Last piece of the puzzle: $+$ is left continuous: we show that $U \mapsto U + V$ is continuous (for any fixed $V$ ).

Proof.

Note

U + V \in C_{A}

(for some

A

) if and only if

A \in U + V

, which happens if and only if

\forall_{U} x, \forall_{V} y, x + y \in A

. This is equivalent to saying

{x : \forall_{V} y, x + y \in A} \in U,

which is equivalent to $U \in C_{{x : \forall_{V} y, x + y \in A}}$ , so the pre-image is open.

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