(i) There exists some cardinal number κ for which there exists an order-embedding $\psi :(X,\precsim )⟶({\{0,1\}}^{\kappa },{\le }_{lex})$.

(ii) There exists some cardinal number κ and a (complete) preorder ≲ on ${\{0,1\}}^{\kappa }$ that is coarser than ${\le }_{lex}$ for which there exists a continuous order-embedding $\vartheta :(X,\precsim ,t)⟶({\{0,1\}}^{\kappa },\lesssim ,t^{\lesssim })$.

The validity of assertion (i) is (at least implicitly) well known. It holds without assuming $(X_{\mid \sim },t_{\mid \sim })$ to be a Hausdorff-space. Nevertheless we must repeat its proof here in order to preparing the proof of assertion (ii). Let, therefore, $\kappa$ be the cardinality $\mid X_{\mid \sim }\mid$ of $X_{\mid \sim }$. Then we arbitrarily choose some bijective function $\beta :[0,\kappa [⟶X_{\mid \sim }$ in order to define the desired order-embedding $\psi :(X,\precsim )⟶({\{0,1\}}^{\kappa },{\le }_{lex})$ by identifying $\psi \left(x\right)$ for every $x\in X$ with the tuple ${\left(x_{\alpha }\right)}_{\alpha <\kappa }\in {\{0,1\}}^{\kappa }$ that is defined by setting

for all ordinal numbers $\alpha <\kappa$. Since for every point $x\in X$ and every point $y\in X$ the validity of the equivalence $x\precsim y\iff d\left(x\right)\subseteq d\left(y\right)$ holds it follows that $\psi$ is an order-embedding.

In order to now verify assertion (ii) let S be an arbitrary subset of ${\{0,1\}}^{\kappa }$. Then a lacuna of S is a non-degenerate (non-trivial) interval of ${\{0,1\}}^{\kappa }$ that is disjoint from S and has an upper and lower bound in S. A gap of S is a maximal lacuna. Although ≾ is not necessarily complete it is easily to be seen that the inclusion $t\supseteq t^{\precsim }$ would imply the continuity of $\psi$ considered as a function $\psi :(X,\precsim ,t)⟶({\{0,1\}}^{\kappa },\lesssim ,t^{\lesssim })$ if $\psi \left(X\right)$ would not have any gaps of the form $[u,v[$ or $]u,v]$, i.e. half-closed half-open or half-open half-closed gaps (cf. Beardon [1]). But in order to eliminate these gaps a difficulty appears. Indeed, let ${\left(x_i\right)}_{i\in I}$ be a net that converges to some point $x\in X$. Then it may happen that ${\displaystyle \underset{i\in I,\psi \left(x_i\right){\le }_{lex}\psi \left(x\right)}{lim}{x}_i=x}$ or that ${\displaystyle \underset{i\in I,\psi \left(x_i\right){\ge }_{lex}\psi \left(x\right)}{lim}{x}_i=x}$.

This means that a (crucial) gap to be eliminated in order to guarantee continuity of $\psi$ at x is of the form or or or

