(Received 1 August 1984: received for publication 29 July 1985)
Key words and phrases: Ordered regular vector spaces, nonconvex vector minimization. axiom of choice.

In ORDER to produce a vector minimization principle which contains Ekeland's variational theorem [8] as well as the results of [13, 14], we have to introduce cone valued metrics. Working with these metrics provides noteworthy technical facilities in some application oriented investigations, the most relevant one from our point of view being the fixed point theory. From the results of Eisenfeld and Lakshmikantham [5-7] comes the idea of application of regular cone valued metrics, which will play an important role in this paper.

The normal cone valued metrics induce uniformizable topologies and every uniformizable Hausdorff topology can be induced by a normal and regular cone valued metric according to a result due to Antonovskij, Boltjanskij and Sarymsakov [1]. The above considerations, largely exposed in our preprint [15] will only be summarized in Section 5. Problems concerning a special sort of relativized regularity considered first in [14] will be given in Sections 3 and 4, after showing by some examples in Section 2 the consistency of the notion introduced. The principal result of this paper is theorem 6.1, which constitutes a general nonconvex vector minimization principle containing Ekeland's variational principle [8] and the results of [13, 14]. The principle comprised by theorem 6.1 is in fact a criterion for the above-mentioned relativized regularity of a cone. This is asserted by theorem 6.2. This paper ends with the deduction from theorem 6.1 generalizations of some results due to Eisenfeld and Lakshmikantham [7], among others an ordered vector space variant of the Kirk-Caristi fixed point theorem [10].

The proofs use the axiom of choice. In [16] we have shown that a denumerable variant of this axiom suffices when $E$$E$EE$E$ is a Fréchet space. This is the case also when $E=R$$E=R$E=RE=R$E=R$ with the natural ordering. The relations of this axiom with the fixed point theorem cited above and with other ordering principles of the analysis were considered by Brønsted [2] (see also the assertion in [16. Section 1]).

Let $E$$E$EE$E$ be a vector space over the reals and let $K$$K$KK$K$ be an acute convex cone in it, i.e. a set having the properties: (i) $K+K\subset K$$K+K\subset K$K+K sub KK+K \subset K$K+K\subset K$; (ii) $tK\subset K$$tK\subset K$tK sub Kt K \subset K$tK\subset K$ for each nonnegative real number $t$$t$tt$t$; (iii) $K\cap (-K)=\{0\}$$K\cap (-K)=\{0\}$K nn(-K)={0}K \cap(-K)=\{0\}$K\cap (-K)=\{0\}$. We shall refer to $K$$K$KK$K$ with these properties simply as to a cone in $E$$E$EE$E$, and shall suppose throughout that $K\ne \{0\}$$K\ne \{0\}$K!={0}K \neq\{0\}$K\ne \{0\}$. Putting $x⩽y$$x⩽y$x <= yx \leqslant y$x⩽y$ whenever $y-x$$y-x$y-xy-x$y-x$ is in $K$$K$KK$K$, we obtain a reflexive, transitive and antisymmetric order relation on $E$$E$EE$E$, which is translation invariant and invariant with respect to the multiplication with nonnegative reals. It is called the order induced by $K$$K$KK$K$ or simply the $K$$K$KK$K$-order in $E$$E$EE$E$. The vector space $E$$E$EE$E$ endowed with an order relation as defined
above is called ordered vector space and the cone $K$$K$KK$K$ inducing the order in it is called its positive cone. $K$$K$KK$K$ bounded, $K$$K$KK$K$ monotone etc. will mean bounded, monotone etc, with respect to the order induced by $K$$K$KK$K$.
The set $A$$A$AA$A$ in the ordered vector space $E$$E$EE$E$ with the positive cone $K$$K$KK$K$ will be called full if $A=(A+K)\cap (A-K)$$A=(A+K)\cap (A-K)$A=(A+K)nn(A-K)A= (A+K) \cap(A-K)$A=(A+K)\cap (A-K)$.
Suppose that $E$$E$EE$E$ is an ordered vector space endowed with a locally convex vector space topology which is Hausdorff. The positive cone $K$$K$KK$K$ in $E$$E$EE$E$ is said to be normal if the zero element of $E$$E$EE$E$ has a neighbourhood basis $B(0)$$B(0)$B(0)B(0)$B(0)$ consisting of full sets.

