#### Proof

It is obvious that (1) ⇒ (3).

(2) ⇔ (4). By Theorem 1 in [15], *U*_{fin}(*B*_{Γ}, *B*_{Γ}) = *S*_{1}(*B*_{Γ}, *B*_{Γ}) = *S*_{fin}(*B*_{Γ}, *B*_{Γ}).

(3) ⇒ (2). Let {𝓕_{i}} ⊂ *B*_{Γ} and 𝓢 = {*h*_{m}}_{m∈ℕ} be a countable sequentially dense subset of *B*(*X*). For each *i* ∈ ℕ we consider a countable sequentially dense subset 𝓢_{i} of *B*(*X*) and 𝓕_{i} =
$\begin{array}{}{\displaystyle \{{F}_{i}^{m}{\}}_{m\in \mathbb{N}}}\end{array}$ where

$\begin{array}{}{\displaystyle {\mathcal{S}}_{i}=\{{f}_{i}^{m}\}:=\{{f}_{i}^{m}\in B(X):{f}_{i}^{m}\upharpoonright {F}_{i}^{m}={h}_{m}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{f}_{i}^{m}\upharpoonright (X\setminus {F}_{i}^{m})=1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{for}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}m\in \mathbb{N}\}.}\end{array}$

Since 𝓕_{i} =
$\begin{array}{}{\displaystyle \{{F}_{i}^{m}{\}}_{m\in \mathbb{N}}}\end{array}$ is a Borel *γ*-cover of *X* and 𝓢 is a countable sequentially dense subset of *B*(*X*), we have that 𝓢_{i} is a countable sequentially dense subset of *B*(*X*) for each *i* ∈ ℕ. Indeed, let *h* ∈ *B*(*X*), there is a sequence {*h*_{s}}_{s∈ℕ} ⊂ 𝓢 such that {*h*_{s}}_{s∈ℕ} converges to *h*. We claim that
$\begin{array}{}{\displaystyle \{{f}_{i}^{s}{\}}_{s\in \mathbb{N}}}\end{array}$ converges to *h*. Let *K* = {*x*_{1}, …, *x*_{k}} be a finite subset of *X*, *ϵ* > 0 and let *W* = 〈*h*, *K*, *ϵ*〉 := {*g* ∈ *B*(*X*) : |*g*(*x*_{j}) – *h*(*x*_{j})| < *ϵ* for *j* = 1, …, *k*} be a base neighborhood of *h*, then there is *m*_{0} ∈ ℕ such that *K* ⊂
$\begin{array}{}{\displaystyle {F}_{i}^{m}}\end{array}$ for each *m* > *m*_{0} and *h*_{s} ∈ *W* for each *s* > *m*_{0}. Since
$\begin{array}{}{\displaystyle {f}_{i}^{s}}\end{array}$ ↾ *K* = *h*_{s} ↾ *K* for every *s* > *m*_{0},
$\begin{array}{}{\displaystyle {f}_{i}^{s}}\end{array}$ ∈ *W* for every *s* > *m*_{0}. It follows that {
$\begin{array}{}{\displaystyle {f}_{i}^{s}}\end{array}$}_{s∈ℕ} converges to *h*.

Since *B*(*X*) satisfies *S*_{fin}(𝓢, 𝓢), there is a sequence
$\begin{array}{}{\displaystyle ({F}_{i}=\{{f}_{i}^{{m}_{1}},...,{f}_{i}^{{m}_{s(i)}}\}:i\in \mathbb{N})}\end{array}$ such that for each *i*, *F*_{i} ⊂ 𝓢_{i}, and ⋃_{i∈ℕ} *F*_{i} is a countable sequentially dense subset of *B*(*X*).

For 0 ∈ *B*(*X*) there is a sequence
$\begin{array}{}\{{f}_{{i}_{j}}^{{m}_{s({i}_{j})}}{\}}_{j\in \mathbb{N}}\subset \bigcup _{i\in \mathbb{N}}{F}_{i}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{such that}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\{{f}_{{i}_{j}}^{{m}_{s({i}_{j})}}{\}}_{j\in \mathbb{N}}\end{array}$ converges to 0. Consider a sequence (
$\begin{array}{}{F}_{{i}_{j}}^{{m}_{s({i}_{j})}}\end{array}$ : *j* ∈ ℕ). Then

$\begin{array}{}{F}_{{i}_{j}}^{{m}_{s({i}_{j})}}\end{array}$ ∈ 𝓕_{ij};

{$\begin{array}{}{F}_{{i}_{j}}^{{m}_{s({i}_{j})}}\end{array}$ : *j* ∈ ℕ} is a *γ*-cover of *X*.

