We give the main results of this paper by the following two subsections.
2.1
\(p_{b}\)Open balls in partial bmetric spaces
The following partial bmetric spaces were introduced by Shukla in [1].
Definition 2.1
[1]
Let X be a nonempty set. A mapping \(p_{b}:X\times X\longrightarrow \mathbb {R}^{+}\) is called a partial bmetric with coefficient \(s\ge1\) and \((X,p_{b})\) is called a partial bmetric space with coefficient \(s\ge1\) if the following are satisfied for all \(x,y,z\in X\):

(1)
\(x=y \Longleftrightarrow p_{b}(x,x)=p_{b}(y,y)=p_{b}(x,y)\).

(2)
\(p_{b}(x,y)=p_{b}(y,x)\).

(3)
\(p_{b}(x,x)\le p_{b}(x,y)\).

(4)
\(p_{b}(x,y)\le s(p_{b}(x,z)+p_{b}(z,y))p_{b}(z,z)\).
Remark 2.2
If \(s=1\) in Definition 2.1, then \((X,p_{b})\) is a partial metric space, which was introduced by Matthews (for example, see [3]). Further, put \(d_{p_{b}}:X\times X\longrightarrow \mathbb {R}^{+}\) by \(d_{p_{b}}(x,y)=2p_{b}(x,y)p_{b}(x,x)p_{b}(y,y)\) for all \(x,y\in X\), then \(d_{p_{b}}\) is a metric on X and \((X,d_{p})\) is a metric space.
However, if \(s>1\), then we cannot guarantee that each partial bmetric can induce a bmetric by the method in Remark 2.2. So Mustafa et al. gave the following partial bmetric \(p_{b}\) by modifying Definition 2.1(4) and proved that the \(p_{b}\) induces a bmetric by the method in Remark 2.2.
Definition 2.3
([3])
Let X be a nonempty set. A mapping \(p_{b}:X\times X\longrightarrow \mathbb {R}^{+}\) is called a partial bmetric with coefficient \(s\ge1\) and \((X,p_{b})\) is called a partial bmetric space with coefficient \(s\ge1\) if the following are satisfied for all \(x,y,z\in X\):

(1)
\(x=y \Longleftrightarrow p_{b}(x,x)=p_{b}(y,y)=p_{b}(x,y)\).

(2)
\(p_{b}(x,y)=p_{b}(y,x)\).

(3)
\(p_{b}(x,x)\le p_{b}(x,y)\).

(4)
\(p_{b}(x,y)\le s(p_{b}(x,z)+p_{b}(z,y)p_{b}(z,z))+\frac {1s}{2}(p_{b}(x,x)+p_{b}(y,y))\).
Remark 2.4
If x, y, z satisfy Definition 2.3(1), (2), (3) and are different from each other, then it is easy to check that x, y, z Definition 2.3(4) holds.
As a known fact, Proposition 1.1 and Claim 1.2 are not true if \((X,p_{b})\) is a partial bmetric space in the sense of Definition 2.1 ([6]). So it is important to check whether Proposition 1.1 and Claim 1.2 are true if \((X,p_{b})\) is a partial bmetric space in the sense of Definition 2.3. The following example shows that the result of the check is negative, which comes from [6]. In the following, all partial bmetric spaces are in the sense of Definition 2.3.
Example 2.5
Let \(X=\{u,v,w\}\) and put \(p_{b}:X\times X\longrightarrow \mathbb {R}^{+}\) as follows:

(i)
\(p_{b}(u,u)=p_{b}(w,w)=1\) and \(p_{b}(v,v)=0.5\).

(ii)
\(p_{b}(u,w)=p_{b}(w,u)=1.5\).

(iii)
\(p_{b}(v,w)=p_{b}(w,v)=1\).

(iv)
\(p_{b}(u,v)=p_{b}(v,u)=3\).
Let \(B_{p_{b}}(u,\varepsilon)\) be described in Proposition. Then the following hold:

(1)
\(p_{b}\) is a partial bmetric with coefficient \(s=3\).

(2)
\(w\in B_{p_{b}}(u,1)\) and for any \(\varepsilon>0\), \(B_{p_{b}}(w,\varepsilon)\nsubseteq B_{p_{b}}(u,1)\).
Proof
(1) It is not difficult to check that \(p_{b}\) satisfies Definition 2.3(1), (2), (3). In order to check that \(p_{b}\) satisfies Definition 2.3(4), we only need to consider the following three cases by Remark 2.4.