The cone $K_0$$K_0$K_(0)K_{0}$K_0$ in $K$$K$KK$K$ is called $K$$K$KK$K$ bound regular (sequentially $K$$K$KK$K$ bound regular) if each $K_0$$K_0$K_(0)K_{0}$K_0$ increasing and $K$$K$KK$K$ order bounded net (sequence) in $K_0$$K_0$K_(0)K_{0}$K_0$ converges to an element of $K_0$$K_0$K_(0)K_{0}$K_0$. If $K$$K$KK$K$ is itself $K$$K$KK$K$ bound regular (sequentially $K$$K$KK$K$ bound regular) then it is called regular (sequentially regular).

If $K_0$$K_0$K_(0)K_{0}$K_0$ is a $K$$K$KK$K$ bound regular subcone of $K$$K$KK$K$ then it is obviously a regular cone.
The cone $K$$K$KK$K$ is called fully regular if each of its $K$$K$KK$K$ increasing and topologically bounded nets is convergent to one of its elements.

A stronger notion of relativized regularity was considered in [11, 1.8.4]. Various concepts of bound regularity were introduced and used in [14].

2.1. Let $K$$K$KK$K$ be a normal cone of the locally convex Hausdorff space $E$$E$EE$E$. Suppose that $K_0$$K_0$K_(0)K_{0}$K_0$ is subcone of $K$$K$KK$K$ which is complete and has the property that its linear span is complete, metrizable and does not contain any subspace isomorphic with the Banach space $c_0$$c_0$c_(0)c_{0}$c_0$ of the real sequences converging to 0 in its usual norm. According to theorem 1 of McArthur in [12], $K_0$$K_0$K_(0)K_{0}$K_0$ is fully regular. Since $K$$K$KK$K$ is normal, every $K_0$$K_0$K_(0)K_{0}$K_0$ increasing $K$$K$KK$K$ order bounded net in $K_0$$K_0$K_(0)K_{0}$K_0$ is topologically bounded too (see for example [17, II, proposition 1.4]) and hence convergent by the full regularity of $K_0$$K_0$K_(0)K_{0}$K_0$. The limit of such a net is in $K_0$$K_0$K_(0)K_{0}$K_0$, the latter being complete by hypothesis. We have thus shown that $K_0$$K_0$K_(0)K_{0}$K_0$ is a $K$$K$KK$K$ bound regular subcone of $K$$K$KK$K$.
2.2. Suppose that $B$$B$BB$B$ is a complete, convex and bounded subset of the locally convex Hausdorff space $E$$E$EE$E$. Suppose $B$$B$BB$B$ does not contain 0 . Let $K_0$$K_0$K_(0)K_{0}$K_0$ be the cone generated by $B$$B$BB$B$. Then $K_0$$K_0$K_(0)K_{0}$K_0$ is a $K$$K$KK$K$ bound regular subcone of every normal cone $K$$K$KK$K$ containing it. Indeed, if we consider the subspace sp $K_0$$K_0$K_(0)K_{0}$K_0$ of $E$$E$EE$E$, then this space can be endowed with a norm considering the unit ball in it the convex circled hull of $B$$B$BB$B$. This topology on sp $K_0$$K_0$K_(0)K_{0}$K_0$ is finer than that induced from $E$$E$EE$E$ and sp $K_0$$K_0$K_(0)K_{0}$K_0$ is a Banach space with respect to the norm. In this space $K_0$$K_0$K_(0)K_{0}$K_0$ admits a plastering and hence it is fully regular according to theorem 1.12 of Krasnosel'skij in [11]. Now, since $K$$K$KK$K$ is a normal cone containing $K_0$$K_0$K_(0)K_{0}$K_0$ the reasoning in 2.1 applies to conclude $K_0$$K_0$K_(0)K_{0}$K_0$ is $K$$K$KK$K$ bound regular.