But, of course, there may exist another net ${\left(y_j\right)}_{j\in J}$ that converges to x. It, thus, follows that a possibility to eliminate these (crucial) gaps only exists if or in case that and that in case that i.e. the precise form of the (crucial) gaps must be independent of the considered net that converges to x. In order to guarantee independence of (crucial) gaps from particularly chosen nets that converge to x the assumption $t_{\mid \sim }^{\precsim }$ to be a Hausdorff-topology is needed (cf. Example 1). Indeed, if $t_{\mid \sim }^{\precsim }$ is Hausdorff independence of (crucial) gaps from particularly considered nets that converge to x follows by distinguishing between four different cases (cf. the above described possibilities of (crucial) gaps) each of which can be done by applying always the same indirect argument that is based upon condition SB and the definition of $\psi$ applied to the equations ${\displaystyle \underset{i\in I}{lim}{x}_i=x=\underset{j\in J}{lim}{y}_j}$. Since this argument is routine and obvious in nature it may be omitted for the sake of brevity. The independence of (crucial) gaps from particularly chosen nets that converge to x allows us to now proceed by considering the collection $\mathbb{H}$ of all half-closed half-open and all half-open half-closed gaps of $\psi \left(X\right)$. In accordance with Beardon [1] we define as follows an equivalence relation ∼ on ${\{0,1\}}^{\kappa }$ with respect to $\mathbb{H}$. If $[r,s[$ and $[s,t[$ are adjacent gaps of $X\setminus \bigcup \mathbb{H}$ then $[r,t]$ is an equivalence class of ∼. In addition, if $[u,v[$ and $]w,z]$ respectively are gaps that do not belong to pairs of adjacent gaps of $X\setminus \bigcup \mathbb{H}$ then $[u,v]$ and $[w,z]$ respectively define the corresponding equivalence classes of ∼. All the other equivalence classes of ∼ are defined to be singletons. Since the equivalence classes of ∼ are closed intervals of $({\{0,1\}}^{\kappa },{\le }_{lex})$ we may define the desired preorder ≲ on ${\{0,1\}}^{\kappa }$ that is coarser than ${\le }_{lex}$ by setting for all $x\in {\{0,1\}}^{\kappa }$ and all $y\in {\{0,1\}}^{\kappa }$. Since for all $x\in {\{0,1\}}^{\kappa }$ and all $y\in {\{0,1\}}^{\kappa }$ the inequality $x{\le }_{lex}y$ implies that $x\lesssim y$ it follows that ≲, actually, is coarser than ${\le }_{lex}$. Hence, we now consider the (continuous) identity-function $id:({\{0,1\}}^{\kappa },{\le }_{lex},t^{{\le }_{lex}})⟶({\{0,1\}}^{\kappa },\lesssim ,t^{\lesssim })$. Furthermore, the definition of ∼ implies that the image of the composition of the functions $\psi :(X,\precsim )⟶({\{0,1\}}^{\kappa },{\le }_{lex})$ and $id:({\{0,1\}}^{\kappa },{\le }_{lex},t^{{\le }_{lex}})⟶({\{0,1\}}^{\kappa },\lesssim ,t^{\lesssim })$ has with respect to ≲ neither half-closed half-open nor half-open half-closed gaps.

Since $\psi :(X,\precsim )⟶({\{0,1\}}^{\kappa },{\le }_{lex})$ is an order-embedding we, therefore, may conclude that $\vartheta :(X,\precsim ,t)⟶({\{0,1\}}^{\kappa },\lesssim ,t^{\lesssim })$ is a continuous order-embedding. This last observation finishes the proof of the theorem. □

Let X be the real unit interval $[0,1]$. Then we endow X with the order

It follows that $t^{⊴}=t_{nat\mid [0,1]}\cup \left\{[0,1]\right\}$ which means that $t^{⊴}$ is not Hausdorff. Let $\psi :(X,⊴)⟶({\{0,1\}}^{2^{{\aleph }_0}},{\le }_{lex})$ be the order-embedding that has been described in Theorem 1. Then we may conclude that the nets (sequences) ${\left(\frac{1}{2}-\frac{1}{2n}\right)}_{n\ge 1}$ and ${\left(\frac{1}{4}-\frac{1}{4n}\right)}_{n\ge 1}$ converge with respect to $t^{⊴}$ to 1 but that $\left[{\displaystyle \underset{n\ge 1,\psi \left(\frac{1}{2}-\frac{1}{2n}\right){<}_{lex}\psi \left(1\right)}{sup}\psi \left(\frac{1}{2}-\frac{1}{2n}\right),\psi \left(1\right)}\right[⫋\left[{\displaystyle \underset{n\ge 1,\psi \left(\frac{1}{4}-\frac{1}{4n}\right){<}_{lex}\psi \left(1\right)}{sup}\psi \left(\frac{1}{4}-\frac{1}{4n}\right),\psi \left(1\right)}\right[$.

