Georgian Mathematical Journal 1(94), No. 5, 567-573

ON PERFECT MAPPINGS FROM ℝ TO ℝ

I. TSERETELI

Abstract. Perfect mappings from ℝ to ℝ and also mappings close to perfect ones are considered. Some of their properties are given.

This paper deals with some properties of perfect mappings from the space of real numbers with natural topology ℝ to ℝ. For example, it is shown that no perfect mapping from ℝ to ℝ has finite partial limits at +∞ and −∞, that no bounded mapping from ℝ to ℝ is perfect, and so on. We also consider mappings from ℝ to ℝ (not necessarily continuous) with closed, compact and bounded fibers and mapppings under which the image of any closed subset of ℝ is closed in ℝ.

Recall that a mapping f : X → Y of topological spaces is called a closed mapping [1] if f is continuous and the following condition is satisfied:

(CL ↓) for every closed subset F of the space X the image f(F) is closed in Y.

Recall also that a mapping f : X → Y of topological spaces is said to be perfect [1] if X is a Hausdorff space, f is a closed mapping and f satisfies the following condition:

(CM_{p}^{−1}) for any point y ∈ Y the fiber f^{−1}(y) is compact.

Note that if f : X → Y is a mapping from the Hausdorff space X to the space Y (not necessarily continuous), satisfying the conditions (CL ↓) and (CM_{p}^{−1}) (see Example 1 below), then for every compact subspace Z of the space Y the inverse image f^{−1}(Z) is compact. (The proof of this fact repeats the proof of Theorem 3.7.2 [11].) This implies that every function f : X → ℝ from the Hausdorff space X to ℝ, satisfying the conditions (CL ↓) and (CM_{p}^{−1}), is a Borel function.

In the sequel K will denote the class of mappings (not necessarily continuous) from ℝ to ℝ.

We shall define the following classes of mappings:

K(C) ≡ {f ∈ K|f is continuous }; 567

1991 Mathematics Subject Classification. 54C10.

1072-947X/94/0900-0567$07.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1994 Plenum Publishing Corporation

K(CL ↓) ≡ {f ∈ K| for any closed subset F of K the image f(F) is closed in ℝ};

K(CM_{p}^{−1}) ≡ {f ∈ K| for any point y ∈ ℝ the fiber f^{−1}(y) is compact }; K(CL_{p}^{−1}) ≡ {f ∈ K| for any point y ∈ ℝ the fiber f^{−1}(y) is closed in ℝ}; K(B_{p}^{−1}) ≡ {f ∈ K| for any point y ∈ ℝ the fiber f^{−1}(y) is a bounded

subset of ℝ}.

Theorem 1. Let f : ℝ → ℝ be a closed mapping and let

lim_{x→+∞} f(x) = a ∈ ℝ ( lim_{x→−∞} f(x) = a ∈ ℝ).

Then there exists a real number M such that for any x > M (for any x < M) we have f(x) = a.

Proof. Let lim_{x→+∞} f(x) = a ∈ ℝ (the other case is analogous). Assume that our assertion is not true. Then for any real number M there must exist a real number x_{M} > M such that f(x_{M}) ≠ a. In particular, ∃x_{1} : x_{1} > 1 : f(x_{1}) ≠ a; ∃x_{2} > x_{1} + 1 : f(x_{2}) ≠ a and so on ∃x_{k} : x_{k−1} + 1 : f(x_{k}) ≠ a; ∃x_{2} > x_{1} + 1 : f(x_{2}) ≠ a and so on. Thus we have a sequence (x_{k})_{k≥1} of real numbers such that x_{k} > k and f(x_{k}) ≠ a. for any natural number k.

It is obvious that lim_{k→∞} x_{k} = +∞, the set {x_{k}}_{k=1}^{∞} is closed in ℝ and a ∈ {f(x_{k})}_{k=1}^{∞}. But, by assumption, lim_{x→+∞} f(x) = a. Hence, by virtue of the continuity of f, lim_{k→∞} f(x_{k}) = a. Therefore a belongs to the closure of the set {x_{k}}_{k=1}^{∞} in ℝ. Since a ∈ {f(x_{k})}_{k=1}^{∞}, the set {f(x_{k})}_{k=1}^{∞} is not closed in ℝ.

