LINEAR OPENNESS OF MULTIFUNCTIONS IN METRIC SPACES

Here, S stands for the closure of the set S. Various openness and almost openness notions are build on the inclusions (1.1) and (1.2). Every openness property implies the corresponding almost openness property, for S⊆ S. It is expected for the converse implication to be true in case that the multifunction F has a closed graph. In this regard, we recall the result in [12, Chapter 6, Lemma 36, page 202] (see also [7, Lemma 6.4.1, page 435]), where Y is a more general space, namely, a uniform space. The metric version of this result revolves around the composite inclusion

Theorem 2.3. Let the metric space X be complete and let the metric space Y resemble normed spaces. Let the multifunction F have a closed graph. Let ω > 0 be a real number. The following two conditions are equivalent: (i) for every (x, y) ∈ graph(F) and for every > 0, there holds the inclusion (1.4); (ii) for every ζ > 0 there exists ∈ (0,ζ), such that for every (x, y) ∈ graph(F), there holds the inclusion (1.5).
The proof of Theorem 2.3 is given in Section 3. Theorem 2.3 may fail if the metric space Y does not resemble normed spaces. A counterexample is given in Section 5.
The best linear openness result states the equivalence of the same openness condition to the weakest almost openness condition based on the inclusion (1.5). It is no longer a corollary of Theorem 1.1. It can be derived in case that Y is a complete metric space which resembles normed spaces.
Theorem 2.4. Let the metric space X be complete, and let the metric space Y be complete and resemble normed spaces. Let the multifunction F have a closed graph. Let ω > 0 be a real number. The following two conditions are equivalent: (i) for every (x, y) ∈ graph(F) and for every > 0, there holds the inclusion (1.4); (ii) for every (x, y) ∈ graph(F) and for every ζ > 0, there exists ∈ (0,ζ) such that there holds the inclusion (1.5).
The proof of Theorem 2.4 is given in Section 4. Theorem 2.4 may fail if either the metric space Y does not resemble normed spaces or it is not complete. Counterexamples are given in Section 5.
Lemma 3.2 may fail if the metric space Y does not resemble normed spaces. A counterexample is given in Section 5.

Proof of Theorem 2.4
In contrast with Theorems 2.1 and 2.3, the almost openness condition of Theorem 2.4 does not state anything about the modulus ∆, so Lemma 3.1 cannot be applied.
Since the condition Ꮽ always implies the condition ᐆ (recall the elementary flowchart at the end of Section 2), Theorem 2.4 is a straightforward corollary of Lemma 4.1 and Theorem 4.2 below. Lemma 4.1 prefigures a new condition, which is based on the rather elaborate relation and which is weaker than the condition ᐆ provided that Y resembles normed spaces. Theorem 4.2 states the equivalence of the condition Ꮽ to the new condition no matter whether Y resembles normed spaces. Let (x, y) ∈ graph(F). Suppose that for every ζ > 0, there exists ∈ (0,ζ) such that there holds the inclusion (1.5). Then for every v ∈ Y \ {y} and for every θ ∈ (0,1), there exists ∈ (0,d(v, y)/(θω)) such that there holds the relation (4.1).
Finally, let the latter condition be satisfied, let (x, y) ∈ graph(F) such that y = v, and let > d(v, y)/ω. We have to prove that (4.2) holds. Let θ ∈ (d(v, y)/(ω ),1). Following the spirit of some ideas in [2, page 195 in order to get a point (a,b) ∈ graph(F) such that  ((a,b),(x, y)) < ω , so a ∈ B(x, ). Note that b ∈ F(a) ⊆ F(B(x, )). We claim that v = b. Suppose, to the contrary, that b = v. By hypothesis, there exists α ∈ (0,d(v,b)/(θω)) such that the set is nonempty. Let q ∈ S. Since q ∈ F(B(a,α)), it follows that there exists p ∈ B(a,α) such that q ∈ F(p). Since d(q,b) < ωα, it follows that d ((p, q),(a,b) Further, it follows from the latter inequality of the Eke-

Counterexamples
Our first counterexample shows that if the metric space X is not complete, then Theorems 2.1, 2.3, and 2.4 may fail, namely, the almost openness condition ᐄ does not imply the openness condition Ꮽ (recall again the elementary flowchart at the end of Section 2).
Counterexample 5.1. Let Q be the metric space of rational numbers and let R be the metric space of real numbers. Let F : Q → R be the multifunction given by graph(F) = {(x, y); x = y}. Let ω = 1.
Then Q is not complete, whereas R is complete and resembles normed spaces, and F has a closed graph. On the one hand, if (x, y) ∈ graph(F) and > 0, then F(B Q (x, )) = B R (y, ) ∩ Q, hence ∆( ) = 0. Here, B Q introduces the balls in Q, whereas B R introduces the balls in R. On the other hand, B R (y, ) ⊆ F(B Q (x, )), hence ∆( ) = .
Our next counterexample shows that if the metric space Y does not resemble normed spaces, then Theorems 2.3 and 2.4 may fail, namely, the almost openness condition ᐅ does not imply the openness condition Ꮽ. Neither does the following openness condition, which is implied by Ꮽ and which implies Ꮾ: (i) there exists ζ > 0 such that for every ∈ (0,ζ) and for every (x, y) ∈ graph(F), there holds the inclusion (1.4).
The counterexample shows also that Lemma 3.2 may fail, that is, the moduli ∆ and ∆ are not superadditive.

