Question 6.T.17: A function f : D → R is continuous if and only if, for every...
A function f : D → \mathbb{R} is continuous if and only if, for every open set G there is an open set H such that f^{−1}(G) = H ∩ D.
(i) Suppose f is continuous and G ⊆ \mathbb{R} is open. If G ∩ f(D) = Ø, then f^{−1}(G) = Ø and we take H to be the open set Ø. Assume, therefore, that c ∈ f^{−1}(G), so that f(c) ∈ G. Since G is open, there is a positive number ε such that
(f(c) − ε, f(c) + ε) ⊆ G. (6.24)
By the continuity of f, there is a δ > 0 such that
x ∈ (c − δ, c + δ) ∩ D ⇒ f(x) ∈ (f(c) − ε, f(c) + ε). (6.25)
Using the notation U_{c} = (c − δ, c + δ), V_{c} = (f (c) − ε, f(c) +ε), (6.24)
and (6.25) are equivalent to
f(U_{c} ∩ D) ⊆ V_{c} ⊆ G,
which implies
U_{c} ∩ D ⊆ f^{−1}(G). (6.26)
Let
H =\cup\left\{U_{c} : c ∈ f^{−1}(G)\right\},
which is clearly open, and, in view of (6.26), satisfies
H ∩ D ⊆ f^{−1}(G).
But since every point c ∈ f^{−1}(G) lies in U_{c} ∩ D, we actually have
H ∩ D ⊆ f^{−1}(G).
(ii) Conversely, suppose that for every open set G there is an open set H such that f^{−1}(G) = H ∩ D. Take any c ∈ D and let ε be any positive number. Since the interval
G = (f(c) − ε, f(c) + ε)
is open, there is an open set H which satisfies H ∩ D = f^{−1}(G). Since f(c) ∈ G,
c ∈ f^{−1}(G) ⊆ H;
and since H is open, there is a δ > 0 such that
(c − δ, c + δ) ⊆ H,
from which we conclude that
x ∈ (c − δ, c + δ) ∩ D ⇒ x ∈ H ∩ D
⇒ x ∈ f^{−1}(G)
⇒ f(x) ∈ G
⇒ f(x) ∈ (f(c) − ε, f(c) + ε).
Thus f is continuous at c. But since c was an arbitrary point in D, f is continuous on D.