Sequential closedness

Let $𝑋$ be a topological space. A subset $𝐹 \subseteq 𝑋$ is sequentially closed iff every convergent sequence $(𝑥 𝑛) \infty 𝑛 = 1$ in $𝐹$ such that $(𝑥 𝑛) \to 𝑥$ has $𝑥 \in 𝐹$ . #m/def/topology Every closed set is sequentially closed.

Proof

Let $𝐹 \subseteq 𝑋$ be closed, $𝑉 = 𝑋 ∖ 𝐹$ , and $(𝑥 𝑛) \infty 𝑛 = 1$ be sequence in $𝐹$ such that $(𝑥 𝑛) \to 𝑥 \in 𝑉$ . But since $𝑉$ is an open neighbourhood of $𝑥$ , there exists some $𝑁$ such that $𝑥 𝑛 \in 𝑉$ for all $𝑛 > 𝑁$ , so $𝑥 𝑛 \notin 𝐹$ for all $𝑛 > 𝑁$ , a contradiction.

Main theorem

Let $𝑋$ be first-countable space. Then $𝐹 \subseteq 𝑋$ is closed iff it is sequentially closed. #m/thm/topology

Proof

The forward direction is given above. For the converse, let $𝐹 \subseteq 𝑋$ such that for every sequence $(𝑥 𝑛) \infty 𝑛 = 1$ , $(𝑥 𝑛) \to 𝑥$ implies $𝑥 \in 𝐹$ . Suppose $𝑉 = 𝑋 ∖ 𝐹$ is not open in $𝑋$ Then there exists some $𝑦 \in 𝑉$ such that for every neighbourhood $𝑈$ of $𝑦$ , $𝑈 \cap 𝐹 \neq \emptyset$ . Since $𝑋$ is first-countable, we may construct a nested neighbourhood basis ${𝑆 𝑛} \infty 𝑛 = 1$ , for which $𝑆 𝑛 \cap 𝐹 \neq \emptyset$ for all $𝑛 \in ℕ$ . One may then construct a sequence $(𝑥 𝑛) \infty 𝑛 = 1$ such that $𝑥 𝑛 \in 𝑆 𝑛 \cap 𝐹$ for all $𝑛 \in ℕ$ . But then $(𝑥 𝑛) \to 𝑦 \in 𝑉$ , contradicting the requirement that $𝐹$ be sequentially closed. Hence $𝑉$ must be open, whence $𝐹$ is closed.

#state/tidy | #lang/en | #SemBr