(1)
\(x=u\), \(y=v\), \(z=w\):
$$\begin{aligned}& p_{b}(u,v)=3, \\& 3\bigl(p_{b}(u,w)+p_{b}(w,v)p_{b}(w,w)\bigr)+ \frac{13}{2}\bigl(p_{b}(u,u)+p_{b}(v,v)\bigr)=3. \end{aligned}$$
So \(p_{b}(u,v)\le3(p_{b}(u,w)+p_{b}(w,v)p_{b}(w,w))+ \frac{13}{2}(p_{b}(u,u)+p_{b}(v,v))\).

(2)
\(x=u\), \(y=w\), \(z=v\):
$$\begin{aligned}& p_{b}(u,w)=1.5 , \\& 3\bigl(p_{b}(u,v)+p_{b}(v,w)p_{b}(v,v)\bigr)+ \frac{13}{2}\bigl(p_{b}(u,u)+p_{b}(w,w)\bigr)=8.5 . \end{aligned}$$
So \(p_{b}(u,w)\le3(p_{b}(u,v)+p_{b}(v,w)p_{b}(v,v))+ \frac{13}{2}(p_{b}(u,u)+p_{b}(w,w))\).

(3)
\(x=v\), \(y=w\), \(z=u\):
$$\begin{aligned}& p_{b}(v,w)=1, \\& 3\bigl(p_{b}(v,u)+p_{b}(u,w)p_{b}(u,u)\bigr)+ \frac{13}{2}\bigl(p_{b}(v,v)+p_{b}(w,w)\bigr)=9. \end{aligned}$$
So \(p_{b}(v,w)\le3(p_{b}(v,u)+p_{b}(u,w)p_{b}(u,u))+ \frac{13}{2}(p_{b}(v,v)+p_{b}(w,w))\).
Thus, \(p_{b}\) is a partial bmetric with coefficient \(s=3\).
(2) Since \(p_{b}(u,w)=1.5<1+1=p_{b}(u,u)+1\), \(w\in B_{p_{b}}(u,1)\). In addition, for any \(\varepsilon>0\), \(p_{b}(w,v)=1<1+\varepsilon =p_{b}(w,w)+\varepsilon\), so \(v\in B_{p_{b}}(w,\varepsilon)\). On the other hand, \(p_{b}(u,v)=3\nless2=1+1=p_{b}(u,u)+1\), so \(v\notin B_{p_{b}}(u,1)\). This shows that \(B_{p_{b}}(w,\varepsilon)\nsubseteq B_{p_{b}}(u,1)\). □
Remark 2.6
Example 2.5 shows that Proposition 1.1 and Claim 1.2 are not true if \((X,p_{b})\) is a partial bmetric space.
However, we have the following.
Proposition 2.7
([7])
Let
\((X,p_{b})\)
be a partial bmetric space and △ be described in Claim
1.2. Then △ is a subbase for some topology on
X. We denote the topology by
\(\mathscr {T}_{p_{b}}\).
It is well known that the space \((X,\mathscr {T}_{p_{b}})\) is \(T_{0}\) but does not need to be \(T_{1}\) ([7]). The following proposition give a sufficient and necessary such that \((X,\mathscr {T}_{p_{b}})\) is a \(T_{1}\)space.
Proposition 2.8
Let
\((X,p_{b})\)
be a partial bmetric space in the sense of Definition
2.3. Then the following are equivalent:

(1)
\((X,\mathscr {T}_{p_{b}})\)
is a
\(T_{1}\)space.

(2)
\(p_{b}(x,y)>\max\{p_{b}(x,x),p_{b}(y,y)\}\)
for each pair of distinct points
\(x,y\in X\).
Proof
(1) ⟹ (2): Let \((X,\mathscr {T}_{p_{b}})\) be a \(T_{1}\)space. If \(x,y\in X\) and \(x\ne y\), then there is a neighborhood U of x such that \(y\notin U\). Since △ is a subbase of \((X,\mathscr {T}_{p_{b}})\) from Proposition 2.7, there are \(\varepsilon_{1},\varepsilon_{2},\ldots,\varepsilon_{k}>0\) such that \(y\notin\bigcap\{B_{p_{b}}(x,\varepsilon_{i}):i=1,2,\ldots,k\}\), and hence there is \(i_{0}\in\{1,2,\ldots,k\}\) such that \(y\notin B_{p_{b}}(x,\varepsilon_{i_{0}})\). So \(p_{b}(x,y)\ge p_{b}(x,x)+\varepsilon_{i_{0}}>p_{b}(x,x)\). In the same way, \(p_{b}(x,y)>p_{b}(y,y)\). So \(p_{b}(x,y)>\max\{ p_{b}(x,x),p_{b}(y,y)\}\).
(2) ⟹ (1): Let \(x,y\in X\) and \(x\ne y\). If \(p_{b}(x,y)>\max\{p_{b}(x,x),p_{b}(y,y)\}\). Then \(p_{b}(x,y)>p_{b}(x,x)\). Put \(\varepsilon=p_{b}(x,y)p_{b}(x,x)>0\), then \(p_{b}(x,y)=p_{b}(x,x)+\varepsilon\), and so \(y\notin B_{p_{b}}(X,\varepsilon)\). In the same way, there is \(\varepsilon'>0\) such that \(x\notin B_{p_{b}}(y,\varepsilon')\). Consequently, \((X,\mathscr {T}_{p_{b}})\) is a \(T_{1}\)space. □
2.2 Balls in cone metric spaces
Definition 2.9
([2, 5])
Let E be a real Banach space. A subset P of E is called a cone of E and \((E,P)\) is called a cone space if the following are satisfied: where θ is zero element in E.

