Transfinite Induction and General Topology
241
A.6.2 Proposition Suppose that f : D → E is continuous in the Scott
topologies on domains D and E. Then whenever x ∈ D, a ∈ D c , and a [ x,
we have f (a) [ f (x).
Proof: Let a ∈ D c . Since f is continuous at a, given any Scott neighbourhood
V of f (a), there is a Scott neighbourhood U of a such that f (U ) ⊆ V . Let
b ∈ approx(f (a)) be arbitrary. Then V = ↑ b is a Scott neighbourhood of
f (a). Furthermore, ↑ a is a Scott neighbourhood of a contained in any Scott
neighbourhood U of a. Therefore, we have f (↑ a) ⊆ ↑ b. Thus, if a [ x, then
x ∈ ↑ a. Therefore, f (x) ∈ ↑ b, that is, b [ f (x). But b ∈ approx(f (a)) is
arbitrary. Therefore, f (a) [ f (x), as required.
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A.6.3 Proposition Suppose that f : D → E is continuous in the Scott
topologies on domains D and E. Then f is monotonic.
Proof: Suppose that x [ y in D. Note that if a ∈ approx(x) is arbitrary,
then a ∈ D c and a [ x, so that a [ y. By Proposition A.6.2, we then have
f (a) [ f (y). Now, approx(x) can be thought of as a net approx(x) = {a i | i ∈
I}, as in Proposition A.6.1, and moreover a i → x. Therefore, f (a i ) → f (x).
Hence, by Theorem 3.2.4, for each b ∈ approx(f (x)) there is i 0 such that
b [ f (a i ) whenever i 0 ≤ i. But a i [ x [ y, for each i, and so a i [ y and
hence f (a i ) [ f (y) whenever i 0 ≤ i by our first observation. From this we see
that b [ f (y). Finally, we now have f (x) = {b | b ∈ approx(f (x))} [ f (y)
so that f (x) [ f (y), as required.
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A.6.4 Proposition A function f : D → E, where D and E are domains,
is continuous in the Scott topologies on D and E if and only if it is order
continuous in the sense of Definition 1.1.7.
Proof: Suppose that f is continuous in the Scott topologies on D and E.
Then f is monotonic by Proposition A.6.3. Let A ⊆ D be a directed set, and
let A = A. By Proposition A.6.1, A = {a i | i ∈ I} → A as a net, and hence
f (a i ) → f (A) by our hypothesis concerning f . Therefore, by Theorem 3.2.4,
for each b ∈ approx(f (A)), there exists i 0 such that b [ f (a i ) whenever i 0 ≤ i.
From this we obtain f (A) = f ( A) = {b | b ∈ approx(f (A))} [ {f (a i ) |
i ∈ I} = f (A). Thus, f ( A) [ f (A), and it follows that f is order
continuous by the remarks following Definition 1.1.7.
Conversely, suppose that f is order continuous and that s i → s in the Scott
topology on D. Now, f is monotonic. Therefore, on noting that approx(s) is
directed and thinking of it as the net {a j | j ∈ J }, we have that the set
{f (a j ) | j ∈ J } is directed and f (s) = f ( approx(s)) = f (approx(s)) =
{f (a j ) | j ∈ J }. Therefore, given any b ∈ approx(f (s)), there is j ∈ J
such that b [ f (a j ), where a j ∈ approx(s). Since s i → s, it follows from
Theorem 3.2.4 that there is i 0 such that a j [ s i whenever i 0 ≤ i. Hence,
241
A.6.2 Proposition Suppose that f : D → E is continuous in the Scott
topologies on domains D and E. Then whenever x ∈ D, a ∈ D c , and a [ x,
we have f (a) [ f (x).
Proof: Let a ∈ D c . Since f is continuous at a, given any Scott neighbourhood
V of f (a), there is a Scott neighbourhood U of a such that f (U ) ⊆ V . Let
b ∈ approx(f (a)) be arbitrary. Then V = ↑ b is a Scott neighbourhood of
f (a). Furthermore, ↑ a is a Scott neighbourhood of a contained in any Scott
neighbourhood U of a. Therefore, we have f (↑ a) ⊆ ↑ b. Thus, if a [ x, then
x ∈ ↑ a. Therefore, f (x) ∈ ↑ b, that is, b [ f (x). But b ∈ approx(f (a)) is
arbitrary. Therefore, f (a) [ f (x), as required.
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A.6.3 Proposition Suppose that f : D → E is continuous in the Scott
topologies on domains D and E. Then f is monotonic.
Proof: Suppose that x [ y in D. Note that if a ∈ approx(x) is arbitrary,
then a ∈ D c and a [ x, so that a [ y. By Proposition A.6.2, we then have
f (a) [ f (y). Now, approx(x) can be thought of as a net approx(x) = {a i | i ∈
I}, as in Proposition A.6.1, and moreover a i → x. Therefore, f (a i ) → f (x).
Hence, by Theorem 3.2.4, for each b ∈ approx(f (x)) there is i 0 such that
b [ f (a i ) whenever i 0 ≤ i. But a i [ x [ y, for each i, and so a i [ y and
hence f (a i ) [ f (y) whenever i 0 ≤ i by our first observation. From this we see
that b [ f (y). Finally, we now have f (x) = {b | b ∈ approx(f (x))} [ f (y)
so that f (x) [ f (y), as required.
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A.6.4 Proposition A function f : D → E, where D and E are domains,
is continuous in the Scott topologies on D and E if and only if it is order
continuous in the sense of Definition 1.1.7.
Proof: Suppose that f is continuous in the Scott topologies on D and E.
Then f is monotonic by Proposition A.6.3. Let A ⊆ D be a directed set, and
let A = A. By Proposition A.6.1, A = {a i | i ∈ I} → A as a net, and hence
f (a i ) → f (A) by our hypothesis concerning f . Therefore, by Theorem 3.2.4,
for each b ∈ approx(f (A)), there exists i 0 such that b [ f (a i ) whenever i 0 ≤ i.
From this we obtain f (A) = f ( A) = {b | b ∈ approx(f (A))} [ {f (a i ) |
i ∈ I} = f (A). Thus, f ( A) [ f (A), and it follows that f is order
continuous by the remarks following Definition 1.1.7.
Conversely, suppose that f is order continuous and that s i → s in the Scott
topology on D. Now, f is monotonic. Therefore, on noting that approx(s) is
directed and thinking of it as the net {a j | j ∈ J }, we have that the set
{f (a j ) | j ∈ J } is directed and f (s) = f ( approx(s)) = f (approx(s)) =
{f (a j ) | j ∈ J }. Therefore, given any b ∈ approx(f (s)), there is j ∈ J
such that b [ f (a j ), where a j ∈ approx(s). Since s i → s, it follows from
Theorem 3.2.4 that there is i 0 such that a j [ s i whenever i 0 ≤ i. Hence,