On the other hand, since the set {x_{k}}_{k=1}^{∞} is closed in ℝ, the set f({x_{k}})_{k=1}^{∞} must be closed too by the condition. This is the contradiction. ■

Corollary 1. Let f : ℝ → ℝ be a closed mapping and let

lim_{x→+∞} f(x) = a ∈ ℝ ( lim_{x→−∞} f(x) = a ∈ ℝ).

Then the set f^{−1}(a) is not bounded.

Proof. Let lim_{x→+∞} f(x) = a ∈ ℝ (lim_{x→−∞} f(x) = a ∈ ℝ). Then, by the above theorem, there exists a real number M such that f^{−1}(a) ⊃ (M; +∞)(f^{−1}(a) ⊃ (−∞; M)). Therefore f^{−1}(a) is not bounded. ■

Corollary 2. If f : ℝ → ℝ is a perfect mapping, then there are no limits of the function f at +∞ and −∞. In particular, no perfect mapping f : ℝ → ℝ has asymptotes.

Proof. Let lim_{x→+∞} f(x) = a ∈ ℝ (lim_{x→−∞} f(x) = a ∈ ℝ). Then, by the previous corollary, f^{−1}(a) is not bounded. But since f is a perfect mapping, the set f^{−1}(a) must be compact and hence bounded. This is the contradiction. ■

ON PERFECT MAPPINGS FROM ℝ TO ℝ 569

Theorem 2. Let f ∈ K(CL ↓) ∩ K(CL_{p}^{−1}) (see Example 2 below). Assume that a ∈ ℝ. If there exists a left-hand limit lim_{x→a−} f(x) (respectively, if there exists a right-hand limit lim_{x→a+} f(x)), then lim_{x→a−} f(x) = f(a) (respectively, lim_{x→a+} f(x) = f(a)).

Proof. Let lim_{x→a−} f(x) = b ∈ ℝ. Assume that b ≠ f(a). (The other case is analogous.)

We have two possible cases:

1) ∃δ > 0 (δ ∈ ℝ) : ∀x ∈ (a − δ; a) : f(x) = b;

2) ∀δ > 0 (δ ∈ ℝ) ∃x_{δ} ∈ (a − δ; a) : f(x_{δ}) ≠ b.

Let us consider each of them separately.

Case 1. We have f^{−1}(b) ⊃ (a−δ; a). By assumption, f(a) ≠ b. Therefore a ∉ f^{−1}(b). On the other hand, since f^{−1}(b) ⊃ (a−δ; a), the point a belongs to the closure of the set f^{−1}(b). Hence the set f^{−1}(b) is not closed. This is the contradiction. Therefore case 1 is impossible.

Case 2. Denote \frac{|b−f(a)|}{2} ≡ ε. Clearly, ε > 0. Since lim_{x→a−} f(x) = b, there exists a positive real number δ such that for f(x) ∈ (f(a)−ε; f(a)+ε) for any x ∈ (a − δ; a). Therefore b ∉ (f(a) − ε; f(a) + ε).

By the condition ∃x_{1} ∈ (a − δ; a) : f(x_{1}) ≠ b.

Assume that z_{1} ≡ a − δ and z_{2} ≡ max\{a − \frac{δ}{2}; x_{1}\}. Then there exists x_{2} ∈ (z_{2}; a) such that f(x_{2}) ≠ b.

Let the points x_{1}, x_{2}, …, x_{k−1} be already constructed. Denote z_{k} ≡ max\{a − \frac{δ}{k}; x_{k−1}\}. According to the condition there exists a point x_{k} such that x_{k} ∈ (z_{k}; a) and f(x_{k}) ≠ b.

Therefore for any natural k we have a point x_{k} ∈ ℝ such that a − \frac{δ}{k} < x_{k} < a and f(x_{k}) ≠ b.

Since lim_{k→∞}(a − \frac{δ}{k}) = a, we have lim_{k→∞} x_{k} = a.

Since lim_{x→a−} f(x) = b, we have lim_{k→∞} f(x_{k}) = b.

The set {a} ∪ {x_{k}}_{k=1}^{∞} is obviously closed in ℝ.

