$\left(b\right)\;T$
(b) The operator GLP *is upper-half continuous, i.e.*

$$G L_{P}(\bigcup_{n\in N}X_{n})=\bigcap_{n\in N}G L_{P}(X_{n})$$
$$q u e n c e\ \langle X_{n}\rangle_{n\in N}.$$

for every monotonically increasing sequence hXnin∈N .

Proof: Two implications need to be proved: (a) ⇒ (b), and (b) ⇒ (a).

Proof of the implication (a) ⇒ (b). Here, assuming (a), we need to prove two inclusions:
(i) GLP (Sn∈N Xn) ⊆Tn∈N GLP (Xn), and
(ii) Tn∈N GLP (Xn) ⊆ GLP (Sn∈N Xn).

To prove (i), note that since Xn ⊆Sj∈N Xj , we have

$$G L_{P}(\bigcup X_{j})\subseteq G L_{P}(X_{n}).$$
$${\mathrm{As~}}n{\mathrm{~is~arbitrary}},$$
$$G L_{P}(\bigcup X_{j})\subseteq\bigcap_{n\in N}G L_{P}(X_{n}).$$

This proves (i).

To prove (ii), let p ∈Tn∈N GLP (Xn). Then, for every n ∈ N, p ∈ GLP (Xn)
and so, for every n ∈ N, there is an inclusion-minimal support U for p such that U ∩ Xn = ∅.

But by (a) there are only finitely many inclusion-minimal supports for P-proof schemes for p. Therefore there is a support of an inclusion minimal support of a proof scheme of p, U0, such that for infinitely many n's

$$U_{0}\cap X_{n}=\varnothing.$$
d the implication $(a)\,\Rightarrow\,(b)$ fol. 

But the sequence hXnin∈N is monotonically increasing. Therefore for all n ∈ N,
U0 ∩ Xn = ∅. But then

$$U_{0}\cap\bigcup_{n\in N}X_{n}=\varnothing,$$

so that p ∈ GLP (Sn∈N Xn). Thus (ii) holds and the implication (a) ⇒ (b) follows.

To prove that (b) ⇒ (a), assume that the operator GLP is upper-half continuous.

We need to show that for every p, the reduced defining equation for p is finite.

So let us assume that rEqP
p is not finite. This means that there is an infinite set X = {U1, U2**, . . .**}, where U1 ≺ U2 **≺ · · ·** , such that