Proving continuity with a subbasis

Let $(𝑋, T 𝑋)$ and $(𝑌, T 𝑌)$ each be a Topological space, let $S$ be a subbasis of $T 𝑌$ , and let $𝑓 : 𝑋 \to 𝑌$ . Then $𝑓$ is continuous iff the preïmage $𝑓 - 1 (𝑉)$ is open for all $𝑉 \in S$ . #m/thm/topology

Proof

Since a subbasis is a family of open sets, it is clear that given continuous $𝑓$ the preïmage of subbasic open neighbourhoods is open. Let $𝑓 : 𝑋 \to 𝑌$ such that for all $𝑉 \in S$ , the preïmage $𝑓 - 1 (𝑉)$ is open. First consider the completed basis $B$ . Let $𝑉 \in B$ , implying there exists a finite sequence $(𝑆 𝑖) 𝑛 𝑖 = 1 \in S$ where $𝑛 \in ℕ$ such that $𝑉 = ⋂ 𝑛 𝑖 = 1 𝑆 𝑖$ . Then

𝑓 - 1 (𝑉) = 𝑓 - 1 (𝑛 ⋂ 𝑖 = 1 𝑆 𝑖) = 𝑛 ⋂ 𝑖 = 1 𝑓 - 1 (𝑆 𝑖)

which is the finite intersection of open sets and is thus open. Hence for all $𝑉 \in B$ , the preïmage $𝑓 - 1 (𝑉)$ is open. Now consider the entire generated topology $T 𝑌$ . Let $𝑉 \in T 𝑌$ , implying there exists an indexed family $(𝑆 𝑖) 𝑖 \in 𝐼 \in B$ such that $𝑉 = ⋃ 𝑖 \in 𝐼 𝑆 𝑖$ . Then

𝑓 - 1 (𝑉) = 𝑓 - 1 (⋃ 𝑖 \in 𝐼 𝑆 𝑖) = ⋃ 𝑖 \in 𝐼 𝑓 - 1 (𝑆 𝑖)

which is the union of open sets and thus open. Hence the preïmage of every open set is open, wherefore $𝑓$ is continuous.

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