Proposition

Suppose $X$ is a first-countable space, $A$ is any subset of $X$, and $x$ is any point of $X$.

$x \in \overline{A}$ if and only if $x$ is a limit point of a sequence of points in $A$.
$x \in \Int{A}$ if and only if every sequence in $X$ converging to $x$ is eventually in $A$.
$A$ is closed in $X$ if and only if $A$ contains every limit of every convergent sequence of points in $A$.
$A$ is open in $X$ if and only if every sequence in $X$ converging to a point of $A$ is eventually in $A$.

Solution

1

Let $\{ x_n \}$ be a sequence of points in $A$. Suppose $x_n$ converges to $x$. Suppose $x \notin \overline{A}$. Then there exists a closed set $C \subset X$ such that $x \notin C$ and $A \subset C$. Then $C^c$ is a neighborhood of $x$ that contains no points from $\{ x_n \mid n \in \mathbb{N} \}$. This implies that $x_n$ does not converge to $x$. This is a contradiction, so $x \in \overline{A}$.

On the contrary, suppose that $x \in \overline{A}$. Since $X$ is first countable, $x$ has a countable neighborhood basis. Let $\{ B_1, B_2, \cdots \}$ denote such a countable neighborhood basis.

We claim that $A \cap B_1 \cap B_2 \cap \cdots \cap B_n \ne \emptyset$ for each $n \in \mathbb{N}$.

Suppose that $A \cap B_1 \cap \cdots \cap B_n = \emptyset$.

$B_1 \cap \cdots \cap B_n$ is the intersection of finitely many open sets. Thus it is open. Then $(B_1 \cap \cdots \cap B_n)^c$ is a closed set containing $A$. Thus $\overline{A} \subset (B_1 \cap \cdots \cap B_n)^c$.

$x \in \overline{A}$.
Each $B_i$ is a neighborhood of $x$, so $x \in B_1 \cap \cdots \cap B_n$. Therefore, $x \notin (B_1 \cap \cdots \cap B_n)^c$.

This is a contradiction, so $A \cap B_1 \cap \cdots \cap B_n \ne \emptyset$.

For each $n \in \mathbb{N}$, let $x_n \in A \cap B_1 \cap \cdots \cap B_n$. This is possible because we showed that the intersection is nonempty. Then $\{ x_n \}$ is a sequence in $A$.

Let $U$ be a neighborhood of $x$. Then $B_k \subset U$ for some $k \in \mathbb{N}$. For all $n \geq k$, $x_n \in A \cap B_1 \cap \cdots \cap B_k \cap \cdots \cap B_n \subset B_k \subset U$. Thus $x_n$ converges to $x$.

Note that the first part that states a limit of a sequence is in the boundary can be proved without first countability. This is true for any topological space. This lemma is important because of the other direction. When we say $x \in \overline{A}$, we expect that $x$ is in or “right next to” $A$. However, this is not always true. Consider $\mathbb{R}$ with the co-countable topology. Let $A = [0, 1]$. Let $C$ be a closed set containing $A$. Then $C^c$ is an open set disjoint from $A$. Since the complement of an non-empty open set must be countable, $C^c = \emptyset$. Thus the only closed set covering $A$ is $\mathbb{R}$, so the closure of $A$ is $\mathbb{R}$. This is “much bigger” than $A$.

$2 \in \overline{A}$. Let $\{ a_n \}$ be a sequence in $A$. Let $U = \mathbb{R} \setminus \{ a_n \mid n \in \mathbb{N} \}$. Then $U$ is open because its complement is countable. $U$ must contain $2$ because $2 \notin \{ a_n \mid n \in \mathbb{N} \}$. Therefore, we found a neighborhood of $2$ which contains no term of $\{ a_n \}$, so no sequence in $A$ converges to $2$.

2

Suppose that every sequence in $X$ converging to $x$ is eventually in $A$. Let $B = \{ B_1, B_2, \cdots \}$ be a neighborhood basis for $X$ at $x$. We claim that there exists an $n \in \mathbb{N}$ such that $B_1 \cap B_2 \cap \cdots \cap B_n \subset A$. Suppose otherwise.

That means that $\forall n \in \mathbb{N}$, $(B_1 \cap B_2 \cap \cdots \cap B_n) \setminus A \ne \emptyset$. For each $n \in \mathbb{N}$, let $a_n \in (B_1 \cap B_2 \cap \cdots \cap B_n) \setminus A$. Then we have a sequence $\{ a_n \}$. Let $U$ be a neighborhood of $x$. Since $B$ is a neighborhood basis for $X$ at $x$, there exists a $k \in \mathbb{N}$ such that $B_k \subset U$. Then $\forall n \geq k$, $a_n \in B_1 \cap B_2 \cap \cdots \cap B_n \subset B_k \subset U$. Therefore, $\{ a_n \}$ converges to $x$. This implies that $\{ a_n \}$ is eventually in $A$. However, we specially picked each $a_n$ from the complement of $A$. This is a contradiction, so there must exist an $n \in \mathbb{N}$ such that $B_1 \cap B_2 \cap \cdots \cap B_n \subset A$. $B_1 \cap \cdots \cap B_n$ is the intersection of $n$ open sets, so it is open. Thus it is a neighborhood of $x$ that is contained in $A$, so $x \in \Int{A}$.

3

Suppose $A$ is closed in $X$. Then $A = \overline{A}$. Let $\{ a_n \}$ be a convergent sequence in $A$. Let $x$ denote $\{ a_n \}$’s limit. By (1), $x \in \overline{A}$. Thus $x \in A$. Therefore, every limit of every convergent sequence of points in $A$ is in $A$.

Suppose that $A$ contains every limit of every convergent sequence of points in $A$. Let $x \in \overline{A}$. By (1), $x$ is a limit point of a sequence of points in $A$. Thus $x \in A$. This implies that $\overline{A} \subset A$. By definition, $A \subset \overline{A}$, so $A = \overline{A}$. Therefore, $A$ is closed.

4

Suppose $A$ is open in $X$. Let $\{ x_n \}$ be a sequence of points in $X$. Let $a \in A$. Suppose $x_n$ converges to $a$. Since $A$ is open, $A = \Int{A}$. Thus $a \in \Int{A}$. This implies that $\{ x_n \}$ is eventually in $A$ by (2).

On the contrary, suppose that every sequence in $X$ converging to a point of $A$ is eventually in $A$. Let $a \in A$. Then every sequence in $X$ converging to $a$ is eventually in $A$. By (2), $a \in \Int{A}$. This implies that $A = \Int{A}$, so $A$ is open.