(i) There exist some ordinal number λ and order-embeddings

and

(ii) In addition to assertion (i) (total) preorders $\le \subset ⊴$ on D and ${\le }_{lex}\subset \lesssim$ and order-embeddings

and

can be chosen in such a way that the compositions

and

Of course, we may assume without loss of generality that $\kappa :=\mid X_{\mid \sim }\mid \ge 2$. In order to now verify assertion (i) we first notice the validity of the equivalences (**) which mean that for all points $x\in X$ and $y\in X$. Let $\lambda$ be the maximum of $\mid D\mid$ and $\kappa$ and $\beta :[0,\kappa [⟶X_{\mid \sim }$ an arbitrarily chosen bijective function. Then the equivalences (**) imply in combination with condition LR the existence of an order-embedding $\eta :(D,\le )⟶({\{0,1\}}^{\lambda },{\le }_{lex})$ that is defined by identifying $\eta \left(a\right)$ for every $a\in D$ with the tuple , if $a\notin \varphi \left(X\right)$. Let, therefore, $a\in \varphi \left(X\right)$. Then we must at first verify that the subsequent definition is independent of the particular chosen point $x\in X$ such that $\varphi \left(x\right)=a$. This means that we must prove that the equation $\varphi \left(x\right)=\varphi \left(y\right)$ implies that $\left[x\right[=\left[y\right]$. Indeed, if $\varphi \left(x\right)=\varphi \left(y\right)$ then the equivalences (**) allow us to conclude that $l\left(x\right)=l\left(y\right)$ and $r\left(x\right)=r\left(y\right)$. Hence, it follows that assumptions of condition LR are satisfied which implies that $\left[x\right]=\left[y\right]$. Let, consequently, some $x\in X$ such that $a=\varphi \left(x\right)$ be arbitrarily chosen. Then we may identify $\eta \left(a\right)$ with the tuple ${\left(a_{\alpha }\right)}_{\alpha <\lambda }$ that is defined by setting for all ordinal numbers $\alpha <\lambda$ (cf. the definition of $\psi$ in the proof of Theorem 1). Of course, also the order-embedding $\psi :(X,\precsim )⟶({\{0,1\}}^{\lambda },{\le }_{lex})$ has to be defined in the same way, i.e. by identifying $\psi \left(x\right)$ for every $x\in X$ with the tuple ${\left(x_{\alpha }\right)}_{\alpha <\lambda }$ that is defined by setting

for every ordinal number $\alpha <\lambda$. The definitions of the order-embeddings $\eta$ and $\psi$ imply that $\psi =\eta \circ \varphi$ which proves assertion (i).

In order to prove the validity of assertion (ii) we use the notation that has been introduced in the proof of assertion (ii) of Theorem 1. Then we apply the arguments that have been used in the proof of assertion (ii) of Theorem1 in order to verify that the image of the order-embedding $\rho :=i{d}_{\psi \left(X\right)}:$$(\psi \left(X\right),{\le }_{lex})⟶({\{0,1\}}^{\lambda },\lesssim )$ neither has half-closed half-open nor half-open half-closed gaps. The proof of assertion (i) implies that ${\eta }^{-1}:(\psi \left(X\right),{\le }_{lex})⟶(\varphi \left(X\right),\le )$ is an order-isomorphism. Hence, the validity of the following implications that will be abbreviated by (*) holds:

If the crucial gap I of $\varphi \left(X\right)$ is the image of some half-closed half-open or some halfopen half-closed interval J then J is a crucial gap of $\varphi \left(X\right)$. And, conversely:

If the crucial gap J of $\psi \left(X\right)$ is the image of some half-closed half-open or some half-open half-closed interval I then I is a crucial gap of $\psi \left(X\right)$.

In particular, we may conclude that the crucial gaps of $\varphi \left(X\right)$ are independent from particularly chosen converging nets. Hence, we may apply the Beardon construction that has been described in the proof of assertion (ii) of Theorem 1 in order to define a preorder ⊴ on D that is coarser than ≤ in such a way that neither the image of the order-embedding $\nu :=i{d}_{\varphi \left(X\right)}:(\varphi \left(X\right),\le )⟶(D,⊴)$ nor the image of the order-embedding $\eta :\left(\varphi X\right),⊴)⟶({\{0,1\}}^{\lambda },\lesssim )$ has any half-closed half-open or half-open half closed gaps. Hence, the validity of the implications (*) allows us to conclude that the compositions $\theta :=\nu \circ \varphi :(X,\precsim ,t)⟶(D,⊴,t^{⊴})$, $\vartheta :=\rho \circ \psi :(X,\precsim ,t)⟶({\{0,1\}}^{\lambda },\lesssim ,t^{\lesssim })$ and $\chi :=\rho \circ \eta :(D,⊴,t^{⊴})⟶({\{0,1\}}^{\lambda },\lesssim )$ are continuous (cf. the proof of assertion (ii) of Theorem 1). In addition, the validity of assertion (i) guarantees that $\vartheta =\chi \circ \theta$ , which finishes the proof of the theorem. □

HD: Let ≾ have a continuous multi-utility representation. Then $(X_{\mid \sim },{\precsim }_{\mid \sim })$ is a Hausdorff space.

