(e_{i})| \Big] + 2 \delta n. \end{align*} $$

We get $ \mathrm {diam}(U \cap B_{c_{0}^{*}}) \leq 4 \delta n + 2 [ 1 - \sum _{i=1}^{n} |x^{*}(e_{i})| ] < 2 \varepsilon $ .

Now, let us assume that $ X $ satisfies $ (B_{X^{*}})^{\prime }_{2\varepsilon } = (1 - \varepsilon ) B_{X^{*}} $ . Clearly, $ X $ is infinite-dimensional, as $ (B_{X^{*}})^{\prime }_{2\varepsilon } $ is nonempty. Moreover, $ X^{*} $ is separable because the Szlenk index of X is $ \omega $ (see, e.g. [Reference Lancien39, Proposition 3 and Theorem 1]). Thus, by Theorem 4.12, the dual $ X^{*} $ is isometric to $ \ell _{1} $ . Let $ e^{*}_{1}, e^{*}_{2}, \dots $ be a basis of $ X^{*} $ that is $ 1 $ -equivalent to the canonical basis of $ \ell _{1} $ , and let $ e^{**}_{1}, e^{**}_{2}, \dots $ be the dual basic sequence in $ X^{**} $ . We claim that the functionals $ e^{**}_{n} $ are $ w^{*} $ -continuous. Suppose that $ e^{**}_{n} $ is not $ w^{*} $ -continuous for some $ n $ . It means that $ \{ x^{*} \in X^{*} : e^{**}_{n}(x^{*}) = 0 \} $ is not $ w^{*} $ -closed. By the Banach-Dieudonné theorem, the set $ \{ x^{*} \in B_{X^{*}} : e^{**}_{n}(x^{*}) = 0 \} $ is not $ w^{*} $ -closed, too. The space $ (B_{X^{*}}, w^{*}) $ is metrizable, so there is a sequence $ x^{*}_{k} $ in $ B_{X^{*}} $ with $ e^{**}_{n}(x^{*}_{k}) = 0 $ , which $ w^{*} $ -converges to some $ x^{*} $ with $ e^{**}_{n}(x^{*}) \neq 0 $ . Without loss of generality, let us assume that $ e^{**}_{n}(x^{*})> 0 $ and that $ e^{**}_{i}(x^{*}_{k}) $ converges to some $ a_{i} $ for every $ i $ . Then, clearly, $ a_{n} = 0 $ . Note that $ \sum _{i=1}^{\infty } |a_{i}| \leq 1 $ , which follows from the fact that $ \sum _{i=1}^{\infty } |e^{**}_{i}(x^{*}_{k})| = \Vert x^{*}_{k} \Vert \leq 1 $ for every $ k $ . Let us put $ a^{*} = \sum _{i=1}^{\infty } a_{i} e^{*}_{i} $ , $ y^{*}_{k} = x^{*}_{k} - a^{*} $ and $ y^{*} = x^{*} - a^{*} $ . Then, $ e^{**}_{n}(y^{*}_{k}) = 0, e^{**}_{n}(y^{*})> 0 $ , the sequence $ y^{*}_{k} $ is $ w^{*} $ -convergent to $ y^{*} $ and, moreover, $ e^{**}_{i}(y^{*}_{k}) $ converges to $ 0 $ for every $ i $ . Choosing a subsequence and making a small perturbation, we can find a sequence $ z^{*}_{l} $ which is a block sequence with respect to the basis $ e^{*}_{i} $ and which still $ w^{*} $ -converges to $ y^{*} $ . Without loss of generality, let us assume that $ \Vert z^{*}_{l} \Vert $ converges to some $ \lambda $ , clearly, with $ \lambda \geq \Vert y^{*} \Vert> 0 $ , and let us consider $ u^{*}_{l} = \frac {1}{\Vert z^{*}_{l} \Vert } z^{*}_{l} $ and $ u^{*} = \frac {1}{\lambda } y^{*} $ .