(1)
P is closed, \(P\ne\emptyset\), and \(P\ne\{\theta\}\).

(2)
\(a,b\in \mathbb {R}^{+}\) and \(\alpha,\beta\in P \Longrightarrow a\alpha +b\beta\in P\).

(3)
\(\alpha,\alpha\in P \Longrightarrow \alpha=\theta\).
Definition 2.10
([2, 5])
Let \((E,P)\) be a cone space. Some partial orderings ≤, <, and ≪ on E with respect to P are defined as follows, respectively, where \(P^{\circ}\) denotes the interior of P. Let \(\alpha,\beta\in E\).

(1)
\(\alpha\le\beta\) if \(\beta\alpha\in P\).

(2)
\(\alpha<\beta\) if \(\alpha\le\beta\) and \(\alpha\ne\beta\).

(3)
\(\alpha\ll\beta\) if \(\beta\alpha\in P^{\circ}\).
Remark 2.11
Let \((E,P)\) be a cone space. For the sake of conveniences, we also use notations ‘≥’, ‘>’, and ‘≫′’ on E with respect to P. The meanings of these notations are clear and the following hold:

(1)
\(\alpha\ge\theta\) if and only if \(\alpha\in P\).

(2)
\(\alpha\gg\theta\) if and only if \(\alpha\in P^{\circ}\).

(3)
\(\alpha\beta\gg\theta\) if and only if \(\alpha\gg\beta\).

(4)
\(\alpha\beta\ge\theta\) if and only if \(\alpha\ge\beta\).

(5)
\(\alpha\gg\beta \Longrightarrow \alpha>\beta \Longrightarrow \alpha\ge\beta\).
In addition, in order to guarantee the existence of elements \(\varepsilon\gg\theta\), we always assume that the cone P has nonempty interior ([5]).
Definition 2.12
([2, 5])
Let X be a nonempty set and let \((E,P)\) be a cone space. A mapping \(d:X\times X\longrightarrow E\) is called a cone metric on X, and \((X,d)\) is called a cone metric space if the following are satisfied:

(1)
\(d(x,y)\ge\theta\) for all \(x,y\in X\) and \(d(x,y)=\theta\) if and only if \(x=y\).

(2)
\(d(x,y)=d(y,x)\) for all \(x,y\in X\).

(3)
\(d(x,y)\le d(x,z)+d(z,y)\) for all \(x,y,z\in X\).
Notation 2.13
Let \((X,d)\) be a cone metric space, \(x\in X\), and \(\varepsilon\gg\theta\). In this section, we use the following notations for balls in \((X,d)\):

(1)
\(B(x,\varepsilon)=\{y\in X:d(x,y)\ll\varepsilon\}\).

(2)
\(B_{1}(x,\varepsilon)=\{y\in X:d(x,y)<\varepsilon\}\).

(3)
\(B_{2}(x,\varepsilon)=\{y\in X:d(x,y)\le\varepsilon\}\).
Proposition 2.14
([5])
Let
\((X,d)\)
be a cone metric space. Put
\(\mathscr {B}=\{B(x,\varepsilon):x\in X\textit{ and}\ \varepsilon\gg\theta\}\). Then
\(\mathscr {B}\)
is a base for some topology
\(\mathscr {T}\)
on
X.
In this section, just as the investigation in [5], we always suppose that each cone metric space is a topological space described in Proposition 2.14.
Proposition 2.15
Let
\((X,d)\)
be a cone metric space. For each
\(x\in X\)
and each
\(\varepsilon\gg\theta\), the following hold:

(1)
\(B(x,\varepsilon)\subseteq B_{1}(x,\varepsilon)\subseteq B_{2}(x,\varepsilon)\).

(2)
\(\overline{B(x,\varepsilon)}\subseteq\overline{B_{1}(x,\varepsilon )}\subseteq\overline{B_{2}(x,\varepsilon)}\).