2.3. Let $S$$S$SS$S$ be a nonempty set and let $R^S$$R^S$R^(S)R^{S}$R^S$ be the vector space of all the real functions defined on $S$$S$SS$S$, endowed with the topology of the pointwise convergence. This topology is in fact the direct product topology of $R^S$$R^S$R^(S)R^{S}$R^S$ and converts this space in a locally convex Hausdorff space. The cone $R_{+}^S$$R_{+}^S$R_(+)^(S)R_{+}^{S}$R_{+}^S$ of all the nonnegative functions in $R^S$$R^S$R^(S)R^{S}$R^S$ is normal since the generating family of seminorms $p_s$$p_s$p_(s)p_{s}$p_s$ on $R^S$$R^S$R^(S)R^{S}$R^S$ defined by $p_s(x)=|x(s)|$$p_s(x)=|x(s)|$p_(s)(x)=|x(s)|p_{s}(x)=|x(s)|$p_s(x)=|x(s)|$ is monotone on it ([17, II, proposition 1.5]). It is also regular, since the convergence in $R^S$$R^S$R^(S)R^{S}$R^S$ is the pointwise one and hence the regularity of $R_{+}^S$$R_{+}^S$R_(+)^(S)R_{+}^{S}$R_{+}^S$ is a direct consequence of the regularity of $R_{+}$$R_{+}$R_(+)R_{+}$R_{+}$, the nonnegative half line.
2.4. Let $R^S$$R^S$R^(S)R^{S}$R^S$ be the vector space considered at 2.3. Denote by $c_0(S)$$c_0(S)$c_(0)(S)c_{0}(S)$c_0(S)$ the subspace consisting of the functions $x$$x$xx$x$ having the property that for every positive real $\epsilon$$\epsilon$epsi\varepsilon$\epsilon$ the set $\{s\in S:\mid x(s)>\epsilon \}$$\{s\in S:\mid x(s)>\epsilon \}${s in S:∣x(s) > epsi}\{s \in S: \mid x(s)>\varepsilon\}$\{s\in S:\mid x(s)>\epsilon \}$ is finite. Let us define a norm on $c_0(S)$$c_0(S)$c_(0)(S)c_{0}(S)$c_0(S)$ by putting $\Vert x\Vert =\underset{s}{max}|x(s)|$$\Vert x\Vert =\underset{s}{max} |x(s)|$||x||=max_(s)|x(s)|\|x\|=\max _{s}|x(s)|$\Vert x\Vert =\underset{s}{max}|x(s)|$. Equipped with this norm,
$c_0(S)$$c_0(S)$c_(0)(S)c_{0}(S)$c_0(S)$ becomes a Banach space ([3, II.2]). The cone $c_0^{+}(S)$$c_0^{+}(S)$c_(0)^(+)(S)c_{0}^{+}(S)$c_0^{+}(S)$ of the nonnegative functions in $c_0(S)$$c_0(S)$c_(0)(S)c_{0}(S)$c_0(S)$ is normal and regular.
2.5. Consider the Banach space $c$$c$cc$c$ of all the convergent sequences of real numbers endowed with the sup norm. Then $c^{+}$$c^{+}$c^(+)c^{+}$c^{+}$, the cone of sequences with nonnegative terms in normal. Let $c_0$$c_0$c_(0)c_{0}$c_0$ be the subspace in $c$$c$cc$c$ of the sequences converging to 0 . Then $c_0^{+}=c^{+}\cap c_0$$c_0^{+}=c^{+}\cap c_0$c_(0)^(+)=c^(+)nnc_(0)c_{0}^{+}=c^{+} \cap c_{0}$c_0^{+}=c^{+}\cap c_0$ is a regular cone which is neither $c^{+}$$c^{+}$c^(+)c^{+}$c^{+}$bound regular, nor fully regular.
2.6. Let $K$$K$KK$K$ be a generating closed normal cone in the barelled space $E$$E$EE$E$. The set $L^{+}$$L^{+}$L^(+)L^{+}$L^{+}$of all the linear and continuous operators $A$$A$AA$A$ with the property that $A(K)\subset K$$A(K)\subset K$A(K)sub KA(K) \subset K$A(K)\subset K$ forms a regular cone in the space $L(E)$$L(E)$L(E)L(E)$L(E)$ of all the linear and continuous operators acting in $E$$E$EE$E$ and equipped with the topology of simple convergence (see [15, 9.1]).
2.7. Let $H$$H$HH$H$ be a real Hilbert space. The linear and continuous operator $A$$A$AA$A$ acting in $H$$H$HH$H$ is called positive if $(Ax,x)⩾0$$(Ax,x)⩾0$(Ax,x) >= 0(A x, x) \geqslant 0$(Ax,x)⩾0$ for every $x$$x$xx$x$ in $H$$H$HH$H$. Let the vector space $L(H)$$L(H)$L(H)L(H)$L(H)$ of all the linear and continuous operators acting in $H$$H$HH$H$ be endowed with the topology of simple convergence. Then the set $L^{+}$$L^{+}$L^(+)L^{+}$L^{+}$of all the positive operators forms a regular cone in $L(H)([15,9.2])$$L(H)([15,9.2])$L(H)([15,9.2])L(H)([15,9.2])$L(H)([15,9.2])$.

