() @ Appl. Gen. Topol. 17, no. 2(2016), 83-91doi:10.4995/agt.2016.4116 c© AGT, UPV, 2016 On monotonic fixed-point free bijections on subgroups of R Raushan Z. Buzyakova New York, USA (Raushan Buzyakova@yahoo.com) Abstract We show that for any continuous monotonic fixed-point free automor- phism f on a σ-compact subgroup G ⊂ R there exists a binary operation +f such that 〈G, +f 〉 is a topological group topologically isomorphic to 〈G, +〉 and f is a shift with respect to +f . We then show that mono- tonicity cannot be replaced by the property of being periodic-point free. We explore a few routes leading to generalizations and counterexam- ples. 2010 MSC: 06F15; 54H11; 26A48. Keywords: ordered group; topological group; homeomorphism; shift; mono- tonic function; fixed point; periodic point. 1. Introduction In this paper we present a few results in the direction of the following general problem. Problem. Let G be a topological group and let f be a continuous automorphism on G. Is it possible to restructure the algebra of G without changing the topology so that f is a shift, or taking the inverse, or possibly some other function nicely defined in terms of the new binary operation? We show that any fixed-point free monotonic bijection on a σ-compact subgroup G ⊂ R is a shift with respect to some group structure on G topologically isomorphic to G. In particular, any continuous fixed-point free bijection on the reals R is a shift with respect to some group operation on R compatible with the Euclidean topology of R. This result can be used in particular to Received 12 September 2015 – Accepted 01 June 2016 http://dx.doi.org/10.4995/agt.2016.4116 R. Z. Buzyakova show that any fixed-point free continuous bijection f on R can be colored in three colors. In other words, there exists a cover {Fi : i = 1, 2, 3} such that f(Fi)∩Fi = ∅ for each i = 1, 2, 3. This colorability fact and an ǫ− δ argument for shifts was communicated to the author by Carlos Nicolas. Our theorem shows that an argument for shifts covers the bijection case as any bijective fixed-point free map is a shift with respect to some topology-compatible group operation on R. For a reader interested in coloring, a possible tricolor for the shift f(x) = x + 3 is A = ⋃ n∈Z[6n, 6n + 2], B = ⋃ n∈Z[6n + 2, 6n + 4], and C = ⋃ n∈Z[6n + 4, 6n + 6]. We then discuss failures and successes of certain natural generalization routes. In particular, we show that in our main theorem monotonicity cannot be replaced by the property of being periodic-point free. We use standard notations and terminology. For topological basic facts and terminology one can consult [2]. Since we do not use any intricate algebraic facts, any abstract algebra textbook is a sufficient reference. We consider only continuous maps. All group shifts under discussion are shifts by a non-neutral element. 2. Study By R we denote the set of reals endowed with the Euclidean topology. If a different binary operation or topology is used, it will be specified. Let us agree that given a bijection f and a positive integer n, by fn we denote the composition of n copies of f and by f−n we denote the composition of n copies of f−1. The expression f0 is the identity map. We will use the following folklore facts: Facts: (1) If 〈G, +〉 is a group and f : G → G is a bijection, then 〈G, ⊕f 〉 is a group, where f(x) ⊕f f(y) = f(x + y). (2) The groups in Fact 1 are isomorphic by virtue of f. (3) If 〈G, TG, +〉 is a topological group and f : 〈G, TG〉 → 〈G, TG〉 is a homeomorphism, then 〈G, TG, +〉 and 〈G, TG, ⊕f 〉 are topologically iso- morphic by virtue of f. Theorem 2.1. Let f : R → R be a continuous fixed-point free bijection. Then there exists a binary operation +f on R such that 〈R, +f 〉 is a topological