> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]]. $(X, \topo)$ is **first countable** if every point $x \in X$ has a [[Cardinality|countable]] [[Neighbourhood Base|neighbourhood base]]. > [!theorem] Lemma > > Let $(X, \topo)$ be a first countable topological space. Then for any $x \in X$, there exists a neighbourhood base $\seq{U_j}$ such that $U_{j} \supset U_{j + 1}$ for all $j \in \nat$. > > *Proof*. Neighbourhood bases don't have to cover everything around $x$, they just have to contain sufficiently small sets. Let $\seq{U_j}$ be a countable neighbourhood base at $x$. Take $U'_1 = U_1$, and for each $k \in \nat$, let $U'_{k + 1} = U_k' \cap U_{k + 1}$, then $U'_{k + 1} \subset U_k'$ is an [[Open Set|open]] [[Neighbourhood|neighbourhood]] of $x$. > > Let $V \in \topo$, $x \in V$ be an open set containing $x$. Then there exists $k \in \nat$ such that $x \in U_k \subset V$. From our construction, we have $x \in U'_k \subset U_k \subset V$. Since every set in $\seq{U'_j}$ contains $x$, and for any open neighbourhood of $x$ there is a set $U'_k$ contained in it, $\seq{U'_j}$ is a neighbourhood base. > [!theorem] > > Let $(X, \topo)$ be a first countable topological space and $A \subset X$. Then $x \in \ol{A}$ if and only if there exists a [[Sequence|sequence]] $\seq{x_n} \subset A$ [[Limit|converging]] to $x$. > > *Proof*. Let $x \in \ol{A}$ and $\seq{U_j}$ be a countable neighbourhood base at $x$ such that $U_j \supset U_{j + 1}$ for all $j \in \nat$. > > For any $x \in A$, a neighbourhood $V$ of $x$ intersects $A$ at $x$, and for any [[Accumulation Point|accumulation point]] $x \in \text{acc}(A)$, Since $x \in \ol{A} = A \cup \text{acc}(A)$, any neighbourhood $V$ of $x$ intersects $A$. > > Choose $x_j \in U_j \cap A$ for each $j \in \nat$. Then for any neighbourhood $V$ of $x$, there exists $U_k$ such that $x_j \in U_j \subset U_k$ for all $j \ge k$. Therefore $x_j \to x$. > > Now let $x \in \ol{A}^c$. Since $\ol{A}^c$ is open, it is an open neighbourhood of $x$ that does not intersect $A$. For any sequence $\seq{x_j} \subset A$, there exists no $k \in \nat$ such that $x_j \in \ol{A}^c$ for all $j \ge k$ as $x_j \not\in \ol{A}^c$ for all $j \in \nat$. Therefore there exists no sequence $\seq{x_j} \subset A$ such that $x_j \to x$.