(3)
\(\overline{B_{2}(x,\varepsilon)}=B_{2}(x,\varepsilon)\).
Proof
(1) It holds by Remark 2.11(5).
(2) It holds by the above item (1).
(3) Let \(y\in\overline{B_{2}(x,\varepsilon)}\). Then, whenever \(\eta\gg \theta\), \(B(y,\eta)\cap B_{2}(x,\varepsilon)\ne\emptyset\). Pick \(z\in B(y,\eta)\cap B_{2}(x,\varepsilon)\). Then \(d(x,z)\le\varepsilon\) and \(d(z,y)\ll\eta\). It follows that \(d(x,y)\le d(x,z)+d(z,y)\ll\varepsilon+\eta\). Let \(\eta\rightarrow\theta\). Then \(d(x,y)\le\varepsilon\). So \(y\in B_{2}(x,\varepsilon)\). This proves that \(\overline{B_{2}(x,\varepsilon)}\subseteq B_{2}(x,\varepsilon)\). On the other hand, it is clear that \(\overline{B_{2}(x,\varepsilon )}\supseteq B_{2}(x,\varepsilon)\). So \(\overline{B_{2}(x,\varepsilon)}=B_{2}(x,\varepsilon)\). □
The following example shows that any ‘⊆’ in Proposition 2.15(1), (2) cannot be replaced by ‘=’.
Example 2.16
Let the cone space \((E,P)\) be defined as in [5], Example 1, i.e., \(E=\mathbb {R}^{2}=\{(r,s):r,s\in \mathbb {R}\}\) is the Euclidean plane and \(P=\{(r,s)\in E:r,s\ge0\}\). Let \(X=\{x,y,z\}\). Define \(d:X\times X\longrightarrow E\) as follows: \(d(x,x)=d(y,y)=d(z,z)=(0,0)\), \(d(x,y)=d(y,x)=d(y,z)=d(z,y)=(1,1)\), and \(d(x,z)=d(z,x)=(1,0)\). It is not difficult to check that \((X,d)\) is a cone metric space. Let \(\varepsilon=(1,1)\gg\theta\).

(1)
Note that \(d(x,y)=(1,1)=\varepsilon\). By Remark 2.11, \(d(x,y)\le\varepsilon\), \(d(x,y)\nless\varepsilon \), and \(d(x,y)\not\ll\varepsilon\). So \(y\notin B(x,\varepsilon)\), \(y\notin B_{1}(x,\varepsilon)\), and \(y\in B_{2}(x,\varepsilon)\). Also, \(\varepsilond(x,z)=(1,1)(1,0)=(0,1)\in P(\{\theta\}\cup P^{\circ})\). By Remark 2.11, \(\varepsilond(x,z)>\theta\), and \(\varepsilon d(x,z)\not\gg\theta\), hence \(d(x,z)<\varepsilon\) and \(d(x,z)\not\ll\varepsilon\). So \(z\notin B(x,\varepsilon)\), \(z\in B_{1}(x,\varepsilon)\), and \(y\in B_{2}(x,\varepsilon)\). It follows that \(B(x,\varepsilon)= \{x\}\), \(B_{1}(x,\varepsilon)=\{x,z\}\), and \(B_{2}(x,\varepsilon)=\{x,y,z\}\). So any ‘⊆’ in Proposition 2.15(1) cannot be replaced by ‘=’.

(2)
Note that \((X,d)\) is Hausdorff ([5]). In fact, each cone metric space is metrizable ([8]). So \(\overline{B(x,\varepsilon)}=B(x,\varepsilon)\), \(\overline{B_{1}(x,\varepsilon)}=B_{1}(x,\varepsilon)\), and \(\overline{B_{2}(x,\varepsilon)}=B_{2}(x,\varepsilon)\). By the above item (1), any ‘⊆’ in Proposition 2.15(2) cannot be replaced by ‘=’.
Remark 2.17

(1)
By Example 2.16, Equality 1.3 is not true. Indeed, in Example 2.16, \(\overline{B(x,\varepsilon)}=\{x\}\) and \(B_{2}(x,\varepsilon)=\{x,y,z\}\). So \(\overline{B(x,\varepsilon)}\ne B_{2}(x,\varepsilon)\).

(2)
Let \((X,d)\) be a cone metric space. In [5], the authors showed that \(\overline{B(x,\varepsilon)}\) and \(B_{2}(x,\varepsilon)\) are sequentially closed in \((X,d)\) ([5], Proposition 2). In fact, \((X,d)\) is metrizable, and hence \((X,d)\) is a sequential space, i.e., closed and sequentially closed in \((X,d)\) are equivalent. On the other hand, indeed, the closure \(\overline{B(x,\varepsilon)}\) of \(B(x,\varepsilon)\) is closed and \(B_{2}(x,\varepsilon)\) is closed by Proposition 2.15(2).