Indeed, let *K* be a finite subset of *X* and *U* = 〈0, *K*,
$\begin{array}{}\frac{1}{2}\end{array}$〉 be a base neighborhood of 0, then there is *j*_{0} ∈ ℕ such that
$\begin{array}{}{f}_{{i}_{j}}^{{m}_{s({i}_{j})}}\end{array}$ ∈ *U* for every *j* > *j*_{0}. It follows that *K* ⊂
$\begin{array}{}{F}_{{i}_{j}}^{{m}_{s({i}_{j})}}\end{array}$ for every *j* > *j*_{0}. We thus get that *X* satisfies *U*_{fin}(*B*_{Γ}, *B*_{Γ}), and, hence, by Theorem 1 in [15], *X* satisfies *S*_{1}(*B*_{Γ}, *B*_{Γ}).

(2) ⇒ (1). Let {*S*_{i}} ⊂ 𝓢 and *S* = {*d*_{n} : *n* ∈ ℕ} ∈ 𝓢. Consider the topology *τ* generated by the family 𝓟 = {*f*^{–1}(*G*) : *G* is an open set of ℝ and *f* ∈
$\begin{array}{}{\displaystyle S\cup \bigcup _{i\in \mathbb{N}}{S}_{i}}\end{array}$}. Since *P* =
$\begin{array}{}{\displaystyle S\cup \bigcup _{i\in \mathbb{N}}{S}_{i}}\end{array}$ is a countable dense subset of *B*(*X*) and *X* is Tychonoff, we have that the space *Y* = (*X*, *τ*) is a separable metrizable space. Note that a function *f* ∈ *P*, considered as mapping from *Y* to ℝ, is a continuous function i.e. *f* ∈ *C*(*Y*) for each *f* ∈ *P*. Note also that an identity map *φ* from *X* on *Y*, is a Borel bijection. By Corollary 12 in [6], *Y* is a *QN*-space and, hence, by Corollary 20 in [17], *Y* has the property *S*_{1}(*B*_{Γ}, *B*_{Γ}). By Corollary 21 in [17], *B*(*Y*) is an *α*_{2} space.

Let *q* : ℕ ↦ ℕ × ℕ be a bijection. Then we enumerate {*S*_{i}}_{i∈ℕ} as {*S*_{q(i)}}_{q(i)∈ℕ×ℕ}. For each *d*_{n} ∈ *S* there are sequences *s*_{n,m} ⊂ *S*_{n,m} such that *s*_{n,m} converges to *d*_{n} for each *m* ∈ ℕ. Since *B*(*Y*) is an *α*_{2} space, there is {*b*_{n,m} : *m* ∈ ℕ} such that for each *m*, *b*_{n,m} ∈ *s*_{n,m}, and, *b*_{n,m} → *d*_{n} (*m* → ∞). Let *B* = {*b*_{n,m} : *n*, *m* ∈ ℕ}. Note that *S* ⊂ [*B*]_{seq}.

Since *X* is a *σ*-set (that is, each Borel subset of *X* is *F*_{σ})(see [17]), *B*_{1}(*X*) = *B*(*X*) and *φ*(*B*(*Y*)) = *φ*(*B*_{1}(*Y*)) ⊆ *B*(*X*) where *φ*(*B*(*Y*)) := {*p* ∘ *φ* : *p* ∈ *B*(*Y*)} and *φ*(*B*_{1}(*Y*)) := {*p* ∘ *φ* : *p* ∈ *B*_{1}(*Y*)}.

Since *S* is a countable, sequentially dense subset of *B*(*X*), for any *g* ∈ *B*(*X*) there is a sequence {*g*_{n}}_{n∈ℕ} ⊂ *S* such that {*g*_{n}}_{n∈ℕ} converges to *g*. But *g* we can consider as a mapping from *Y* into ℝ and a set {*g*_{n} : *n* ∈ ℕ} as subset of *C*(*Y*). It follows that *g* ∈ *B*_{1}(*Y*). We get that *φ*(*B*(*Y*)) = *B*(*X*).