Now let us consider the set f({a} ∪ {x_{k}}_{k=1}^{∞}) = {f(a)} ∪ {f(x_{k})}_{k=1}^{∞}. Since for any k ≥ 1 f(x_{k}) ≠ b and, by assumption, f(a) ≠ b, we obtain b ∉ f({a}∪{x_{k}}_{k=1}^{∞}). On the other hand, since lim_{k→∞} f(x_{k}) = b, b belongs to the closure of the set f({a} ∪ {x_{k}}_{k=1}^{∞}) = {f(a)} ∪ {f(x_{k})}_{k=1}^{∞}. Hence the set f({a} ∪ {x_{k}}_{k=1}^{∞}) is not closed in ℝ.

But the set {a} ∪ {x_{k}}_{k=1}^{∞} is closed in ℝ and, by the condition, the set f({a} ∪ {x_{k}}_{k=1}^{∞}) must be closed in ℝ. This is the contradiction. ■

Corollary 3. Let f ∈ K(CL ↓) ∩ K(CL_{p}^{−1}). Then any point of discontinuity of the mapping f must be only of the second kind.

Theorem 3. Let f ∈ K(CL ↓) ∩ K(B_{p}^{−1}) (see Example 3 below). Then the function f does not have finite partial limits at +∞ and −∞.

Proof. Assume, conversely, that there is a sequence of real numbers x_{1}, x_{2}, …, x_{n}, … such that lim_{n→∞} x_{n} = +∞ and lim_{n→∞} f(x_{n}) exists. Let lim_{n→∞} f(x_{n}) = a ∈ ℝ. (The case concerning −∞ is analogous.) Here we have two possible cases:

(a) the set { n | f(x_{n}) = a } is empty or finite;

(b) the set { n | f(x_{n}) = a } is infinite.

Consider each of them separately.

Case (a). There exists a natural number n_{0} such that f(x_{n}) ≠ a for any n ≥ n_{0}. Then a ≠ f({x_{n}}_{n=n_{0}}^{∞}) = {f(x_{n})}_{n=n_{0}}^{∞}. But lim_{n→∞} f(x_{n}) = a. Therefore the set f({x_{n}}_{n=n_{0}}^{∞}) is not closed in ℝ. On the other hand, since lim_{n→∞} x_{n} = +∞, the set {x_{n}}_{n=n_{0}}^{∞} is closed in ℝ and, hence, by the condition, the set f({x_{n}}_{n=n_{0}}^{∞}) must be closed in ℝ. This is the contradiction.

Case (b). There exists a subsequence (x_{n_{i}})_{i} ≥ 1 of the sequence (x_{n})_{n≥1} such that f(x_{n_{i}}) = a for any i ≥ 1. Since lim_{n→∞} x_{n} = +∞, we have lim_{i→∞} x_{n_{i}} = +∞. Hence the set {x_{n_{i}}}_{i=1}^{∞} is not bounded. On the other hand, f^{−1}(a) ⊃ {x_{n_{i}}}_{i=1}^{∞} and therefore the set f^{−1}(a) will not be bounded. This is the contradiction. ■

Remark 1. Corollary 2 is a consequence of Theorem 3.

Corollary 4. No perfect mapping from ℝ to ℝ has finite partial limits at +∞ and −∞.

Corollary 5. Let f : ℝ → ℝ be a function such that at least one of the following two conditions is satisfied:

1) ∃M, L ∈ ℝ : ∀x > M : |f(x)| < L;

2) ∃M, L ∈ ℝ : ∀x < M : |f(x)| < L.

Then the function f is not perfect. in particular, no bounded function from ℝ to ℝ is perfet.

Proof. If f : ℝ → ℝ, satisfies for example, the first condition, then there exist at least two (possibly equal to each other) partial limits of f at +∞: \overline{lim}_{x→+∞}f(x) and \underline{lim}_{x→+∞}f(x). But this contradicts Corollary 4. ■

Corollary 6. If f : ℝ → ℝ is a perfect mapping, then for any positive real numbers M and L there exist points x′′_{M} < −M and x′_{M} > M such that |f(x′′_{M})| > L and |f(x′_{M})| > L.

Theorem 4. Let f ∈ K(CL ↓) ∩ K(CL_{p}^{−1}) (see Example 2 below). Assume that a ∈ ℝ and let (x_{n})_{n≥1} be a sequence of real numbers such that lim_{n→∞} x_{n} = a. Assume also that the limit lim_{n→∞} f(x_{n}) exists. Then lim_{n→∞} f(x_{n}) = f(a).