group topologically isomorphic to R and f is a shift with respect to +f . Proof. Since f is fixed-point free, we conclude that either f(x) > x for all x or f(x) < x for all x. We will carry out our argument assuming the former but will make necessary comments for the latter case. To define our new binary operation, we will first put each x ∈ R into correspondence with xf ∈ R. Definition of xf . We define xf for each x ∈ R in three steps a follows: (1) Put 0f = 0 and nf = f n(0) for each n ∈ Z \ {0}. (2) Let h : [0, 1) → [0, f(0)) be an order preserving bijection (hence home- omorphism). For each x ∈ [0, 1), put xf = h(x). c© AGT, UPV, 2016 Appl. Gen. Topol. 17, no. 2 84 On monotonic fixed-point free bijections on subgroups of R Remark. For ”f(x) < x”-case, h is onto (f(0), 0] and order-reversing. Note that h(0) = 0. Thus, this definition agrees with the first step. (3) Fix any x ∈ R. Then there exist a unique integer n and a unique x′ in [0, 1) such that x = x′ + n. Put xf = f n(h(x′)). Note, this definition agrees with steps (1) and (2). Indeed, if x ∈ (0, 1), then x′ = x and n = 0. Then xf = f 0(h(x′)) = h(x), which agrees with step (2). If x is an integer, then x′ = 0 and x = 0 + n. Then nf = xf = f n(h(0)) = fn(0), which agrees with step (1). For further reference, we denote by g the correspondence x 7→ xf . Claim 1. The correspondence g is an order-preserving bijection on R and, hence, a homeomorphism. For ”f(x) < x”-case, g is order-reversing. Let us first show that g is surjective. Fix y ∈ R. Since f is a fixed-point free continuous bijection, R = ⋃ n∈ω([f −(n+1)(0), f−n(0)] ∪ [fn(0), fn+1(0)]). Therefore, there exists x ∈ [0, f(0)) such that y = fn(x) for some integer n. By part (2) of xf -definition, x = h(z ′) for some z′ ∈ [0, 1). Put z = z′ + n. Then, by part (3), g(z) = zf = f n(h(z′)) = fn(x) = y. Let us show that g is order-preserving. Fix a, b ∈ R such that a < b. Let a = a′ + na and b = b ′ + nb, where a ′, b′ ∈ [0, 1) and na, nb ∈ Z. We assume that na, nb are non-negative. Other cases are treated similarly. Case (na < nb): By part (2) of the definition of xf , we have h(a ′), h(b′) ∈ [0, f(0)). Since f is a fixed-point free homeomorphism, it is order- preserving. Therefore, fna(0) ≤ fna(h(a′)) ≤ fna+1(0). Therefore, af ∈ [fna(0), fna+1(0)]. Similarly, bf ∈ [fnb(0), fnb+1(0)]. Since nb > na, we conclude that af ≤ bf. Since f and h are one-to-one, we conclude that af < bf . Case ¬(na < nb): Re-write b − a > 0 as (b′ − a′) + (nb − na) > 0. Since |b′ − a′| < 1, we conclude that nb − na ≥ 0. By the case’s assumption, n = na = nb. Hence, a ′ < b′. By part (2) of the definition, h(a′) < h(b′). Since f is order-preserving, af = f n(h(a′)) < fn(h(b′)) = bf . The claim is proved. Claim 1 and Fact 3 imply that 〈R, ⊕g〉 is a topological group topologically isomorphic to R by virtue of g. The next claim completes the proof of the theorem Claim 2. f is an ⊕g-shift and f(xf ) = xf ⊕g 1f for all xf ∈ R. Fix xf ∈ R. Let x = x′ + n, where x′ ∈ [0, 1) and n is an integer. Then xf = f n(h(x′)). Therefore, f(xf ) = f n+1(h(x′)). Put pf = f n+1(h(x′)). Then p = x′ + (n + 1) = (x′ + n) + 1. By the definition of ⊕g, we have g(p) = g(x′ + n) ⊕g g(1), that is, pf = xf ⊕g 1f . Since pf = f(xf ), the claim is proved. To stress the dependence of ⊕g on f we put +f = ⊕g, which completes our proof. � c© AGT, UPV, 2016 Appl. Gen. Topol. 