[proofplan] We prove by contradiction. If the weak topology were first-[countable](/page/Countable%20Set) at the origin, a countable neighbourhood base would be determined by countably many functionals $\{f_n\} \subset X^*$. We show $Y := \bigcap_{n=1}^\infty \ker f_n \neq \{0\}$ by a Baire category argument: an infinite-dimensional Banach space has uncountable Hamel dimension (since it cannot be written as a countable union of finite-dimensional, hence nowhere [dense](/page/Dense%20Subset), subspaces), and countably many functionals can remove at most countable codimension. Any nonzero $y_0 \in Y$ is invisible to every basic neighbourhood, contradicting the Hausdorff property of the weak topology. For the second claim, we exhibit a concrete weakly sequentially [closed set](/page/Closed%20Set) that is not weakly closed. [/proofplan]

[step:Assume first-countability and extract a countable family of functionals determining the neighbourhood base]Suppose for contradiction that the weak topology $\sigma(X, X^*)$ is first-countable at $0$. Then there exists a countable neighbourhood base $\{U_n\}_{n=1}^\infty$ at $0$. Each $U_n$ contains a basic weak neighbourhood \begin{align*} V_n = \{x \in X : |g_{n,1}(x)| < \varepsilon_{n,1}, \ldots, |g_{n,k_n}(x)| < \varepsilon_{n,k_n}\} \end{align*} for some $g_{n,1}, \ldots, g_{n,k_n} \in X^*$ and $\varepsilon_{n,j} > 0$. Define the countable collection \begin{align*} \mathcal{F} = \{g_{n,j} : n \in \mathbb{N}, \; 1 \le j \le k_n\} \subset X^*. \end{align*} Enumerate $\mathcal{F} = \{f_1, f_2, f_3, \ldots\}$ (repeating elements if $\mathcal{F}$ is finite).[/step]

[guided]The weak topology $\sigma(X, X^*)$ is the coarsest topology making every $f \in X^*$ [continuous](/page/Continuity). At the origin, a basic weak neighbourhood is a finite intersection of sets $\{x : |f(x)| < \varepsilon\}$ for various $f \in X^*$ and $\varepsilon > 0$. If the weak topology were first-countable at $0$, a countable base of such neighbourhoods would exist, each determined by finitely many functionals. The entire base is therefore controlled by the countable collection $\mathcal{F}$. The key insight is that in an infinite-dimensional space, countably many functionals cannot control the entire weak topology — there are "too many directions" for countably many linear conditions to detect.[/guided]

[step:Show that $\bigcap_{n=1}^\infty \ker f_n \neq \{0\}$ via uncountable Hamel dimension]We claim that $Y := \bigcap_{n=1}^\infty \ker f_n \neq \{0\}$. The argument has two parts. **Part 1: An infinite-dimensional Banach space has uncountable Hamel dimension.** Suppose for contradiction that $X$ has a countable Hamel basis $\{v_1, v_2, \ldots\}$. Then $X = \bigcup_{N=1}^\infty \operatorname{span}\{v_1, \ldots, v_N\}$. Each $\operatorname{span}\{v_1, \ldots, v_N\}$ is a finite-dimensional subspace, hence closed (finite-dimensional subspaces of a [normed space](/page/Normed%20Vector%20Space) are complete, hence closed). Each is a proper subspace of $X$ (for $N$ smaller than $\dim X$), so it has empty interior: if a proper closed subspace $V \subsetneq X$ contained a ball $B(x_0, r)$, then for any $x \in X$, the vector $x_0 + \frac{r(x - x_0)}{2\|x - x_0\|_X + 1}$ would lie in $B(x_0, r) \subset V$, and by linearity and the fact that $x_0 \in V$, we could solve for $x$ to get $x \in V$, giving $V = X$, a contradiction. Therefore each $\operatorname{span}\{v_1, \ldots, v_N\}$ is nowhere dense. By the [Baire Category Theorem](/theorems/630) (applicable since $X$ is a Banach space, hence a complete [metric space](/page/Metric%20Space)), $X$ cannot be a countable union of nowhere dense sets. This contradicts $X = \bigcup_{N=1}^\infty \operatorname{span}\{v_1, \ldots, v_N\}$. Therefore $\dim X$ (the Hamel dimension) is uncountable. **Part 2: Countably many functionals remove at most countable codimension.** Each functional $f_n \in X^*$ has $\ker f_n$ of codimension at most $1$ in $X$. For $F_N := \bigcap_{n=1}^N \ker f_n$, we have $\operatorname{codim} F_N \le N$ (each successive kernel intersection removes at most one dimension). The intersection $Y = \bigcap_{n=1}^\infty \ker f_n$ therefore has codimension at most $\aleph_0$. Since $\dim X$ is uncountable and $\operatorname{codim} Y \le \aleph_0$, subtracting a countable cardinal from an uncountable one leaves an uncountable remainder: $\dim Y$ is uncountable. In particular, $Y \neq \{0\}$.[/step]