OC: Let $(X_{\mid \sim },{\precsim }_{\mid \sim })$ be a Hausdorff space and let ≾ be closed. Then $d\left(y\right)$ is open (and closed) for every point $y\in {\mathbf{N}}_{\precsim }\left(x\right)$ that is maximal with respect to $(X,\precsim )$.

Proof. HD: Let ≾ have a continuous multi-utility representation and let $x\in X$ and $y\in X$ be arbitrarily chosen points such that $not(x\precsim y)$. Then there exists a continuous and increasing function $f_{xy}:(X,\precsim ,t^{\precsim })⟶(\mathbb{R},\le ,t_{nat})$ such that $f_{xy}\left(y\right)<f_{xy}\left(x\right)$. Hence, the desired conclusion follows.

OC: Let $y\in {\mathbf{N}}_{\precsim }\left(x\right)$ be a maximal element of $(X,\precsim )$, which means that $r\left(y\right)=\varnothing$. Then the assumption $t_{\mid \sim }^{\precsim }$ to be Hausdorff implies with help of condition SB that $l\left(y\right)\ne \varnothing$. Hence, we may distinguish between the cases $\left(l\right(y),\precsim )$ to have a maximal element and $\left(l\right(y),\precsim )$ to have no maximal element. Let us, therefore, assume at first that $\left(l\right(y),\precsim )$ to have a maximal element m. Then the interval $]m,y[$ is empty. This means, in particular, that there exists no net ${\left(m\left(s\right)\right)}_{s\in S}$ of points $m\left(s\right)\in r\left(m\right)$ that converges to y. Hence, the set $\mathbf{U}\left(y\right):=\{t\in r(m)\mid t\in r(m)\setminus [y\left]\right\}$ must be closed and we may conclude that $\left[y\right]=r\left(m\right)\setminus \mathbf{U}\left(y\right)$ is open and closed. We, thus, proceed by showing that both sets $l\left(y\right)$ as well as $d\left(y\right)$ are open and closed. In order to verify these properties of $l\left(y\right)$ and $d\left(y\right)$ respectively it suffices to prove that $l\left(y\right)$ is closed and that $d\left(y\right)$ is open. Let, therefore, in a first step some point $p\in \overline{l\left(y\right)}$ be arbitrarily chosen. Then we have to show that $p\in l\left(y\right)$. We, thus, consider some net ${\left(p_o\right)}_{o\in O}$ of points $p_o\in l\left(y\right)$ that converges to p. Since ≾ is closed and $p_o\prec y$ for all $o\in O$ it follows that $p\precsim y$ and it remains to verify that the equivalence $p\sim y$ can be excluded. Indeed, if $p\sim y$ then the just proved property of $\left[y\right]$ to be open (and closed) implies that there exists some index $o_y$ such that $p_o\sim y$ for all $o\in O$ that are at least as great as $o_y$. This contradiction implies that $l\left(y\right)$ must be closed. For later use, in particular in the proof of Theorem 3, we abbreviate this conclusion by (*). Since $l\left(y\right)$ is open and $\left[y\right]$ is open it follows in a second step that $d\left(y\right)=l\left(y\right)\cup \left[y\right]$ is open, which completes the discussion of the case $\left(l\right(y),\precsim )$ to have a maximal element. We now still must think of the situation $\left[{\displaystyle \underset{q\in l\left(y\right)}{sup}q}\right]$ to coincide with $\left[y\right]$. Let, in this situation, $(C,\precsim )$ be some sub-chain of $\left(l\right(y),\precsim )$ such that $\left[{\displaystyle \underset{c\in C}{sup}c}\right]=\left[y\right]$ Because of (*) we may assume without loss of generality $\left[y\right]$ to be not open (and closed). We, thus, may arbitrarily choose some point $c\in C$ in order then consider some net ${\left(y_i\right)}_{i\in I}$ of points $y_i\in r\left(c\right)$ that converges to y. Because of the maximality of y with respect to $(X,\precsim )$ it follows that for every $i\in I$ there exist points $c^{\prime }\in C$ and $c^{\prime \prime }\in C$ such that $c\precsim c^{\prime }\precsim y_i\precsim c^{\prime \prime }$. Indeed, otherwise the definition of $t^{\precsim }$ implies that $\left[y\right]$ is the meet of two open intervals and, thus, open (and closed), which contradicts our assumption $\left[y\right]$ to not being open (and closed). This argument will be abbreviated it by (M). But this consideration allows us to conclude that for every point $l\in l\left(y\right)$ the set $r\left(l\right)\cap d\left(y\right)$ is an open neighborhood of y. Hence, it follows for every point $l\in l\left(y\right)$ that $d\left(y\right)=l\left(y\right)\cup \left(r\right(l)\cap d(y\left)\right)$ is open (and closed), which still was to be shown. □