We claim that *B* ∈ 𝓢, i.e. that [*B*]_{seq} = *B*(*X*). Let *f* ∈ *B*(*Y*) and {*f*_{k} : *k* ∈ ℕ} ⊂ *S* such that *f*_{k} → *f* (*k* → ∞). For each *k* ∈ ℕ there is {
$\begin{array}{}{\displaystyle {f}_{k}^{n}}\end{array}$ : *n* ∈ ℕ} ⊂ *B* such that
$\begin{array}{}{\displaystyle {f}_{k}^{n}}\end{array}$ → *f*_{k} (*n* → ∞). Since *Y* is a *QN*-space (Theorem 16 in [6]), there exists an unbounded *β* ∈ ℕ^{ℕ} such that {
$\begin{array}{}{\displaystyle {f}_{k}^{\beta (k)}}\end{array}$} converges to *f* on *Y*. It follows that {
$\begin{array}{}{\displaystyle {f}_{k}^{\beta (k)}}\end{array}$ : *k* ∈ ℕ} converge to *f* on *X* and [*B*]_{seq} = *B*(*X*).

(5) ⇒ (6). By Velichko’s Theorem ([18]), a space *B*_{1}(*X*) is sequentially separable for any separable metric space *X*.

Let {𝓕_{i}} ⊂ *F*_{Γ} and 𝓢 = {*h*_{m}}_{m∈ℕ} be a countable sequentially dense subset of *B*_{1}(*X*).

Similarly implication (3) ⇒ (2) we get *X* satisfies *U*_{fin}(*F*_{Γ}, *F*_{Γ}), and, hence, by Lemma 13 in [17], *X* satisfies *S*_{1}(*F*_{Γ}, *F*_{Γ}).

(6) ⇒ (5). By Corollary 20 in [17], *X* satisfies *S*_{1}(*B*_{Γ}, *B*_{Γ}). Since *X* is a *σ*-set (see [17]), *B*_{1}(*X*) = *B*(*X*) and, by implication (2) ⇒ (1), we get *B*_{1}(*X*) satisfies *S*_{1}(𝓢, 𝓢).□

#### Proof

(1) ⇒ (2). Let *X* be a set of reals satisfying the hypotheses and *β* be a countable base of *X*. Consider a sequence {𝓑_{i}}_{i∈ℕ} of countable Borel *ω*-covers of *X* where 𝓑_{i} =
$\begin{array}{}{\displaystyle \{{W}_{i}^{j}{\}}_{j\in \mathbb{N}}}\end{array}$ for each *i* ∈ ℕ.

Consider a topology *τ* generated by the family 𝓟 = {
$\begin{array}{}{\displaystyle {W}_{i}^{j}}\end{array}$ ∩ *A* : *i*, *j* ∈ ℕ and *A* ∈ *β*} ⋃ {(*X* ∖
$\begin{array}{}{\displaystyle {W}_{i}^{j}}\end{array}$) ∩ *A* : *i*, *j* ∈ ℕ and *A* ∈ *β*}.

Note that if *χ*_{P} is a characteristic function of *P* for each *P* ∈ 𝓟, then a diagonal mapping *φ* = *Δ*_{P∈𝓟}*χ*_{P} : *X* ↦ 2^{ω} is a Borel bijection. Let *Z* = *φ*(*X*).

Note that {𝓑_{i}} is countable open *ω*-cover of *Z* for each *i* ∈ ℕ. Since *B*(*Z*) is a dense subset of *B*(*X*), then *B*(*Z*) also has the property *S*_{1}(𝓓, 𝓓). Since *C*_{p}(*Z*) is a dense subset of *B*(*Z*), *C*_{p}(*Z*) has the property *S*_{1}(𝓓, 𝓓), too.

By Theorem 3.2, the space *Z* has the property *S*_{1}(*Ω*, *Ω*). It follows that there is a sequence
$\begin{array}{}{\displaystyle \{{W}_{i}^{j(i)}{\}}_{i\in \mathbb{N}}}\end{array}$ such that
$\begin{array}{}{\displaystyle {W}_{i}^{j(i)}}\end{array}$ ∈ 𝓑_{i} and {
$\begin{array}{}{\displaystyle {W}_{i}^{j(i)}}\end{array}$ : *i* ∈ ℕ} is an open *ω*-cover of *Z*. It follows that {
$\begin{array}{}{\displaystyle {W}_{i}^{j(i)}}\end{array}$ : *i* ∈ ℕ} is Borel *ω*-cover of *X*.