The reduction of definitions of regularity to criterions which use a denumerable set of terms is very desirable from a technical point of view. In this direction we observe first of all that:

Proof of (3.1). The "only if" part is immediate. For the converse implication let us assume that $K_0$$K_0$K_(0)K_{0}$K_0$ is sequentially $K$$K$KK$K$ bound regular but is not $K$$K$KK$K$ bound regular. Then there exists the $K_0$$K_0$K_(0)K_{0}$K_0$ increasing $K$$K$KK$K$ bounded net ${\left(x_i\right)}_{i\in I}$${\left(x_i\right)}_{i\in I}$(x_(i))_(i in I)\left(x_{i}\right)_{i \in I}${\left(x_i\right)}_{i\in I}$ in $K_0$$K_0$K_(0)K_{0}$K_0$ which is not Cauchy. That is, there exists a neighbourhood $U$$U$UU$U$ of 0 such that for any $i$$i$ii$i$ in $I$$I$II$I$ there exists $x_j$$x_j$x_(j)x_{j}$x_j$ and $x_k$$x_k$x_(k)x_{k}$x_k$ with $j⩾i$$j⩾i$j >= ij \geqslant i$j⩾i$ and $k⩾i$$k⩾i$k >= ik \geqslant i$k⩾i$ such that $x_j-x_k\notin U$$x_j-x_k\notin U$x_(j)-x_(k)!in Ux_{j}-x_{k} \notin U$x_j-x_k\notin U$. We can suppose $x_k-x_j\in K_0$$x_k-x_j\in K_0$x_(k)-x_(j)inK_(0)x_{k}-x_{j} \in K_{0}$x_k-x_j\in K_0$ since ( $x_i$$x_i$x_(i)x_{i}$x_i$ ) is $K_0$$K_0$K_(0)K_{0}$K_0$ increasing. Fix $i$$i$ii$i$ and consider $x_j$$x_j$x_(j)x_{j}$x_j$ and $x_k$$x_k$x_(k)x_{k}$x_k$ as above. Put $x_1=x_j$$x_1=x_j$x_(1)=x_(j)x_{1}=x_{j}$x_1=x_j$ and $x_2=x_k$$x_2=x_k$x_(2)=x_(k)x_{2}=x_{k}$x_2=x_k$. Starting with $x_k$$x_k$x_(k)x_{k}$x_k$ instead of $x_i$$x_i$x_(i)x_{i}$x_i$, we can determine $x_3$$x_3$x_(3)x_{3}$x_3$ and $x_4$$x_4$x_(4)x_{4}$x_4$ in ( $x_i$$x_i$x_(i)x_{i}$x_i$ ) so as to have $x_3-x_2\in K_0,x_4-x_5\in K_0$$x_3-x_2\in K_0,x_4-x_5\in K_0$x_(3)-x_(2)inK_(0),x_(4)-x_(5)inK_(0)x_{3}-x_{2} \in K_{0}, x_{4}-x_{5} \in K_{0}$x_3-x_2\in K_0,x_4-x_5\in K_0$ and $x_4-x_3\notin U$$x_4-x_3\notin U$x_(4)-x_(3)!in Ux_{4}-x_{3} \notin U$x_4-x_3\notin U$. Continuing this procedure we obtain a sequence which is $K_0$$K_0$K_(0)K_{0}$K_0$ increasing and $K$$K$KK$K$ bounded, but is not Cauchy, contradicting the hypothesis.