Let ≾ be semi-closed. Then in order for $t_{\mid \sim }^{\precsim }$ to be Hausdorff it is necessary and sufficient that ≾ satisfies the conditions SB and LR.

As it already has been mentioned in the introduction the validity of the conditions SB and LR are necessary in order to guarantee $t_{\mid \sim }^{\precsim }$ to be Hausdorff. Hence, we may concentrate on the sufficiency part of the lemma. In order to verify that the assumption ≾ to be semi-closed implies in combination with validity of the conditions SB and LR that $t_{\mid \sim }^{\precsim }$ is a Hausdorff topology on $X_{\mid \sim }$ we notice at first that condition SB is equivalent to condition LU which states that for every point $x\in X$ at least one of the sets $l\left(x\right)$ or $r\left(x\right)$ is not empty. Let now points $x\in X$ and $y\in X$ such that $not(x\precsim y)$ be arbitrarily chosen. Then the cases $y\prec x$ and $not(y\precsim x)$ are possible. Therefore, we have to distinguish between these possible cases.

Case 1: $y\prec x$. In this situation we distinguish between two more cases.

Case 1.1: There exist points $u\in X$ and $v\in X$ such that the interval $]u,v[$ is empty and $y\precsim u\prec v\precsim x$. In this case $l\left(v\right)$ is an open set that contains y and $r\left(u\right)$ is an open set that contains x. Therefore, the equation $l\left(v\right)\cap r\left(u\right)=\varnothing$ settles 1.1.

Case 1.2: The closed interval $[y,x]$ does not contain any jump. In this situation there exists some point $z\in X$ such that $y\prec z\prec x$. Hence $l\left(z\right)$ and $r\left(z\right)$ respectively are disjoint open sets that contain y and x respectively.

Case 2:  $not(y\precsim x)$. In this situation condition LU implies that the lemma will be shown if the cases $l\left(x\right)\ne \varnothing$ or $r\left(x\right)\ne \varnothing$ successfully have been handled. Since both cases can be settled by completely analogous arguments it suffices to concentrate on the case $l\left(x\right)$ to be not empty. The inequality $l\left(x\right)\ne \varnothing$ implies with help of condition LR that there exists some $z\in l\left(x\right)$ such that $y\notin i\left(z\right)$ in case that $r\left(x\right)=\varnothing$ or that there exist points $v\in l\left(x\right)$ and $z\in r\left(x\right)$ such that $y\notin [v,z]$ in case that $r\left(x\right)\ne \varnothing$. Since ≾ is semi-closed it, thus, follows that $r\left(z\right)$ and $X\setminus i\left(z\right)$ respectively or $]v,z[$ and $X\setminus [v,z]$ respectively are disjoint open sets that contain the point x and the point y respectively, which still was to be shown. □

