Sequential continuity

A function between topological spaces $𝑓 : 𝑋 \to 𝑌$ is sequentially continuous at a point $𝑎 \in 𝑋$ iff any convergent sequence $(𝑎 𝑛) \infty 𝑛 = 1$ in $𝑋$ with limit $𝑐$ maps to a convergent sequence $(𝑓 (𝑎 𝑛)) \infty 𝑛 = 1$ in $𝑌$ with limit $𝑓 (𝑎)$ , i.e.

(𝑎 𝑛) \to 𝑎 ⟹ (𝑓 (𝑎 𝑛)) \to 𝑓 (𝑎)

All continuous maps are sequentially continuous. In case $𝑋$ is first-countable, then $𝑓$ is sequentially continuous at a point $𝑐$ iff. it is continuous at that point. #m/thm/topology

Proof for first-countable equivalence

Let $(𝑋, T 𝑋)$ and $(𝑋, T 𝑌)$ be topological space, $𝑓 : 𝑋 \to 𝑌$ be a map, and $𝑎 \in 𝑋$ be a point. Let $𝑏 = 𝑓 (𝑎)$ , and $𝑋$ be first-countable.

First, assume $𝑓$ is continuous at $𝑎$ . Let $(𝑎 𝑛) \infty 𝑛 = 1$ be a sequence in $𝑋$ with $(𝑎 𝑛) \to 𝑎$ , and $𝑇 \in T 𝑌 (𝑏)$ . Then $𝑓 - 1 (𝑇) \in T 𝑋 (𝑎)$ , and thus there exists $𝑁$ such that $𝑎 𝑛 \in 𝑓 - 1 (𝑇)$ for all $𝑛 > 𝑁$ , whence $𝑓 (𝑎 𝑛) \in 𝑇$ for all $𝑛 > 𝑁$ . Therefore $𝑓 (𝑎 𝑛) \to 𝑏$ , without invoking the First countability axiom.

For the converse, assume $𝑓$ is sequentially continuous at $𝑎$ . Let $(𝑆 𝑛) 𝑛 \in ℕ$ a countable nested open neighbourhood basis of $𝑎$ . Assume $𝑓$ is not continuous at $𝑎$ , i.e. there exists $𝑇 \in T 𝑌 (𝑏)$ such that $𝑆 𝑛 ⊈ 𝑓 - 1 (𝑇)$ for all $𝑛 \in ℕ$ . We can then construct a sequence such that $𝑎 𝑛 \in 𝑆 𝑛$ for all $𝑛 \in ℕ$ , where clearly $(𝑎 𝑛) \to 𝑎$ , but $𝑓 (𝑎 𝑛) \notin 𝑇$ for all $𝑛 \in ℕ$ . whence $(𝑓 (𝑎 𝑛)) ↛ 𝑓 (𝑎)$ , contradicting our requirement that $𝑓$ be sequentially continuous. Therefore, $𝑓$ is continuous at $𝑎$ .

Another topological property that can be shown using sequences for metric spaces is Sequential closedness.

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