17, no. 2 85 R. Z. Buzyakova Our discussion prompts a question of whether Theorem 2.1 can be genera- lized to any subgroup G of R and any continuous periodic-point free bijection f on G. We will show later that a generalization of such a magnitude is not possible. However, certain relaxations on hypotheses can be made. We will next show that the conclusion of Theorem 2.1 holds if we replace R by any σ-compact subgroup of R and f by any fixed-point free monotonic bijection. We believe that ”σ-compact subgroup” can be replaced by ”any subgroup”. We will identify the single statement in our argument that requires additional work for a desired generalization. To make argument clearer, let us handle a few cases informally. If G is a discrete subgroup of R, then it is order-isomorphic to Z. Therefore, any monotonic bijection of G is necessarily a shift. If G has a non-trivial connected component, then G = R and Theorem 2.1 applies. To handle the case when G is zero-dimensional and dense in R let us recall a few classical facts. It is due to Sierpienski [4] (see also [5, 1.9.6]) that any countable metric space with no isolated points is homeomorphic to the space of rationals Q. It is due to Alexandroff and Urysohn [1] that any σ-compact zero-dimensional metric space which is nowhere countable is homeomorphic to the product Q × C of the rationals Q and the Cantor Set C. The immediate applications of these characterizations are the following useful fact: Fact: Let X and Y be homeomorphic to Q (or both homeomorphic to Q × C). Let a, b ∈ X be distinct and let c, d ∈ Y be distinct. Then there exists a homeomorphism h : X → Y such that h(a) = c and h(b) = d. This fact, zero-dimensionality, and homogeneity of G imply the following state- ment. Lemma 2.2. Let G be a zero-dimensional dense σ-compact subgroup of R and U a non-empty open subset of G. Let a, b ∈ U be distinct and c, d ∈ G satisfy c < d. Then there exists a homeomorphism h : U → [c, d]∩G such that h(a) = c and h(b) = d. Proof. Since G is a group, it is either countable or nowhere countable. There- fore, G is homeomorphic to Q or Q×C. Since both cases are handled similarly, we assume that the latter is the case. Then any non-empty open subset of G as well as any infinite closed interval of G are σ-compact and nowhere countable. Therefore, both U and [c, d] ∩ G are homeomorphic to Q × C. Next apply Fact. � The argument of our next result follows that of Theorem 2.1. To avoid un- necessary repetition, we will reference the already presented argument in a few places. Even though we could have put both theorems under one umbrella, for readability purpose, the author decided to present them separately. Theorem 2.3. Let G be a σ-compact subgroup of R and let f : G → G be a fixed-point free monotonic bijection. Then there exists a binary operation +f on G such that 〈G, +f 〉 is a topological group topologically isomorphic to G and f is a shift with respect to +f . c© AGT, UPV, 2016 Appl. Gen. Topol. 17, no. 2 86 On monotonic fixed-point free bijections on subgroups of R Proof. We now may assume that G is a non-discrete zero-dimensional subgroup of R. Since G is closed under addition and contains a nontrivial sequence converging to 0, we conclude that G is dense in R. Since f is a monotonic bijection and G is dense in R, there exists a continuous bijective extension f̄ : R → R of f. Let F be the set of all fixed points of f̄. Let J = {Jn : n = 0, 1, ...} consist of all maximal convex sets of R \ F . Note that if F is empty, then J = {J0 = R}. Put I = {In = Jn ∩ G : n = 0, 1, ...