Georgian Mathematical Journal Volume 10 (2003), Number 1, 127–131

ON MAXIMAL ot-SUBSETS OF THE EUCLIDEAN PLANE

A. KHARAZISHVILI

Abstract. We say that a subset X of the plane R^{2} is an ot-set if any three points of X form an obtuse triangle. Some properties of ot-sets are investigated. It is shown that no finite ot-subset of R^{2} is maximal, but there exists a countable maximal ot-subset of R^{2}. Several related problems are formulated and discussed.

2000 Mathematics Subject Classification: 52A10, 52A20.

Key words and phrases: Euclidean plane, obtuse triangle, maximal ot-set, strip.

There are many interesting problems and open questions in the geometry of Euclidean spaces, which have a set-theoretic flavour and are closely connected with point set theory and convexity (see, e.g., [1]–[6]). A number of such problems was raised by P. Erdös (cf. [1]). Similar questions frequently arise in combinatorial geometry, classical measure theory and equidecomposability theory.

Here we would like to discuss one problem of this kind for the Euclidean plane R^{2}. We begin with some definitions.

Let X be a subset of the plane. We shall say that X is an ot-set if any three-element subset of X forms an obtuse triangle.

We shall say that an ot-set X ⊂ R^{2} is maximal if there is no ot-set on the plane properly containing X.

Each ot-set in R^{2} is contained in some maximal ot-subset of the plane (this fact follows directly from the Kuratowski–Zorn lemma).

As far as we know, the following problem remains unsolved.

Problem. Give a characterization of all maximal ot-subsets of the plane.

Here are some simple examples of maximal ot-sets in R^{2}.

Example 1. Let X be a semi-circumference on the plane without one of its end-points. It can be easily verified that X is a maximal ot-set in R^{2}.

Example 2. Let f : R → R be a function, let X_{f} denote the graph of f and suppose that the following conditions are satisfied:

(1) f is increasing and continuous;

(2) no three points of X_{f} belong to a straight line, i.e. all points of X_{f} are in general position.

Then it is not difficult to show that X_{f} is a maximal ot-set in R^{2}. Note also that conditions (1) and (2) imply at once that the given function f is strictly increasing.

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Let x and y be any two distinct points in R^{2}. We denote by l(x, y) the straight line passing through x and y.

Let l and l′ be any two distinct parallel straight lines on R^{2}. We recall that all points of R^{2} lying between l and l′ form the strip determined by these two lines. This strip is denoted by S(l, l′). In the sequel, only closed strips are under consideration, i.e. we assume throughout the paper that S(l, l′) contains both lines l and l′. The width of S(l, l′) is equal (by definition) to the distance between l and l′.

A single straight line may be regarded as a strip whose width is equal to zero. Let x and x′ be any two distinct points of R^{2}, let l be the straight line passing

through x and perpendicular to the segment [x, x′], let l′ be the straight line passing through x′ and perpendicular to the same segment. We shall say in the sequel that the strip S(l, l′) is determined by the points x and x′. In this case we also denote S(x, x′) = S(l, l′).

Let {S_{i} : i ∈ I} be a family of strips in R^{2}. We shall say that these strips are in general position if, for any two distinct strips S_{i} and S_{j}, a boundary line of S_{i} is not parallel to a boundary line of S_{j}.

It is not hard to show that the plane cannot be covered by finitely many strips and straight lines. From this circumstance it readily follows that no finite ot-subset of R^{2} can be maximal.

Note that the maximal ot-sets described in Examples 1 and 2 above are of cardinality continuum. In this connection, it is natural to ask whether there exists a countable maximal ot-set in R^{2}.

The main goal of this paper is to demonstrate that there are countable locally finite maximal ot-subsets of R^{2}. In order to establish this fact, we need some auxiliary notions and statements.

Let {S_{i} : i ∈ I} and {S_{j} : j ∈ J} be two families of strips in R^{2}. We shall say that these two families are in general position if all strips of the extended family {S_{i} : i ∈ I} ∪ {S_{j} : j ∈ J} are in general position.

Let X be a subset of R^{2} and let {S_{i} : i ∈ I} be a family of strips. We shall say that X and {S_{i} : i ∈ I} are in general position if the families {S_{i} : i ∈ I} and {S(x, x′) : x ∈ X, x′ ∈ X, x ≠ x′} are in general position.