The above assertion is used in establishing the following summation criterion:

Proof of (3.2). Suppose that there is a sequence $\left(x_n\right)$$\left(x_n\right)$(x_(n))\left(x_{n}\right)$\left(x_n\right)$ in $K_0$$K_0$K_(0)K_{0}$K_0$ with $x_n\notin U$$x_n\notin U$x_(n)!in Ux_{n} \notin U$x_n\notin U$ for every $n$$n$nn$n$ and for some neighbourhood $U$$U$UU$U$ of 0 such that $\{s_m=\sum _{n=1}^m{x}_n:m\in N\}$$\left\{s_m=\sum _{n=1}^m x_n:m\in N\right\}${s_(m)=sum_(n=1)^(m)x_(n):m in N}\left\{s_{m}=\sum_{n=1}^{m} x_{n}: m \in N\right\}$\{s_m=\sum _{n=1}^m{x}_n:m\in N\}$ is $K$$K$KK$K$ order bounded. Then ( $s_m$$s_m$s_(m)s_{m}$s_m$ ) forms a $K_0$$K_0$K_(0)K_{0}$K_0$ increasing $K$$K$KK$K$ bounded sequence which is not convergent, that is, $K_0$$K_0$K_(0)K_{0}$K_0$ cannot be sequentially $K$$K$KK$K$ bound regular.

Assume now that $K_0$$K_0$K_(0)K_{0}$K_0$ does not be $K$$K$KK$K$ bound regular, and hence neither sequentially $K$$K$KK$K$ bound regular by (3.1). Hence $K_0$$K_0$K_(0)K_{0}$K_0$ contains a $K_0$$K_0$K_(0)K_{0}$K_0$ increasing $K$$K$KK$K$ order bounded sequence $\left(z_n\right)$$\left(z_n\right)$(z_(n))\left(z_{n}\right)$\left(z_n\right)$ with the property that there exists a neighbourhood $U$$U$UU$U$ of 0 such that $z_{n+1}-z_n\notin U$$z_{n+1}-z_n\notin U$z_(n+1)-z_(n)!in Uz_{n+1}-z_{n} \notin U$z_{n+1}-z_n\notin U$ for every $n$$n$nn$n$. We can assume $z_1=0$$z_1=0$z_(1)=0z_{1}=0$z_1=0$. Then the elements $x_n=z_{n+1}-z_n$$x_n=z_{n+1}-z_n$x_(n)=z_(n+1)-z_(n)x_{n}=z_{n+1}-z_{n}$x_n=z_{n+1}-z_n$ form a sequence in $K_0$$K_0$K_(0)K_{0}$K_0$ such that

is $K$$K$KK$K$ order bounded. This contradicts the condition of the criterion and completes the proof.
Summation criteria for various types of regularity were considered by McArthur [12] and by the author in [13, 14]. The summation techniques in some problems concerning the sequentially regular cones had already been used by Krasnosel'skij in [11, theorems 1.6 and 1.7].

Let $E$$E$EE$E$ be a vector space ordered by the cone $K$$K$KK$K$. Suppose that $M$$M$MM$M$ is a set in $E$$E$EE$E$ and $H$$H$HH$H$ is a subset of $K$$K$KK$K$. The point $x$$x$xx$x$ in $M$$M$MM$M$ will be said an $H$$H$HH$H$ near to minimum point of $M$$M$MM$M$ if

Another regularity criterion for a cone can be stated in terms of near to minimality. It is:

Proof of (4.1). Suppose that $M$$M$MM$M$ is a set of $E$$E$EE$E$ which is $K$$K$KK$K$ lower bounded by $b$$b$bb$b$. Assume that $H$$H$HH$H$ is as in (4.1) and there are no $H$$H$HH$H$ near to minimum points in $M$$M$MM$M$. Then

for each $x$$x$xx$x$ in $M$$M$MM$M$. Let us consider $x_1\in M$$x_1\in M$x_(1)in Mx_{1} \in M$x_1\in M$ arbitrarily and choose