Proof. Assume that lim_{n→∞} f(x_{n}) = b ∈ ℝ and b ≠ f(a). The following two cases are possible:

1) ∃n_{0} : ∀n ≥ n_{0} f(x_{n}) = b;

2) ∀n ∃m(n) > n : f(x_{m(n)}) ≠ b.

Case 1. Clearly, f^{−1}(b) ⊃ {x_{n}}_{n=n_{0}}^{∞} and since, by assumption, f(a) ≠ b, we have a ∉ f^{−1}(b). From f^{−1}(b) ⊃ {x_{n}}_{n=n_{0}}^{∞} and lim_{n→∞} x_{n} = a it follows that the point a belong to the closure of the set f^{−1}(b). hence the set f^{−1}(b) is not closed in ℝ. This is the contradiction.

Case 2. There exists a subsequence (x_{n_{i}})_{i≥1} of the sequence (x_{n})_{n≥1} such that for any natural i ≥ 1 we have f(x_{n_{i}}) ≠ b. Clearly, lim_{i→∞} f(x_{n_{i}}) = b. Since lim_{n→∞} x_{n} = a, we have lim_{i→∞} x_{n_{i}} = a. Hence the set {a}∪{x_{n_{i}}}_{i=1}^{∞} is closed in ℝ. Consider the image f({a}∪{x_{n_{i}}}_{i=1}^{∞}) = {f(a)}∪{f(x_{n_{i}})}_{i=1}^{∞}. Since f(x_{n_{i}}) ≠ b and f(a) ≠ b for any i ≥ 1, we obtain b ∉ {f(a)} ∪ {f(x_{n_{i}})}_{i=1}^{∞}. On the other hand, since lim_{i→∞} f(x_{n_{i}}) = b, b belongs to the closure of the set {f(a)} ∪ {f(x_{n_{i}})}_{i=1}^{∞}. Hence this set is not closed. This is the contradiction. ■

We shall conclude the paper with several examples:

Example 1. Let f_{1} : ℝ → ℝ be a function defined by the formula

f_{1}(x) = \begin{cases}log_{\frac{1}{2}} |x|, & for x ≠ 0, \\ 0, & for x = 0. \end{cases}

We shall show that f_{1} ∈ K(CL ↓) ∩ K(CM_{p}^{−1}) but f_{1} ∉ K(C). Indeed, it is obvious that f_{1} ∈ K(CM_{p}^{−1}) and f_{1} ∉ K(C). Therefore it remains to show that f_{1} ∈ K(CL ↓).

Let F be a closed subset of ℝ and a be an accumulation point of the set f(F). Then there exists a sequence of real numbers (y_{k})_{k≥1} such that for every k ≥ 1 we have y_{k} ∈ f(F), y_{k} ≠ a, y_{k} ≠ 0, and lim_{k→∞} y_{k} = a. Clearly, for every k ≥ 1 there exists a point x_{k} ∈ F such that f(x_{k}) = y_{k}. We have a = lim_{k→∞} y_{k} = lim_{k→∞} f(x_{k}) = lim_{k→∞} log_{\frac{1}{2}} |x_{k}| ⇒ lim_{k→∞} |x_{k}| = (\frac{1}{2})^{a}. Then either there exists a subsequence (x_{n_{i}})_{i≥1} of the sequence (x_{n})_{n≥1} such that lim_{i→∞} x_{n_{i}} = (\frac{1}{2})^{a} or there exists a subsequence (x_{n_{j}})_{j≥1} of the sequence (x_{n})_{n≥1} such that lim_{j→∞} x_{n_{j}} = −(\frac{1}{2})^{a}.

Let (x_{n_{j}})_{j≥1} be a subsequence of the sequence (x_{n})_{n≥1} such that

lim_{j→∞} x_{n_{j}} = −(\frac{1}{2})^{a}.

(If such a subsequence does not exist, one may consider a subsequence (x_{n_{i}})_{i≥1} of the sequence (x_{n})_{n≥1} with lim_{i→∞} x_{n_{i}} = (\frac{1}{2})^{a}).