If Z is a subset of R^{2} and ε > 0, then the symbol V (Z, ε) denotes, as usual, the ε-neighbourhood of Z. Accordingly, we utilize the notation V (z, ε) for a point z ∈ R^{2} and ε > 0.

We shall say that a set C is an open circle in R^{2} if

C = {x ∈ R^{2} : x_{1}^{2} + x_{2}^{2} < r},

where r is a strictly positive real number.

Lemma 1. Let X be a finite set of points in R^{2}, such that all the strips S(x, x′) (x ∈ X, x′ ∈ X, x ≠ x′) are in general position, and let C ⊂ R^{2} be an open circle containing X. Then, for every ε > 0, there exists a finite family of strips {S_{i} : i ∈ I} satisfying the relations:

1) for each i ∈ I, the width of S_{i} is less than ε;

ON MAXIMAL ot-SUBSETS OF THE EUCLIDEAN PLANE 129

2) X and {S_{i} : i ∈ I} are in general position; 3) X ⊂ C \ ∪{S_{i} : i ∈ I} ⊂ V (X, ε). We omit an easy proof of Lemma 1. Lemma 2. Let C be an open circle in R^{2}, let X be a finite ot-subset of

R^{2} and let {S_{1}, S_{2}, …, S_{m}} be a finite family of strips in R^{2} such that X and {S_{1}, S_{2}, …, S_{m}} are in general position. Then there exist pairwise distinct points y_{1}, z_{1}, y_{2}, z_{2}, …, y_{m}, z_{m} satisfying the relations:

1) {y_{1}, z_{1}, y_{2}, z_{2}, …, y_{m}, z_{m}} ⊂ R^{2} \ C;

2) X^{∗} = X ∪ {y_{1}, z_{1}, y_{2}, z_{2}, …, y_{m}, z_{m}} is an ot-subset of R^{2};

3) for each natural number i ∈ [1, m], we have S_{i} = S(y_{i}, z_{i});

4) the family of strips {S(x, x′) : x ∈ X^{∗}, x′ ∈ X^{∗}, x ≠ x′} is in general position.

The proof of Lemma 2 can be obtained by induction on m. We omit the corresponding technical details (which are not difficult).

Theorem 1. There exists a countable locally finite maximal ot-subset of R^{2}.

Proof. Fix a decreasing sequence {ε_{k} : k ≥ 1} of strictly positive real numbers, such that lim_{k→∞} ε_{k} = 0.

Define, by induction on k, two increasing (with respect to the inclusion relation) families {X_{k} : k ≥ 1} and {C_{k} : k ≥ 1} of subsets of R^{2}.

For k = 1, take any finite ot-set X_{1} ⊂ R^{2} with the property that all the strips from {S(x, x′) : x ∈ X_{1}, x′ ∈ X_{1}, x ≠ x′} are in general position, and take also an open circle C_{1} ⊃ X_{1} with radius greater than 1.

Suppose that the sets X_{k} and C_{k} are already defined for a natural number k ≥ 1.

Applying Lemma 1 to X_{k}, C_{k} and ε_{k}, we can find a finite family of strips {S_{i} : i ∈ I_{k}} satisfying the following relations:

1) for each i ∈ I_{k}, the width of S_{i} is less than ε_{k};

2) X_{k} and {S_{i} : i ∈ I_{k}} are in general position;

3) X_{k} ⊂ C_{k} \ ∪{S_{i} : i ∈ I_{k}} ⊂ V (X_{k}, ε_{k}). Now, we apply Lemma 2 to the circle C_{k}, the set X_{k} and the family of strips

{S_{i} : i ∈ I_{k}}. According to the above-mentioned lemma, there exists a finite ot-set X_{k}^{∗} ⊂ R^{2} such that:

a) X_{k} ⊂ X_{k}^{∗};

b) X_{k}^{∗} \ X_{k} ⊂ R^{2} \ C_{k};

c) the family of strips S(x, x′) (x ∈ X_{k}^{∗}, x′ ∈ X_{k}^{∗}, x ≠ x′) is in general position and contains the family {S_{i} : i ∈ I_{k}}.

We put X_{k+1} = X_{k}^{∗}. Besides, let C_{k+1} be an open circle in R^{2} containing X_{k+1} ∪ C_{k}, whose radius is greater than k + 1.

Proceeding in this manner, we are able to construct the desired families of sets {X_{k} : k ≥ 1} and {C_{k} : k ≥ 1}.