$n=1,2,\dots$$n=1,2,\dots$n=1,2,dotsn=1,2, \ldots$n=1,2,\dots$. Then we have

for some $h_n$$h_n$h_(n)h_{n}$h_n$ in $H$$H$HH$H$, where $⩽$$⩽$<=\leqslant$⩽$ denotes the $K$$K$KK$K$ order in $E$$E$EE$E$. Summing these relations from $n=1$$n=1$n=1n=1$n=1$ to $n=m$$n=m$n=mn=m$n=m$, we get

By the hypothesis on $H$$H$HH$H$, there exists a neighbourhood $U$$U$UU$U$ of 0 in $E$$E$EE$E$ so as to have $h_n\notin U$$h_n\notin U$h_(n)!in Uh_{n} \notin U$h_n\notin U$ for every $n$$n$nn$n$. This, together with the above relation show, via the criterion 3.2, that $K_0$$K_0$K_(0)K_{0}$K_0$ cannot be $K$$K$KK$K$ bound regular.

To prove the "if" part of the assertion, assume that every subset of $E$$E$EE$E$ which is $K$$K$KK$K$ bounded from below has $H$$H$HH$H$ near to minimum points for each $H$$H$HH$H$ as in 4.1, but $K_0$$K_0$K_(0)K_{0}$K_0$ is not sequentially $K$$K$KK$K$ bound regular. Using the criterion 3.2 again, it follows that there exist the elements $x_n$$x_n$x_(n)x_{n}$x_n$ in $K_0$$K_0$K_(0)K_{0}$K_0$ without some neighbourhood $U$$U$UU$U$ of 0 such that the sums

have a $K$$K$KK$K$ upper bound $b$$b$bb$b$. This means that the set

is $K$$K$KK$K$ order bounded from below by $-b$$-b$-b-b$-b$. Let $x$$x$xx$x$ be an arbitrarily chosen element of $M$$M$MM$M$. Then $x=-\sum _{n=1}^m{x}_n$$x=-\sum _{n=1}^m x_n$x=-sum_(n=1)^(m)x_(n)x=-\sum_{n=1}^{m} x_{n}$x=-\sum _{n=1}^m{x}_n$ for some $m$$m$mm$m$. We have

and

where we have denoted by $H$$H$HH$H$ the set $\{x_n:n\in N\}$$\left\{x_n:n\in N\right\}${x_(n):n in N}\left\{x_{n}: n \in N\right\}$\{x_n:n\in N\}$. That is. we have got

for every $x$$x$xx$x$ in $M$$M$MM$M$. Since $H$$H$HH$H$ satisfies the hypothesis in the proposition, this is a contradiction.
We shall use criterion 4.1 in the following slightly modified form:

is $K$$K$KK$K$ order bounded, has $H$$H$HH$H$ near to minimum points.

Let $E$$E$EE$E$ be an ordered vector space with the positive cone $K$$K$KK$K$. A $K$$K$KK$K$ metric $r$$r$rr$r$ on a nonempty set $V$$V$VV$V$ is a mapping $r$$r$rr$r$ of $V\times V$$V\times V$V xx VV \times V$V\times V$ into $K$$K$KK$K$ which satisfies for arbitrary elements $u,v$$u,v$u,vu, v$u,v$ and $z$$z$zz$z$ in $V$$V$VV$V$ the following conditions: (i) $r(u,u)=0$$r(u,u)=0$r(u,u)=0r(u, u)=0$r(u,u)=0$; (ii) $r(u,v)=0$$r(u,v)=0$r(u,v)=0r(u, v)=0$r(u,v)=0$ implies $u=v$$u=v$u=vu=v$u=v$; (iii) $r(u,v)=r(v,u)$$r(u,v)=r(v,u)$r(u,v)=r(v,u)r(u, v)=r(v, u)$r(u,v)=r(v,u)$; (iv) $r(u,z)⩽r(u,v)+r(v,z)$$r(u,z)⩽r(u,v)+r(v,z)$r(u,z) <= r(u,v)+r(v,z)r(u, z) \leqslant r(u, v)+r(v, z)$r(u,z)⩽r(u,v)+r(v,z)$, where $⩽$$⩽$<=\leqslant$⩽$ stands for the $K$$K$KK$K$ ordering.