In order to verify the proposition we must show that for any two points $x\in X$ and $y\in X$ such that $not(x\precsim y)$ there exist (open) neighborhoods U of x and V of y such that for every point $u\in U$ and every point $v\in V$ the relation $not(u\precsim v)$ holds. Indeed, having proved the existence of U and V it follows that $(U\times V)\cap \precsim =\varnothing$ and we are done. An analysis of the proof of Lemma 2 allows us to concentrate on the case that also the relation $not(y\precsim x)$ holds and that neither $l\left(x\right)$ nor $r\left(x\right)$ is empty. Let us, therefore, assume in contrast that every open neighborhood $]p,q[$ of x and every open neighborhood $]r,s[$ of y that is disjoint from $]p,q[$ contains points $h\in ]p,q[$ and $k\in ]r,s[$ respectively such that $h\precsim k$. In order to proceed we set $\mathbf{I}\left(x\right):=\left\{\right]a,b[\mid x\in ]a,b[\subseteq ]p,q\left[\right\}$ and $\mathbf{I}\left(y\right):=\left\{\right]c,d[\mid y\in ]c,d[\subseteq ]r,s\left[\right\}$. Since $t_{\mid \sim }^{\precsim }$ is Hausdorff we may conclude that ${\displaystyle \left[x\right]=\bigcap _{]a,b[\in \mathbf{I}\left(x\right)}]a,b[}$ and ${\displaystyle \left[y\right]=\bigcap _{]c,d[\in \mathbf{I}\left(y\right)}]c,d[}$. Then we distinguish between the cases $\left[x\right]$ as well as $\left[y\right]$ to be open and closed, $\left[x\right]$ to be open and closed and $\left[y\right]$ to only be closed, $\left[x\right]$ to only be closed and $\left[y\right]$ to be open and closed and $\left[x\right]$ as well as $\left[y\right]$ to only be closed. The case $\left[x\right]$ as well as $\left[y\right]$ to be open and closed is trivial. Indeed, in this case we may set $U:=\left[x\right]$ and $V:=\left[y\right]$. The remaining three cases can be done by analogous arguments. Hence, we may concentrate without loss of generality on the case $\left[x\right]$ as well as $\left[y\right]$ to only be closed. Let now in every interval $]a,b[\in \mathbf{I}\left(x\right)$ and every interval $]c,d[\in \mathbf{I}\left(y\right)$ points $h_a\in ]a,b[$ and $k_c\in ]c,d[$ such that $h_a\precsim k_c$ be arbitrarily chosen. Then we may assume without loss of generality that $h_a\precsim h_{a^{\prime }}$ for all intervals $]a^{\prime },b^{\prime }[\in \mathbf{I}\left(x\right)$ such that $]a^{\prime },b^{\prime }[\subseteq ]a,b[$ or that $h_a\succsim h_{a^{\prime }}$ for all intervals $]a^{\prime },b^{\prime }[\in \mathbf{I}\left(x\right)$ such that $]a^{\prime },b^{\prime }[\subseteq ]a,b[$ and that $k_c\precsim k_{c^{\prime }}$ for all intervals $]c^{\prime },d^{\prime }[\in \mathbf{I}\left(y\right)$ such that $]c^{\prime },d^{\prime }[\subseteq ]c,d[$ or that $k_c\succsim k_{c^{\prime }}$ for all intervals $]a^{\prime },b^{\prime }[\in \mathbf{I}\left(y\right)$ such that $]c^{\prime },d^{\prime }[\subseteq ]c,d[$. The symmetry of the cases to be considered now allows us to concentrate on the case $h_a\precsim h_{a^{\prime }}$ for all intervals $]a^{\prime },b^{\prime }[\in \mathbf{I}\left(x\right)$ such that $]a^{\prime },b^{\prime }[\subseteq ]a,b[$ and $k_c\precsim k_{c^{\prime }}$ for all intervals $]c^{\prime },d^{\prime }[\in \mathbf{I}\left(y\right)$ such that $]c^{\prime },d^{\prime }[\subseteq ]c,d[$ and on the case $h_a\precsim h_{a^{\prime }}$ for all intervals $]a^{\prime },b^{\prime }[\in \mathbf{I}\left(x\right)$ such that $]a^{\prime },b^{\prime }[\subseteq ]a,b[$ and $k_c\succsim k_{c^{\prime }}$ for all intervals $]c^{\prime },d^{\prime }[\in \mathbf{I}\left(y\right)$ such that $]c^{\prime },d^{\prime }[\subseteq ]c,d[$. Since ≾ is assumed to be semi-closed it follows in the first case that $d\left(x\right)\subseteq d\left(y\right)$ which means that $x\precsim y$ and, thus, contradicts our assumption x to be not smaller or equivalent to y. The assumptions of the second case imply that ${\displaystyle d\left(x\right)\subseteq \bigcap _{]c,d[\in \mathbf{I}\left(y\right)}d\left(k_c\right)}$ But ≾ is semi-closed. Hence, we may conclude that the smallest closed increasing set that contains ${\displaystyle \bigcup _{]c,d[\in \mathbf{I}\left(y\right)}i\left(k_c\right)}$ is $i\left(y\right)$. It, thus, follows that ${\displaystyle \bigcap _{]c,d[\in \mathbf{I}\left(y\right)}d\left(k_c\right)=d\left(y\right)}$ which again implies that $x\precsim y$ and, therefore, contradicts the relation $not(x\precsim y)$. This conclusion, finally, proves the validity of the proposition. □

(i)≾ admits a continuous multi-utility representation.

