Page 16 - MATINF Nr. 8

16 2 RESULTS

For the converse, consider an arbitrary a ∈ R. In order to prove that f −1 ((−∞, a]) ⊆ R
is closed, consider a sequence x = (x n ) ⊆ f −1 ((−∞, a]) such that x n → γ ∈ R. Then
n≥1
f (x n ) ≤ a for each n ≥ 1. Also, x is bounded, and then by our hypothesis, there exists
a real number α ∈ L (x) and β ∈ L (f (x)) such that f (α) ≤ β. Since x is convergent,
then L (x) = {lim (x)}, and therefore γ = α. In particular, f (γ) ≤ β. Now observe that
f (x) ⊆ (−∞, a], and therefore L (f (x)) ⊆ [−∞, a]. Since β ∈ L (f (x)), then β ≤ a. Since we
already have proved that f (γ) ≤ β, then f (γ) ≤ a. Thus γ ∈ f −1 ((−∞, a]), as needed. □

As a corollary of Theorem 1, we obtain the result which corresponds to the property which
is implied by (4).

Theorem 2. Let f : R → R be a function. Then f is upper semi-continuous if and only if it
has the property that given any bounded sequence of real numbers x, there exist a real number
α ∈ L (x) and β ∈ L (f (x)) such that f (α) ≥ β.

Proof. Just observe that f has the property from the statement of Theorem 2 if and only if
the function −f has the property from the statement of Theorem 1. □
As a corollary of Theorem 1 and Theorem 2, we obtain the characterization of functions f
satisfying (6).

Theorem 3. A function f : R → R satisfies (6) for each bounded sequence x if and only if it is
continuous.

Proof. Suppose first that f is continuos on R, and consider a bounded sequence of real
numbers x = (x n ) )
n≥1 . By the Bolzano–Weierstrass theorem, there exists a subsequence (x k n n≥1
) → f (α)
and α ∈ R such that x k n → α as n → +∞. The continuity of f at α gives then f (x k n
as n → +∞. Thus α ∈ L (x) and f (α) ∈ L (f (x)), which gives f (α) ∈ (f (L (x)) ∩ L (f (x))) .
Now if (6) holds for each bounded sequence x, by Theorem 1 we obtain that f is lower
semi-continuous on R, and by Theorem 2 we obtain that f is upper semi-continuous on R, as
well. Thus, f is continuous on R. □
Of course, the characterization of functions f satisfying (5), which was obtained at [1,
Theorem 6], was both topological and related to the order relation on R; this came from the
fact that (5) involves the topology on R and its order relation. Since (6) is related only to the
topology of R, the characterization on f obtained at Theorem 3 is also purely topological.
Using now Theorem 3, one can obtain different characterizations for the continuity of the
function f obtained by comparing sets of limits of subsequences.
Theorem 4. For a function f : R → R, the following assertions are equivalent:
i) f is continuous on R;
ii) For each bounded sequence x = (x n ) ⊆ R, we have that
n≥1
f (L (x)) ⊆ L (f (x)) ; (7)

iii) For each bounded sequence x = (x n ) ⊆ R, we have that
n≥1
L (f (x)) ⊆ f (L (x)) ; (8)

iv) For each bounded sequence x = (x n ) ⊆ R, we have that
n≥1
f (L (x)) = L (f (x)) . (9)