(2) ⇒ (1). Assume that *X* has the property *S*_{1}(*B*_{Ω}, *B*_{Ω}). Let {*D*_{k}}_{k∈ℕ} be a sequence countable dense subsets of *B*(*X*) and *D*_{k} = {
$\begin{array}{}{\displaystyle {f}_{i}^{k}}\end{array}$ : *i* ∈ ℕ} for each *k* ∈ ℕ. We claim that for any *f* ∈ *B*(*X*) there is a sequence {*f*_{k}} ⊂ *B*(*X*) such that *f*_{k} ∈ *D*_{k} for each *k* ∈ ℕ and *f* ∈ {*f*_{k} : *k* ∈ ℕ}. Without loss of generality we can assume *f* = 0. For each
$\begin{array}{}{\displaystyle {f}_{i}^{k}}\end{array}$ ∈ *D*_{k} let
$\begin{array}{}{\displaystyle {W}_{i}^{k}=\{x\in X:-\frac{1}{k}<{f}_{i}^{k}(x)<\frac{1}{k}\}.}\end{array}$

If for each *j* ∈ ℕ there is *k*(*j*) such that
$\begin{array}{}{\displaystyle {W}_{i(j)}^{k(j)}=X,}\end{array}$, then a sequence
$\begin{array}{}{\displaystyle {f}_{k(j)}={f}_{i(j)}^{k(j)}}\end{array}$ uniformly converges to *f* and, hence, *f* ∈ {*f*_{k}(*j*)} : *j* ∈ ℕ}.

We can assume that
$\begin{array}{}{\displaystyle {W}_{i}^{k}}\end{array}$ ≠ *X* for any *k*, *i* ∈ ℕ.

{
$\begin{array}{}{\displaystyle {W}_{i}^{k}}\end{array}$}_{i∈ℕ} a sequence of Borel sets of *X*.

For each *k* ∈ ℕ, {
$\begin{array}{}{\displaystyle {W}_{i}^{k}}\end{array}$ : *i* ∈ ℕ} is a *ω*-cover of *X*.

By (2), *X* has the property *S*_{1}(*B*_{Ω}, *B*_{Ω}), hence, there is a sequence
$\begin{array}{}{\displaystyle \{{W}_{i(k)}^{k}{\}}_{k\in \mathbb{N}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{such that}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{W}_{i(k)}^{k}\in \{{W}_{i}^{k}{\}}_{i\in \mathbb{N}}}\end{array}$ for each *k* ∈ ℕ and
$\begin{array}{}{\displaystyle \{{W}_{i(k)}^{k}{\}}_{k\in \mathbb{N}}}\end{array}$ is a *ω*-cover of *X*.

Consider
$\begin{array}{}{\displaystyle \{{f}_{i(k)}^{k}\}.}\end{array}$ We claim that
$\begin{array}{}{\displaystyle f\in \overline{\{{f}_{i(k)}^{k}:k\in \mathbb{N}\}}.}\end{array}$ Let *K* be a finite subset of *X*, *ϵ* > 0 and *U* = 〈*f*, *K*, *ϵ*〉 be a base neighborhood of *f*, then there is *k*_{0} ∈ ℕ such that
$\begin{array}{}{\displaystyle \frac{1}{{k}_{0}}<\u03f5\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}K\subset {W}_{i({k}_{0})}^{{k}_{0}}.}\end{array}$ It follows that
$\begin{array}{}{\displaystyle {f}_{i({k}_{0})}^{{k}_{0}}\in U.}\end{array}$

Let *D* = {*d*_{n} : *n* ∈ ℕ} be a dense subspace of *B*(*X*). Given a sequence {*D*_{i}}_{i∈ℕ} of dense subspace of *B*(*X*), enumerate it as {*D*_{n,m} : *n*, *m* ∈ ℕ}. For each *n* ∈ ℕ, pick *d*_{n,m} ∈ *D*_{n,m} so that *d*_{n} ∈ {*d*_{n,m} : *m* ∈ ℕ}. Then {*d*_{n,m} : *m*, *n* ∈ ℕ} is dense in *B*(*X*).□

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