(ii)$t_{\mid \sim }^{\precsim }$ is Hausdorff and ≾ is closed.

(iii)$t_{\mid \sim }^{\precsim }$ is Hausdorff and ≾ is semi-closed.

Proof. (i) ⇒ (ii): It already has been mentioned above that a preorder ≾ that admits a continuous multi-utility representation must be closed. Therefore, Lemma 1 guarantees the validity of the implication $``$(i) ⇒ (ii)".

(ii) ⇒ (iii): Since a closed preorder ≾ is semi-closed nothing has to be proved.

(iii) ⇒ (ii): Proposition 1.

(ii) ⇒ (i): Let assertion (ii) be valid and let points $x\in X$ and $y\in X$ such that $y\in {\mathbf{N}}_{\precsim }\left(x\right)$ be arbitrarily chosen. Then we must prove that there exists some continuous and increasing function $f_{xy}:(X,\precsim ,t)⟶(\mathbb{R},\le ,t_{nat})$ such that $f_{xy}\left(y\right)<f_{xy}\left(x\right)$. In order to verify the existence of $f_{xy}$ we distinguish between the cases y to be contained in $X\setminus l\left(x\right)$ and y to be contained in $l\left(x\right)$.

Case 1:$y\in X\setminus l\left(x\right)$. In this case we must distinguish between the situations y being a maximal element of $(X,\precsim )$ and y to not being a maximal element of$(X,\precsim )$. In the first situation we may apply property OC in order to setting

for all $z\in X$. In the second situation, i.e. y is not maximal element of $(X,\precsim )$ there exists some point $u\in X$ such that $y\prec u$. Of course, u may be a maximal element of $(X,\precsim )$. In this case, however, we may apply the argument that has been applied in the first situation. Hence, we may assume without loss of generality that there exist points $u\in X$ and $v\in X$ such that $y\prec u\prec v$. We proceed by assuming at first that v and, thus, also u is contained in ${\mathbf{N}}_{\precsim }\left(x\right)$. We abbreviate this assumption by (**). In addition, we assume both (open) intervals $]y,u[$ and $]u,v[$ to be empty. These assumptions imply that there does not exist any open interval $]r,t[$ of $(X,\precsim )$ that contains u and is completely contained in $]y,v[$. Hence, there cannot exist any net ${\left(u_i\right)}_{i\in I}$ of points $u_i\in ]y,v[$ that converges to u, which implies that the set $\mathbf{U}\left(u\right)$ of all points $s\in ]y,v[$ the indifference classes $\left[s\right]$ of which are different from $\left[u\right]$ must be closed. This means that we now may apply the conclusion that in the proof of Lemma 1 has been abbreviated by (*) in order to conclude that $l\left(u\right)$ is open and closed. Since both points u and v are contained in ${\mathbf{N}}_{\precsim }\left(x\right)$ these considerations, finally, allow us to define the desired continuous and increasing function $f_{xy}:(X,\precsim ,t)⟶(\mathbb{R},\le ,t_{nat})$ by setting