Note that, in virtue of our construction, we have the equality

R^{2} = ∪{C_{k} : k ≥ 1}.

Finally, we define

X = ∪{X_{k} : k ≥ 1}.

A straightforward verification shows that X is a countable locally finite otsubset of the plane, and we are going to demonstrate that X is maximal. For this purpose, take any point t ∈ R^{2} \ X and check that X ∪ {t} cannot be an ot-set in R^{2}. Since lim_{k→∞} ε_{k} = 0, there exists a natural number k_{0} such that, for all natural numbers k > k_{0}, we have the relation

t ∈ C_{k} \ ∪{V (x, ε_{k}) : x ∈ X_{k}}

and, consequently, t ∈ ∪{S_{i} : i ∈ I_{k}} (see relation 3) above). Hence t ∈ S_{i} for some i ∈ I_{k}. But, in view of our construction, S_{i} = S(y, z) where {y, z} ⊂ X_{k+1} \ X_{k} ⊂ X, and we also remember that the width of S_{i} is less than ε_{k}. Now, taking this circumstance into account, it can easily be seen that if k is large enough, then the values of all three angles in the triangle [t, y, z] do not exceed π/2, whence it follows that X ∪ {t} is not an ot-subset of R^{2}.

The theorem has thus been proved. □

A statement analogous to Theorem 1 can be established for the n-dimensional Euclidean space R^{n} where n > 2. In other words, we have the following

Theorem 2. If n > 2, then there exists a countable locally finite maximal ot-set in the space R^{n}.

The proof of this statement can be carried out in the same manner as for the Euclidean plane R^{2} (some additional purely technical details occur, but they do not represent essential difficulties).

It would be interesting to find a characterization of all maximal ot-subsets of the space R^{n}, where n > 2. In this context, the following example is relevant.

Example 3. Consider the three-dimensional Euclidean space R^{3} and its two-dimensional subspace R^{2} × {0}. Let S be a closed semi-circumference in R^{2} × {0} with end-points y and z. Let l(y, z) denote the straight line passing through y and z. Take any point x on l(y, z) not belonging to the linear segment [y, z] and put

X = (S \ {y}) ∪ {(x, 1)}.

It is not hard to check that X is an ot-set in R^{3}. Also, as we know, S \ {y} is a maximal ot-subset of the plane R^{2} × {0} (see Example 1). Since X properly contains S \ {y}, we conclude that S \ {y} is not a maximal ot-set in the space R^{3}.

Moreover, we cannot even assert that X is a maximal ot-subset of R^{3}. For instance, if the semi-circumference S is such that

(∀t ∈ S)(||t − x|| < 1),

then the set

X′ = (S \ {y}) ∪ {(x, 1)} ∪ {(x, −1)} = X ∪ {(x, −1)}

turns out to be an ot-subset of R^{3} and properly contains X.

Remark 1. In connection with Theorem 1, the following question arises naturally: does there exist a countable bounded maximal ot-subset of R^{2}? This question remains open.

Remark 2. Let H be a pre-Hilbert space (over the field R) and let X be a subset of H. We say that X is an rt-set if each three-element subset of X forms a rectangular triangle. In [5] a characterization of all rt-sets was given (for more details, see [6]). In particular, it was demonstrated there that card(X) ≤ c for any rt-set X, where c denotes the cardinality of the continuum. In the case H = R^{2}, it can easily be shown that X is a maximal rt-subset of H if and only if X coincides with the set of vertices of a rectangle in H.

References

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2. H. Hadwiger, Vorlesungen über Inhalt, Oberfläche und Isoperimetrie. Springer-Verlag, Berlin–Göttingen–Heidelberg, 1957.

3. Z. A. Melzak, Problems connected with convexity. Canad. Math. Bull. 8(1965), 565–573.

4. Z. A. Melzak, More problems connected with convexity. Canad. Math. Bull. 11(1968), 489–494.

5. A. B. Kharazishvili, The characterization of rt-sets in Euclidean spaces. (Russian) Soobshch. Akad. Nauk Gruz. SSR (Bull. Acad. Sci. Georgian SSR) 86(1977), 537-540.

6. A. B. Kharazishvili, Selected questions in the geometry of Euclidean spaces. (Russian) Tbilisi Universithy Press, Tbilisi, 1978.

(Received 22.10.2002)

Author’s address:

I. Vekua Institute of Applied Mathematics Tbilisi State University 2, University St., Tbilisi 380093 Georgia