}. Claim 1. If I ∈ I, then there exists a convex clopen O ⊂ I in G such that fn(O) ∩ fm(O) = ∅ whenever n 6= m and I = ⋃ n∈Z f n(O). To prove the claim, fix any point p in J \ G, where J ∈ J such that I = J ∩ G. Such a point exists due to zero-dimensionality of G. Due to absence of fixed points and monotonicity of f̄ on J, we conclude that {fn(p) : n ∈ Z} is unbounded in J on either side. Let Ip be the closed interval in R with the endpoints p and f(p). We then have O = Ip ∩ G is as desired. The claim is proved. For each In ∈ I, fix On that satisfies the conclusion of the claim. Next for each x ∈ G we will define xf as follows. Definition of xf . Define xf in three steps. (1) Fix p∗ ∈ I0 and c ∈ G such that c > 0. Put 0f = p∗ and (nc)f = fn(p∗) for each n ∈ Z. (2) Let Ip∗ be the interval of G with endpoints p ∗ and f(p∗). Let A and B be convex clopen subsets of Ip∗ such that p ∗ ∈ A, f(p∗) ∈ B, and A ∪ B = Ip∗. By Lemma 2.2, there exists a homeomorphism h of [0, c] ∩ G with A ⊕ O1 ⊕ ... ⊕ On ⊕ ... ⊕ B such that h(0) = p∗ and h(c) = f(p∗). For each x ∈ [0, c] ∩ G, put xf = h(x). Remark. Note that our use of Lemma 2.2 is the only part in which the author could not carry out the argument for an arbitrary zero- dimensional subgroup of R. Note that our definition agrees with step 1 for p∗ and f(p∗). (3) For each x ∈ G, there exist a unique x′ ∈ [0, c) ∩ G and a unique n ∈ Z such that x = x′ + nc. Put xf = f n(h(x′)). Due to uniqueness of x′ and n, xf is well-defined. Note that this definition agrees with our definitions at steps 1 and 2. Denote by g the correspondence x 7→ xf . Claim 2. g is surjective. To prove the claim, fix y ∈ G. Then y ∈ In ∈ I for some n. By the choice of On, there exist m ∈ Z and z ∈ On such that y ∈ fm(z). If n = 0, we may assume that z 6= f(p∗). By the definition of h, there exists x′ ∈ [0, c) such that h(x′) = z. Put x = x′ + mc. Then g(x) = fm(h(x′)) = fm(z) = y. The claim is proved. c© AGT, UPV, 2016 Appl. Gen. Topol. 17, no. 2 87 R. Z. Buzyakova Claim 3. g is a one-to-one. To show that g is one-to-one, fix distinct a, b ∈ G and let a = a′ + nc and b = b′ + mc, where a′, b′ ∈ [0, c) and n, m ∈ Z. Then g(a) = fn(h(a′)) and g(b) = fm(h(b′)). If n = m, then g(a) 6= g(b) because both h and fn are one-to-one. Assume now that n 6= m. Let h(a′) ∈ Oi and h(b′) ∈ Oj. Assume that i = j. Then g(a) ∈ fn(Oi) and g(b) ∈ fm(Oi). By the choice of Oi, we have fn(Oi) ∩ fm(Oi) = ∅. If i 6= j, then g(a) ∈ Ii and g(b) ∈ Ij. By the definition of I, Ii ∩ Ij = ∅, which proves the claim. Claim 4. g is a homeomorphism. Observe that if x ∈ [cn, c(n + 1)] ∩ G, then g(x) = fn(h(x − nc)). Thus, g is a homeomorphism on [cn, c(n + 1)] ∩ G. Since {[cn, c(n + 1)] ∩ G}n forms a locally finite closed cover of G and g is closed on each element of the cover, we conclude that g is a closed map. By Claims 2 and 3, g is a homeomorphism on G. The claim is proved. Denote by +f the operation ⊕g defined in Facts 1-3. By Fact 3, 〈G, +f 〉 is a topological group topologically isomorphic to G. Following the argument of Theorem 2.1, f(xf ) = xf +f cf for all xf ∈ G. � It is natural to wonder if monotonicity of f in Theorem 2.3 can be replaced by a periodic-point free homeomorphism. Since the latter can be of wild nature, a counterexample is expected. Before we present our construction, we prove the following lemma. Lemma 2.4. Suppose f : Q → Q is not an identity map and p ∈ Q satisfy the following property: (∗) ∀n > 0∃m > 0 such that fm+1((p−1/n, p+1/n)Q) meets f−m((p−1/n, p+1/n)Q) If 〈Q, ⊕〉 is a topological group topologically isomorphic to Q, then f is not a ⊕-shift. Proof. Let h : Q → 〈Q, ⊕〉 be a topological isomorphism . Let ≺ be the order on 〈Q, ⊕〉 defined by a ≺ b if and only if h−1(a) < h−1(b). Clearly, 〈Q, ⊕, ≺〉 is an ordered topological group. Let f be a ⊕-shift. Since f is not the identity map, there exists c ∈ Q \ {h(0)} such that f(x) = x ⊕ c. Without loss of generality, assume that h(0) ≺ c. Fix a, b ∈ Q such that a ≺ p ≺ b ≺ (a ⊕ c) ≺ (p ⊕ c). Clearly fm+1((a, b)≺) misses f −m((a, b)≺) whenever m > 0. Since the topology on 〈Q, ⊕〉 is Euclidean, there exists N such that (p− 1/N, p+ 1/N)Q ⊂ (a, b)≺. Therefore, fm+1((p − 1/N, p + 1/N)) misses f−m((p − 1/N, p + 1/N)) for any m > 0. � Note that a non-identity homeomorphism with periodic points cannot be a shift in any group structure isomorphic to Q. Therefore, the importance of mono- tonicity in Theorem 2.3 should be demonstrated by a non-monotonic homeo- morphism with no periodic points. c© AGT, UPV, 2016 Appl. Gen. Topol. 