If $K$$K$KK$K$ is a normal cone in the locally convex Hausdorff space $E$$E$EE$E$, then the sets

with $B(0)$$B(0)$B(0)B(0)$B(0)$ a neighbourhood basis in $E$$E$EE$E$ consisting of full sets, and a runs over $V$$V$VV$V$, form a
neighbourhood basis for a Hausdorff topology on $V$$V$VV$V$. The resulting topology is a uniformizable one; it is called the topology induced by $r$$r$rr$r$ on $V$$V$VV$V$. Cauchy nets and complexity can be defined in a natural way (for details see [15]). Antonovskij, Boltjanskij and Sarymsakov have shown in [ $1,11.4$$1,11.4$1,11.41,11.4$1,11.4$ and 11.5] that a completely regular Hausdorff topological space is $K$$K$KK$K$ metrizable with $K$$K$KK$K$ the positive cone $R_{+}^S$$R_{+}^S$R_(+)^(S)R_{+}^{S}$R_{+}^S$ of the space $R^S$$R^S$R^(S)R^{S}$R^S$ considered in 2.3. We have observed that $R_{+}^S$$R_{+}^S$R_(+)^(S)R_{+}^{S}$R_{+}^S$ is a normal and regular cone. That is, every completely regular Hausdorff topological space is $K$$K$KK$K$ metrizable with $K$$K$KK$K$ a normal and regular cone. Hence by [9, I.15], the Hausdorff topological spaces which are $K$$K$KK$K$ metrizable by a $K$$K$KK$K$ metric of this kind are quite those which are uniformizable.

Regular cone valued metrics were considered by Eisenfeld and Lakshmikantham in [5-7] for the case when $K$$K$KK$K$ is a regular cone of a Banach space. Since criteria using a denumerable set of terms also work for regular cones in nonmetrizable locally convex Hausdorff spaces, we have as an immediate consequence of one of them, the criterion 3.2, the following assertion:

Let $V$$V$VV$V$ be a set endowed with a $K_0$$K_0$K_(0)K_{0}$K_0$ metric $r$$r$rr$r$, where $K_0$$K_0$K_(0)K_{0}$K_0$ is a $K$$K$KK$K$ bound regular 7 subcone of the normal cone $K$$K$KK$K$ in the locally convex Hausdorff space $E$$E$EE$E$. Suppose that $\left(v_n\right)$$\left(v_n\right)$(v_(n))\left(v_{n}\right)$\left(v_n\right)$ is a sequence in $V$$V$VV$V$ such that the set

is $K$$K$KK$K$ order bounded. Then $\left(v_n\right)$$\left(v_n\right)$(v_(n))\left(v_{n}\right)$\left(v_n\right)$ is Cauchy in the topology on $V$$V$VV$V$ induced by the $K_0$$K_0$K_(0)K_{0}$K_0$ metric $r$$r$rr$r$.

Let $V$$V$VV$V$ be a topological space and let $F$$F$FF$F$ be an operator from $V$$V$VV$V$ to the ordered topological vector space $E$$E$EE$E$. We shall say that $F$$F$FF$F$ is submonotone if from the conditions:
(i) $lim{v}_{\nu }=v$$lim{v}_{\nu }=v$limv_(nu)=v\lim v_{\nu}=v$lim{v}_{\nu }=v$, where ${\left(v_{\nu }\right)}_{\nu \in I}$${\left(v_{\nu }\right)}_{\nu \in I}$(v_(nu))_(nu in I)\left(v_{\nu}\right)_{\nu \in I}${\left(v_{\nu }\right)}_{\nu \in I}$ is a net in $V$$V$VV$V$ indexed by the totally ordered set $I$$I$II$I$;
(ii) $F\left(v_{\nu }\right)⩽F\left(v_{\mu }\right)$$F\left(v_{\nu }\right)⩽F\left(v_{\mu }\right)$F(v_(nu)) <= F(v_(mu))F\left(v_{\nu}\right) \leqslant F\left(v_{\mu}\right)$F\left(v_{\nu }\right)⩽F\left(v_{\mu }\right)$ whenever $\nu ⩾\mu$$\nu ⩾\mu$nu >= mu\nu \geqslant \mu$\nu ⩾\mu$, it follows that