[guided]The argument combines two ingredients. The first — that infinite-dimensional [Banach spaces](/page/Banach%20Space) have uncountable Hamel dimension — is a consequence of the [Baire Category Theorem](/theorems/630). The intuition is that a countable Hamel basis would make $X$ a "thin" countable union of finite-dimensional subspaces, but infinite-dimensional Banach spaces are "too large" for this. The formal proof writes $X$ as an increasing union of proper finite-dimensional closed subspaces, each nowhere dense, contradicting Baire. Why is a proper closed subspace nowhere dense? If $V \subsetneq X$ contained a ball $B(x_0, r)$, then linearity forces $V = X$: for any $x \in X$, $x_0 + \frac{r(x - x_0)}{2(\|x-x_0\|+1)} \in B(x_0, r) \subset V$, and since $x_0 \in V$ (as the centre of a ball in $V$), we can recover $x - x_0$ by scaling, giving $x \in V$. The second ingredient — that countably many functionals can remove only countable codimension — follows from a simple dimension count. Each functional $f_n$ constrains one degree of freedom (its kernel has codimension $\le 1$), so $N$ functionals constrain at most $N$ degrees. Even infinitely many functionals constrain at most $\aleph_0$ degrees, which is negligible compared to the uncountable dimension of $X$.[/guided]

[step:Derive a contradiction from the nonzero element in $Y$]Pick any $y_0 \in Y$ with $y_0 \neq 0$. Since $y_0 \in \ker f_n$ for every $n$, we have $f_n(y_0) = 0$ for all $n \in \mathbb{N}$. Since each $g_{N,j} \in \mathcal{F} \subset \{f_n\}$, for every $N \in \mathbb{N}$ and every $t \in \mathbb{R}$, \begin{align*} |g_{N,j}(t y_0)| = |t| \cdot |g_{N,j}(y_0)| = 0 < \varepsilon_{N,j} \quad \text{for each } j = 1, \ldots, k_N. \end{align*} Therefore $t y_0 \in V_N \subset U_N$ for every $t \in \mathbb{R}$ and every $N \in \mathbb{N}$. The entire line $\{t y_0 : t \in \mathbb{R}\}$ lies in every member of the neighbourhood base. However, the weak topology is Hausdorff: by the [Hahn-Banach Theorem](/theorems/879), since $y_0 \neq 0$, there exists $g \in X^*$ with $g(y_0) \neq 0$. The set $W := \{x \in X : |g(x)| < |g(y_0)|/2\}$ is a weak neighbourhood of $0$ satisfying $y_0 \notin W$ (since $|g(y_0)| \not< |g(y_0)|/2$). Since $\{U_N\}$ is a neighbourhood base, there exists $N_0$ with $U_{N_0} \subset W$. But $y_0 \in U_{N_0}$ and $y_0 \notin W$, a contradiction. Therefore the weak topology is not first-countable at $0$. Since the weak topology is translation-invariant ($x \mapsto x + a$ is a homeomorphism for each $a$), it is not first-countable at any point.[/step]

[guided]The contradiction is elementary once $y_0$ is in hand. Every basic neighbourhood $V_N$ tests $x$ against the functionals $g_{N,j} \in \mathcal{F}$, and $y_0$ passes all these tests (it lies in every kernel). So the line through $y_0$ sits inside every $U_N$. But the [Hahn-Banach Theorem](/theorems/879) provides $g \in X^*$ with $g(y_0) \neq 0$, and the set $W = \{x : |g(x)| < |g(y_0)|/2\}$ is a weak neighbourhood of $0$ excluding $y_0$. Since $\{U_N\}$ is a neighbourhood base, some $U_{N_0} \subset W$. But $y_0 \in U_{N_0}$ (established above) and $y_0 \notin W$, which is impossible. The neighbourhood base $\{U_N\}$ cannot exist.[/guided]