Linear openness of multifunctions in metric spaces
Then Y is complete and it follows that Y does not resemble normed spaces. Further, the multifunction F has a closed graph and B(y, ) = F(B(x, )) = F(B(x, )) for all (x, y) ∈ graph(F) and for all > 0, hence  F(B(x, )). Our next counterexample shows that if the metric space Y is not complete, then Theorem 2.4 may fail, namely, the almost openness condition ᐆ does not imply the openness condition Ꮽ. Neither does the following openness condition, which is implied by Ꮽ and which implies Ꮿ: (i) for every (x, y) ∈ graph(F), there exists ζ > 0 such that for every ∈ (0,ζ), there holds the inclusion (1.4).
Counterexample 5.3. Let R be the metric space of real numbers and let Q be the metric space of rational numbers. Let F : R → Q be the multifunction given by graph Then R is complete, Q is not complete but resembles normed spaces, and F has a closed graph. On the one hand, if (x, y) ∈ graph(F) and > 0, then ) for all (x, y) ∈ graph(F) and for all ∈ (0,  Then R is complete, but F does not have a closed graph. Note also that, if (x, y) ∈ graph(F) and > 0, then F(B(x, )) = B(y, ) ∩ (0,+∞). On the one hand, if (x, y) ∈ graph(F), y = v, θ ∈ (0,1), and = d(y,v), then ∈ (0,d(y,v)/θ) and the right-hand side of the relation (4.1) equals the nonempty set B (v,d(y,v) − θ ) ∩ B(y, ). On the other hand, v ∈ range(F). Corneliu Ursescu 211

Metric spaces resembling normed spaces
Every normed space endowed with the metric induced by the norm resembles normed spaces. So does every convex subset of a normed space, but this may be false for nonconvex subsets. Let Y be the unit circle centered at the origin of the Hilbert space R 2 . On the one hand, if Y is endowed with the metric induced by the chord length, d(y, y ) = y − y , then Y does not enjoy the resembling property. For example, if y ∈ Y , then Here, B Y introduces the balls in Y . On the other hand, if Y is endowed with the metric induced by the minimum arc length, d(y, y ) = 2arcsin( y − y /2), then Y enjoys the resembling property. Resuming the preceding example, if y ∈ Y , then Note that the two metrics on Y generate the same topology, so the resembling notion cannot be completely characterized by the topology. Next we give two characterizations of metric spaces Y , which resemble normed spaces. One characterization makes use of chains in Y , that is, finite sequences χ = {χ 0 ,χ 1 ,...,χ n } of points of Y . If χ is a chain in Y , we define the mesh of the chain through mesh(χ) = max d χ 0 ,χ 1 ,...,d χ n−1 ,χ n , ( stands for the set of all dyadic numbers. If λ ≥ 0 is a real number, we say that a function for all s ∈ D and for all t ∈ D. In case that the metric space Y is complete, every λ-Lipschitzian function ψ : D → Y can be uniquely extended to a λ-Lipschitzian function ψ : [0,1] → Y through the equality graph(ψ) = graph(ψ). (6.7) Theorem 6.1. Let Y be a metric space. The following three conditions are equivalent to each other: (1) the metric space Y resembles normed spaces; (2) for every y ∈ Y , for every y ∈ Y , for every µ > 0, and for every λ > d(y, y ), there exists a chain χ in Y such that length(χ) ≤ λ, mesh(χ) ≤ µ, as well as χ 0 = y and χ n = y ; (3) for every y ∈ Y , for every y ∈ Y , and for every λ > d(y, y ), there exists a λ-Lipschitzian function ψ : D → Y such that ψ(0) = y and ψ(1) = y .
To conclude, there exists a unique function ψ : D → Y such that ψ is an extension of every ψ n , hence ψ(0) = y and ψ(1) = y . Moreover, if s ∈ D and t ∈ D, then there exists n such that s ∈ D n and t ∈ D n . Since d(ψ n (τ),ψ n (σ)) < λ|τ − σ| for every pair of successive points τ and σ of D n , it follows d(ψ n (t),ψ n (s)) < λ|t − s|. Therefore, ψ is λ-Lipschitzian, condition (3) is satisfied too, and the proof of the theorem is accomplished. Theorem 6.1 (2) and (3) can be rephrased by using established metric terms. For example, the λ-free variant of the condition (2) states that the metric space Y is well chained (see [19, page 13]), whereas the λ-component states that the metric d is intrinsic (see [1, pages 9, 10] and [6, page 77]).
Further, if the metric space Y is complete, then condition (3) states that Y is connected by rectifiable curves and the distance between two points equals the infimum of the lengths of the rectifiable curves connecting those points. If Y is finitely compact, which means that the bounded, closed subsets of Y are compact (see [5, pages 6, 403]), then there exist connecting curves of shortest lengths (see [5, page 25]). If Y is not finitely compact, the existing property may fail. In this regard, we will construct an appropriate subset Y in the Hilbert space R 2 . First, we consider, in R 2 , the two points y ± = (±1,0) and the sequence of points y (k) = (1/k,0). Finally, we consider the set Y = ∪ k Y (k) , where each Y (k) consists of the two line segments joining y (k) to y ± (see On the one hand, if d Y (y, y ) = y − y , then Y is not a complete metric space and it does not resemble normed spaces. On the other hand, if d Y (y, y ) equals the infimum of the lengths of rectifiable curves in Y joining y to y , then Y is a complete metric 214 Linear openness of multifunctions in metric spaces space which resembles normed spaces, but d Y (y − , y + ) does not equal the minimum of the lengths of rectifiable curves in Y joining y − to y + . Indeed, every such a curve contains a set Y (k) , of which length, 2 √ 1 + 1/k 2 , exceeds 2, whereas d Y (y − , y + ) = 2.