for all $z\in X$. In addition, the above considerations imply that for the moment we may assume without loss of generality that there exist no points $h\in ]y,v[$, $k\in ]y,v[$ and $q\in ]y,v[$ such that $h\prec k\prec q$ and both intervals $]h,k[$ and $]k,q]$ are empty. With help of this assumption we now verify that the preordered set $\left(\right]y,v[,\precsim )$ is not scattered. Indeed, the assumption implies by complete induction, each induction step of which may be settled by some straightforward indirect argument that $]y,v[,\precsim )$ contains some order-dense subset, i.e. a subset that does not contain any jumps, or that the set $\mathbf{I}\left(\right]y,v\left[\right):=\left\{\right[c,g]\subseteq ]y,v[\mid ]c,g[=\varnothing \}$ that is ordered by setting $[c,g]⊴[c^{\prime },g^{\prime }]\iff [c,g]=[c^{\prime },g^{\prime }]$ or $g\prec c^{\prime }$ for all intervals $[c,g]\in \mathbf{I}\left(\right]y,v\left[\right)$ and $[c^{\prime },g^{\prime }]\in \mathbf{I}\left(\right]y,v\left[\right)$ is order-dense. Let $[0,1]$ be the real unit interval. Then our considerations allow us to conclude that in any case there exists an order-embedding $i:\mathbb{Q}\cap [0,1],\le )⟶(]y,v[,\precsim )$. We, thus, proceed by showing that for all rationals $q\in \mathbb{Q}\cap [0,1]$ and $q^{\prime }\in \mathbb{Q}\cap [0,1]$ such that $q<q^{\prime }$ the inclusion $\overline{l\left(i\right(q\left)\right)}\subseteq l\left(i\left(q^{\prime }\right)\right)$ holds. Since ≾ is closed and, therefore, also semi-closed it follows that $\overline{l\left(i\right(q\left)\right)}\subseteq d\left(i\left(q\right)\right)$. The validity of the strong inequality $i\left(q\right)\prec i\left(q^{\prime }\right)$, thus, implies the desired inclusions $\overline{l\left(i\right(q\left)\right)}\subseteq d\left(i\left(q\right)\right)\subseteq l\left(i\left(q^{\prime }\right)\right)$. These considerations imply that the assumptions of Peleg’s Theorem (cf. Peleg [10]) are satisfied or, equivalently, that the family ${\left\{l\left(i\left(q\right)\right)\right\}}_{q\in \mathbb{Q}\cap [0,1]}$ is a (decreasing) separable system in the sense of Herden [8]. Peleg’s theorem or Theorem 4.1 in Herden [8], therefore, implies the existence of some continuous and increasing function $f_{xy}:(X,\precsim ,t)⟶(\mathbb{R},\le ,t_{nat})$ such that $f_{xy}\left(y\right)=0$ and $f_{xy}\left(x\right)=1$. Let us abbreviate these arguments by (***). In order to finish the first case we still must consider the situation v to be not contained in ${\mathbf{N}}_{\precsim }\left(x\right)$, i.e. the situation v to be contained in $i\left(x\right)$. Of course, it is possible that $r\left(y\right)=i\left(x\right)$, which means that also $u\in i\left(x\right)$. In this situation, however, $i\left(x\right)$ is an open and closed subset of X. Hence, we may define the desired continuous and increasing function $f_{xy}:(X,\precsim ,t)⟶(\mathbb{R},\le ,t_{nat})$ by setting

for all $z\in X$. These considerations now allow us to assume that $y\prec u\prec v$ but $u\in {\mathbf{N}}_{\precsim }\left(x\right)$ and $v\in i\left(x\right)$. Since it already has been shown that we may assume without loss of generality that at least one of the (open) intervals $]y,u[$ or $]u,v[$ is not empty we first briefly discuss the case the case $]y,u[$ to be not empty. In this case, however, we may apply the arguments that have been summarized by (***) in order to guarantee the existence of some continuous and increasing function $f_{xy}:(X,\precsim ,t)⟶([0,1],\le ,t_{nat})$ such that $f_{xy}\left(y\right)=0$ and $f_{xy}\left(x\right)=1$. Hence, we now may assume that the (open) interval $]u,v[$ is not empty. In this situation reiteration of the just used argument in combination with an analysis of the arguments that have been summarized by (***) imply that there exists some net ${\left(w_k\right)}_{k\in K}$ of points $w_k\in ]u,v[$ that converges to u. We now proceed by applying an indirect argument. This means precisely that we assume for all indexes $k\in K$ each point $w_k$ to be contained in $i\left(x\right)$. Since ≾ is a closed preorder this assumption allows us to conclude, however, that $u\in i\left(x\right)$, which contradicts our assumption u to be an element of ${\mathbf{N}}_{\precsim }\left(x\right)$. This contradiction guarantees the existence of some point $w\in ]u,v[$ such that $w\in {\mathbf{N}}_{\precsim }\left(x\right)$. Since $y\prec u\prec w$ now the same situation is given as has been described in (**). This reduction to assumption (**), finally, settles the first case.