[step:Exhibit a weakly sequentially closed set that is not weakly closed]We construct a concrete example in $\ell^2(\mathbb{N})$. Let $\{e_k\}_{k=1}^\infty$ denote the standard basis and define \begin{align*} A = \{e_m + m \cdot e_n : m, n \in \mathbb{N}, \; n > m\}. \end{align*} **$A$ is weakly sequentially closed.** Let $\{z_i\}_{i=1}^\infty \subset A$ be a sequence with $z_i \rightharpoonup z$ weakly. Write $z_i = e_{m_i} + m_i \cdot e_{n_i}$ with $n_i > m_i$. By the [Boundedness of Weakly Convergent Sequences](/theorems/983), $M := \sup_i \|z_i\|_{\ell^2} < \infty$. Since $\|z_i\|_{\ell^2}^2 = 1 + m_i^2$, we have $m_i \le M$ for all $i$, so $\{m_i\}$ takes values in the finite set $\{1, \ldots, \lfloor M \rfloor\}$. Passing to a subsequence, we may assume $m_i = m_0$ is constant. Then $z_i = e_{m_0} + m_0 \cdot e_{n_i}$. Testing [weak convergence](/page/Weak%20Convergence) against the functional $e_{n_i}^*$ gives $z_{i,n_i} = m_0 \to z_{n_i}$ for each $i$, where $z_{n_i}$ is the $n_i$-th coordinate of $z$. If infinitely many $n_i$ are distinct, then $z$ has infinitely many coordinates equal to $m_0 \neq 0$, contradicting $z \in \ell^2$. So $n_i$ is eventually constant, say $n_i = n_0$, and $z_i = e_{m_0} + m_0 \cdot e_{n_0} \in A$ for all large $i$, giving $z = e_{m_0} + m_0 \cdot e_{n_0} \in A$. **$0$ lies in the weak closure of $A$ but $0 \notin A$.** Given a basic weak neighbourhood of $0$, \begin{align*} W = \{x \in \ell^2 : |f_j(x)| < \varepsilon_j, \; j = 1, \ldots, k\} \end{align*} with $f_j \in (\ell^2)^* \cong \ell^2$ and $\varepsilon_j > 0$, we must find an element of $A \cap W$. Since $e_m \rightharpoonup 0$ weakly (Bessel's inequality gives $f_j(e_m) = f_{j,m} \to 0$ as $m \to \infty$ for each $j$, because $\sum_m |f_{j,m}|^2 < \infty$), choose $m$ large enough that $|f_j(e_m)| < \varepsilon_j/2$ for all $j = 1, \ldots, k$. Then, since $f_{j,n} \to 0$ as $n \to \infty$, choose $n > m$ large enough that $m \cdot |f_j(e_n)| = m|f_{j,n}| < \varepsilon_j/2$ for all $j$. By the triangle inequality, \begin{align*} |f_j(e_m + m \cdot e_n)| \le |f_j(e_m)| + m|f_j(e_n)| < \frac{\varepsilon_j}{2} + \frac{\varepsilon_j}{2} = \varepsilon_j. \end{align*} So $e_m + m \cdot e_n \in A \cap W$. Since $0 \notin A$ (every element of $A$ has $\ell^2$-norm $\ge 1$), the set $A$ is weakly sequentially closed but not weakly closed.[/step]

[guided]The set $A = \{e_m + m \cdot e_n : n > m\}$ is carefully designed to exploit the gap between sequential and topological closure in the weak topology. Two features make it work: 1. **Sequential closedness via boundedness:** Any [weakly convergent](/page/Weak%20Topology) sequence in $A$ must have bounded norms, which forces the multiplier $m$ to stay bounded (since $\|e_m + m \cdot e_n\|^2 = 1 + m^2$). With $m$ bounded, eventually $m$ is constant. Then distinct $n$-values would put nonzero mass at infinitely many coordinates of the weak limit, contradicting square-summability. So $n$ is also eventually constant, and the sequence is eventually constant — hence its limit lies in $A$. 2. **Topological non-closedness via the two-parameter freedom:** To land in a given weak neighbourhood of $0$, we first choose $m$ large (exploiting $e_m \rightharpoonup 0$) and then choose $n$ large (exploiting $f_{j,n} \to 0$ for the finitely many functionals $f_j$). The multiplier $m$ on $e_n$ is harmless because $m$ is fixed while $n \to \infty$ drives $m \cdot f_{j,n} \to 0$. A sequence cannot exploit this two-parameter freedom (it must choose one pair $(m_i, n_i)$ at a time, and the boundedness constraint locks down $m$), but a net can.[/guided]