[proofplan] We prove both implications using the open-ball definition of the norm topology. If $F$ is closed and a sequence in $F$ converges to $x$, then placing $x$ in the open complement of $F$ would force all sufficiently late sequence terms to lie outside $F$, a contradiction. Conversely, if $F$ is not closed, then the complement is not open, so there is a point outside $F$ whose every norm ball meets $F$; choosing one point of $F$ in each ball of radius $1/n$ produces a sequence in $F$ converging to a point outside $F$. [/proofplan] [step:Show that closed sets contain limits of convergent sequences from the set] Assume that $F$ is closed in the norm topology. Let $(x_n)_{n=1}^{\infty}$ be a sequence in $F$, meaning that $x_n\in F$ for every $n\in\mathbb{N}$, and suppose that $x_n\to x$ in $V$ for some $x\in V$. For $a\in V$ and $r>0$, let $B_V(a,r)$ denote the open norm ball \begin{align*} B_V(a,r)=\{y\in V:\|y-a\|<r\}. \end{align*} Suppose, for contradiction, that $x\notin F$. Since $F$ is closed, its complement $V\setminus F$ is open in the norm topology. Hence there exists $r>0$ such that \begin{align*} B_V(x,r)\subset V\setminus F. \end{align*} Since $x_n\to x$ in $V$, there exists $N\in\mathbb{N}$ such that, for every $n\geq N$, \begin{align*} \|x_n-x\|<r. \end{align*} Thus $x_n\in B_V(x,r)\subset V\setminus F$ for every $n\geq N$. This contradicts the assumption that $x_n\in F$ for every $n\in\mathbb{N}$. Therefore $x\in F$. [/step] [step:Construct a convergent sequence in $F$ from any failure of closedness] Assume that the sequential closure property holds. We prove that $F$ is closed. Suppose, for contradiction, that $F$ is not closed. By the definition of closedness in the norm topology, $V\setminus F$ is not open. Therefore there exists $x\in V\setminus F$ such that no open norm ball centered at $x$ is contained in $V\setminus F$. Equivalently, for every $r>0$, \begin{align*} B_V(x,r)\cap F\neq \varnothing. \end{align*} For each $n\in\mathbb{N}$, apply this with $r=1/n$ and choose a point $x_n\in B_V(x,1/n)\cap F$. This defines a sequence $(x_n)_{n=1}^{\infty}$ in $F$ satisfying \begin{align*} \|x_n-x\|<\frac{1}{n} \end{align*} for every $n\in\mathbb{N}$. Let $\varepsilon>0$. By the [Archimedean property](/theorems/737) of $\mathbb{R}$, there exists $N\in\mathbb{N}$ such that $1/N<\varepsilon$. If $n\geq N$, then $1/n\leq 1/N$, and hence \begin{align*} \|x_n-x\|<\frac{1}{n}\leq \frac{1}{N}<\varepsilon. \end{align*} Thus $x_n\to x$ in $V$. The sequential closure property now implies $x\in F$, contradicting $x\in V\setminus F$. Hence $F$ must be closed. [guided] Assume that every convergent sequence with all terms in $F$ has its limit in $F$. We want to prove that $F$ is closed. Since closedness is defined by openness of the complement in the norm topology, it is enough to show that $V\setminus F$ is open. Suppose instead that $F$ is not closed. Then $V\setminus F$ is not open. By the definition of openness in the norm topology, this means there is a point $x\in V\setminus F$ such that no open norm ball centered at $x$ lies entirely inside $V\setminus F$. In symbols, for every radius $r>0$, \begin{align*} B_V(x,r)\cap F\neq \varnothing. \end{align*} This condition says that points of $F$ occur arbitrarily close to $x$, even though $x$ itself is not in $F$. To turn that into a sequence, choose the radii $1/n$. For each $n\in\mathbb{N}$, since $B_V(x,1/n)\cap F$ is nonempty, choose a point \begin{align*} x_n\in B_V(x,1/n)\cap F. \end{align*} Then $(x_n)_{n=1}^{\infty}$ is a sequence in $F$, because $x_n\in F$ for every $n\in\mathbb{N}$. Also, by membership in the ball $B_V(x,1/n)$, we have \begin{align*} \|x_n-x\|<\frac{1}{n} \end{align*} for every $n\in\mathbb{N}$. We now verify directly that $x_n\to x$ in the norm topology. Let $\varepsilon>0$. By the Archimedean property of $\mathbb{R}$, choose $N\in\mathbb{N}$ such that $1/N<\varepsilon$. For every $n\geq N$, monotonicity of the reciprocal function on positive integers gives $1/n\leq 1/N$, so \begin{align*} \|x_n-x\|<\frac{1}{n}\leq \frac{1}{N}<\varepsilon. \end{align*} This is exactly the definition of convergence $x_n\to x$ in the [normed space](/page/Normed%20Space) $V$. The sequential closure property applies to this sequence because every $x_n$ lies in $F$ and because $x_n\to x$ in $V$. Therefore it gives $x\in F$. But the point $x$ was chosen in $V\setminus F$, so this is a contradiction. Hence the assumption that $F$ is not closed is impossible, and $F$ is closed. [/guided] [/step] [step:Conclude the equivalence] The first step proves that closedness of $F$ implies the sequential closure property. The second step proves that the sequential closure property implies closedness of $F$. Therefore $F$ is closed in the norm topology if and only if it contains the limit of every convergent sequence whose terms all lie in $F$. [/step]