First showing \mathit {apply}(f,\cdot ) is continuous for some continuous f : A \twoheadrightarrow B, using the alternative characterisation

1. \mathit {apply}(f,\cdot ) is monotonic: Given a, b \in A where a \sqsubseteq b, we can see that \mathit {apply}(f,a) = f(a) \sqsubseteq f(b) = \mathit {apply}(f,b) by the monotonicity of f.
2. \mathit {apply}(f,\cdot ) preserves lubs: Given a directed X \subseteq A, we can see that \mathit {apply}(f,\bigsqcup X) = f(\bigsqcup X) = \bigsqcup \{f(x) \mid x \in X\} = \bigsqcup \{\mathit {apply}(f,x) \mid x \in X\} follows from the continuity of f.

Next, showing \mathit {apply}(\cdot ,a) is continuous for some fixed a \in A:

1. \mathit {apply}(\cdot ,a) is monotonic: Given f, g \in A \twoheadrightarrow B where f \sqsubseteq g, we can see that \mathit {apply}(f,a) = f(a) \sqsubseteq g(a) = \mathit {apply}(g,a) by the pointwise ordering of functions.
2. \mathit {apply}(\cdot ,a) preserves lubs: Given a directed X \subseteq A \twoheadrightarrow B, we can see that \mathit {apply}(\bigsqcup X, a) = \left (\bigsqcup X\right )(a) = \bigsqcup \{f(a) \mid f \in X\} = \bigsqcup \{\mathit {apply}(f,a) \mid f \in X\} follows from the lub definition for functions.