17, no. 2 88 On monotonic fixed-point free bijections on subgroups of R Example 2.5. There exists a periodic-point free homeomorphism f : Q → Q such that f is not a shift in any topological group structure on Q topologically isomorphic to Q. Construction. All intervals under consideration are in Q. Therefore, instead of (a, b)Q we write (a, b). We will construct a homeomorphism f : Q → Q that satisfies property (*) of Lemma 2.4 for p = 0. Let ǫ be any irrational number strictly between 0 and 1 2 √ 2 . We put ǫ = 1√ 7 for better visualization. Step 1. Let g1 and g−1 be any two order-preserving homeomorphisms with the following ranges and domains: g1 : [ 0 − 1 21 √ 2 , 0 + 1 21 √ 2 ] → [1 − ǫ, 1 + ǫ] g−1 : [−1 − ǫ, −1 + ǫ] → [ 0 − 1 21 √ 2 , 0 + 1 21 √ 2 ] Such maps exist since the endpoints in all intervals are irrational, and therefore, not in G. Step n > 1. Let gn and g−n be any two order-preserving homeomorphisms with the following ranges and domains: gn : gn−1 ◦ ... ◦ g1 ([ 0 − 1 2n √ 2 , 0 + 1 2n √ 2 ]) → [n − ǫ, n + ǫ] g−n : [−n − ǫ, −n + ǫ] → g−1−(n−1) ◦ ... ◦ g −1 −1 ([ 0 − 1 2n √ 2 , 0 + 1 2n √ 2 ]) Note that due to irrationality of √ 2 and √ 7, ranges and domains of these homeomorphisms are clopen intervals in G. For better visualization, let us write out the domains of the defined function with ǫ = 1√ 7 . We have dom(g1) = [ − 1 2 √ 2 , 1 2 √ 2 ] . If n ≤ −1, then dom(gn) = [ n − 1√ 7 , n + 1√ 7 ] . If n ≥ 2, then dom(gn) ⊂ [ (n − 1) − 1√ 7 , (n − 1) + 1√ 7 ] . Due to smallness of 1√ 7 , these sets are mutually disjoint. Let us summarize this observation for further reference. Claim. If n, m ∈ Z \ {0} are distinct, then the domains of gn and gm are disjoint. Next we select special clopen intervals as follows. Definition of An, Bn for n > 0: Fix any non-empty clopen intervals An ⊂ (n − ǫ, n + ǫ) and Bn ⊂ (−n − ǫ, −n + ǫ) with the following properties: P1: An misses the domain of gn+1. Such a set exists because the domain of gn+1 is a proper clopen interval of (n − ǫ, n + ǫ). P2: Bn misses the range of g−(n+1). Such a set exists because the range of g−(n+1) is a proper clopen interval of (−n − ǫ, −n + ǫ). P3: gn ◦ ... ◦ g1 ◦ g−1 ◦ ... ◦ g−n(Bn) misses An. c© AGT, UPV, 2016 Appl. Gen. Topol. 17, no. 2 89 R. Z. Buzyakova For each n, fix a homeomorphism hn from An onto Bn. Define g as follows: g(x) = { gn(x) x is in the domain of gn for n ∈ Z \ {0} hn(x) x ∈ An To show that g is a function on D = ( ⋃ n≥1 An ) ∪( ⋃ {dom(gn) : n ∈ Z \ {0}}), we need to show that any x ∈ D belongs to exactly one element of {An : n = 1, 2, ...} ∪ {dom(gn) : n ∈ Z \ {0}}. By Claim, we may assume that x ∈ An. By property P1, x is not in the domain of gn+1. Since An ⊂ [n − ǫ, n + ǫ], x is not in the domain of any gm with m 6= n + 1. By Claim and selection of An’s, we conclude that x 6∈ Am for n 6= m. Let us show that g is a homeomorphism between its domain and range. By property P2, g is one-to-one. The domain of g is the union of the clopen discrete family {dom(gn) : n ∈ Z \ {0}} ∪ {An : n ∈ Z+}. Since g is one-to-one and is a homeomorphism on each member of the family, g is a homeomorphism. By property P3, g has no periodic points. The complements of the domain and range of g are clopen proper subsets of Q that are unbounded on both sides. Therefore, g has a homeomorphic extension f : Q → Q that has no periodic points. It remains to show that f and 0 satisfy the hypothesis of Lemma 2.4. Fix any n > 0. We have fn (( 0 − 1 2n √ 2 , 0 + 1 2n √ 2 )) = (n − ǫ, n + ǫ). Since An ⊂ (n − ǫ, n + ǫ), we conclude that Bn ⊂ fn+1 (( 0 − 1 2n √ 2 , 0 + 1 2n √ 2 )) . On the other and, Bn ⊂ (−n, −ǫ, −n + ǫ) = f−n (( 0 − 1 2n √ 2 , 0 + 1 2n √ 2 )) . Therefore, m = n is as desired. � Since zero-dimensional spaces may admit very wildly mannered automor- phisms, one may ask if our theorem for R can be extended to Rn. Alas, an example is in order. Example 2.6. There exists a periodic-point free homeomorphism f : R3 → R3 such that f is not a shift in any group structure of R3 topologically isomorphic to R3. Construction. Let h : R3 → R3 be the rotation of the space by √ 2 degrees about the z-axis in the positive direction. Put S = {〈x, y, z〉 ∈ R3 : x2 + y2 ≤ 1}. Define g : S → S by letting g(x, y, z) = 〈x, y, z +1−x2 −y2〉. In words, g slides vertically every point by the distance equal to the distance from the point to the wall of the cylinder. Thus, the points on the boundary of the cylinder are not moved. Clearly, both h and g are homeomorphisms. Define f : R3 → R3 as follows: f(x, y, z) = { h(x, y, z) 〈x, y, z〉 6∈ S g ◦ h(x, y, z) 〈x, y, z〉 ∈ S Since g is the identity on the boundary of the cylinder, we conclude that g ◦h is equal to h on the boundary of S. Therefore, f is a homeomorphism. Points on c© AGT, UPV, 2016 Appl. Gen. Topol. 17, no. 2 90 On monotonic fixed-point free bijections on subgroups of R the z-axis are slided up by 1 unit. Points off the z-axis undergo a rotation by√ 2 degrees. Thus, f has no periodic points. Since the points on the boundary of the cylinder are only rotated, the set {fn(1, 0, 0) : n ∈ ω} is a an infinite subset of the unit disc in the xy-plane centered at the origin. Therefore, the set has a cluster point. However,{〈x, y, z〉 + n〈a, b, c〉 : n ∈ ω} is a closed discrete subset of R3 for any 〈x, y, z〉, 〈a, b, c〉 ∈ R3. Therefore, f cannot be a shift in any group structure on R3 topologically isomorphic to R3. � We would like to finish this study with a few remarks of categorical nature. Assume that f : R → R is a map that is a shift by a non-neutral element with respect to some group operation +f on R that is compatible with the Euclidean topology. Then, f is a homeomorphism and fixed-point free. Therefore, we have a characterization of all maps on R that are shifts by non-neutral elements after some refitting of algebraic structure of R. It is therefore justifiable to view fixed-point free homeomorphisms on R as generalized shifts. Similarly, we can define generalized polynomial (trigonometric, etc.) functions as maps in form f(x) = an ⋆ x n ⊕ ... ⊕ a0, for some addition ⊕ and multiplication ⋆ compatible with the topology of R. Therefore, it would be interesting to consider the following general problem. Problem. Characterize generalized polynomials, trigonometric functions, and generalized versions of other standard calculus functions. At last, recall that we were unsuccessful in generalizing Theorem 2.1 to a desired extent. Therefore, the question is in order. Question. Let G be a topological subgroup of R and let f : G → G be a fixed- point free monotonic homeomorphism. Does there exist a binary operation ⊕ on G such that 〈G, ⊕〉 is a topological group topologically isomorphic to G and f is a ⊕-shift? Acknowledgements. The author would like to thank the referee for valuable remarks and corrections. References [1] P. Alexandroff and P. Urysohn, Uber nuldimensionale Punktmengen, Math Ann. 98 (1928), 89–106. [2] R. Engelking, General Topology, PWN, Warszawa, 1977. [3] C. Nicolas, private communication, 2009 [4] W. Sierpinski, Sur une propriete topologique des ensembles denombrablesdense en soi, Fund. Math. 1 (1920), 11–16. [5] J. van Mill, The Infinite-Dimensional Topology of Function Spaces, Elsevier, 2001. c© AGT, UPV, 2016 Appl. Gen. Topol. 17, no. 2 91