Since F is closed and for x_{n_{j}} ∈ F, −(\frac{1}{2})^{2} ∈ F for each j ≥ 1. But f(−(\frac{1}{2})^{a}) = a. Therefore a ∈ f(F). Hence the set f(F) is closed in ℝ.

Example 2. Let the mapping f_{2} : ℝ → ℝ be defined as follows:

f_{2}(x) = \begin{cases}1, & for x ≤ 0, \\ log_{\frac{1}{2}} x, & for x > 0. \end{cases}

We shall show that f_{2} ∈ K(CL ↓) ∩ K(CL_{p}^{−1}) but f_{2} ∉ K(C) and f_{2} ∉ K(B_{p}^{−1}).

It is obvious that f_{2} ∉ K(C) and f_{2} ∈ K(CL_{p}^{−1}).

Since f_{2}^{−1}(1) = (−∞; 0] ∪ \{\frac{1}{2}\} , f_{2} ∉ K(B_{p}^{−1}).

Now let us prove that f_{2} ∈ K(CL ↓). Assume that F ⊆ ℝ is a closed subset of ℝ and a ∈ ℝ is an accumulation point of the set f(F). Then there exists a sequence of real numbers (y_{k})_{k≥1} such that for any k ≥ 1 we have y_{k} ∈ f(F), y_{k} ≠ a, y_{k} ≠ 1, and lim_{k→∞} y_{k} = a.

Since for any k ≥ 1 we have y_{k} ≠ 1 and y_{k} ∈ f(F), there exists, for each k ≥ 1, a positive real number x_{k} such that x_{k} ∈ F and f(x_{k}) = log_{\frac{1}{2}} x_{k} = y_{k}. Thus

lim_{k→∞} y_{k} = a ⇒ lim_{k→∞} (\frac{1}{2})^{y_{k}} = (\frac{1}{2})^{a} ⇒ lim_{k→∞} x_{k} = (\frac{1}{2})^{a}.

Since F is closed and for each k ≥ 1, we have x_{k} ∈ F, (\frac{1}{2})^{a} ∈ F. But f((\frac{1}{2})^{a}) = a. Hence a ∈ f(F). Therefore the set f(F) is closed in ℝ.

Example 3. Let the mapping f_{3} : ℝ → ℝ be determined by the formula

f_{3}(x) = \begin{cases}x, & for x ∈ (−∞; 0] ∪ [1; +∞), \\ 0, & for x ∈ (0; 1). \end{cases}

We shall show that f_{3} ∈ K(CL ↓) ∩ K(B_{p}^{−1}) but f_{3} ∉ K(C) and f_{3} ∉ K(CL_{p}^{−1}). That f_{3} ∉ K(C) and f_{3} ∈ K(B_{p}^{−1}) is obvious.

Since f_{3}^{−1}(0) = [0; 1), we have f_{3} ∉ K(CL_{p}^{−1}). Let us show that f_{3} ∈ K(CL ↓). For this take any closed subset F of ℝ and consider an accumulation point a of the set f(F). Since \overline{f(F)} ⊂ \overline{f(ℝ)} = (−∞; 0] ∪ [1; +∞), we have a ∈ (−∞; 0] ∪ [1; +∞).

If a ∈ (−∞; 0] (if a ∈ [1; +∞)), then there exists a sequence of real numbers (y_{k})_{k≥1} such that for every k ≥ 1 we have y_{k} ∈ f(F), y_{k} ≠ a, y_{k} < 0 (respectively, y_{k} > 1) and lim_{k→∞} y_{k} = a.

Since each y_{k} ∈ f(F), for every k ≥ 1 there exists x_{k} ∈ F such that y_{k} = f(x_{k}) = x_{k}. Therefore a = lim_{k→∞} y_{k} = lim_{k→∞} x_{k}. Since F is closed, a ∈ F. But, clearly, f(a) = a.

Hence a ∈ f(F). Therefore f(F) is a closed subset of ℝ.

Reference

1. R. Engelking, General Topology. PWN-Polish Scientific Publishers, Warszawa, 1977.

(Received 07.05.93)

Author’s address: Faculty of Mechanics and Mathematics I. Javakhishvili Tbilisi State University 2, University St., Tbilisi 380043 Republic of Georgia