[proofplan] We prove the cycle $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$. The implication $(2) \Rightarrow (3)$ is immediate. For $(1) \Rightarrow (2)$, we reduce to the separable case (replacing $X$ by the closed span of the sequence), where the dual is weak-$*$ separable, the weak topology on bounded sets is metrizable, and weak compactness becomes weak sequential compactness via metric-space arguments. The deep direction is $(3) \Rightarrow (1)$: we embed $X$ into $X^{**}$ via the canonical embedding, use the [Banach-Alaoglu Theorem](/theorems/212) to extract a weak-$*$ cluster point $\xi \in X^{**}$, and then use the sequential cluster point hypothesis together with [Goldstine's Lemma](/theorems/898) to show $\xi$ lies in $\phi(X)$, so the cluster point is actually in $X$. [/proofplan] [step:Prove $(2) \Rightarrow (3)$: every [weakly convergent](/page/Weak%20Topology) subsequence provides a weak cluster point] Assume (2): every sequence in $A$ has a weakly convergent subsequence with limit in $X$. Let $(x_n)$ be any sequence in $A$. By (2), there exist a subsequence $(x_{n_k})$ and $x \in X$ with $x_{n_k} \rightharpoonup x$. Then $x$ is a weak cluster point of $(x_n)$: for every weak neighbourhood $U$ of $x$ and every $N \in \mathbb{N}$, the subsequence $(x_{n_k})$ is eventually in $U$, so some $x_{n_k}$ with $n_k \ge N$ lies in $U$. This establishes (3). [/step] [step:Prove $(1) \Rightarrow (2)$ — reduce to the separable case] Assume (1): the weak closure $K := \overline{A}^{\,\sigma(X, X^*)}$ is weakly compact. Let $(x_n)$ be a sequence in $A$. Define $Y := \overline{\operatorname{span}}\{x_n : n \ge 1\}$, the norm-closed linear span of the sequence in $X$. Since $Y$ is the closed span of a [countable set](/page/Countable%20Set), $Y$ is a [separable Banach](/page/Separable%20Space) space. Set $A_0 := \{x_n : n \ge 1\} \subset Y$. We will extract a subsequence converging weakly in $X$. [guided] The key idea of the separable reduction is this: the sequence $(x_n)$ lives in the separable subspace $Y$, and separability of $Y$ gives us metrizability of the weak topology on bounded subsets of $Y$. Since sequences behave well in [metric spaces](/page/Metric%20Space) — every sequentially compact set is compact, and vice versa — we can reduce the problem to extracting a convergent subsequence in a metric topology. Why is $Y$ separable? The set $\{x_n : n \ge 1\}$ is countable, so the set of all finite rational-coefficient linear combinations $\{\sum_{k=1}^N q_k x_{n_k} : N \in \mathbb{N}, q_k \in \mathbb{Q}\}$ is a countable [dense subset](/page/Dense%20Subset) of $\operatorname{span}\{x_n\}$ in the norm topology. Its norm-closure is $Y$, and a subspace of a metric space with a countable dense subset is separable. [/guided] [/step] [step:Show that $Y^*$ is weak-$*$ separable and the weak topology on bounded subsets of $Y$ is metrizable] Since $Y$ is a separable [Banach space](/page/Banach%20Space), $Y^*$ contains a countable subset $\{g_m\}_{m=1}^\infty$ that is dense in the unit ball $\overline{B}_{Y^*}$ with respect to the norm of $Y^*$. To construct such a set: let $\{y_k\}_{k=1}^\infty$ be a countable dense subset of the unit sphere of $Y$. For each $y_k$, the [Hahn-Banach Theorem](/theorems/879) (applied with $p = \|\cdot\|_Y$ and the one-dimensional subspace spanned by $y_k$) provides $g_k \in Y^*$ with $\|g_k\|_{Y^*} = 1$ and $g_k(y_k) = \|y_k\|_Y = 1$. The set of all rational-coefficient finite linear combinations of the $g_k$, intersected with $\overline{B}_{Y^*}$, is countable and norm-dense in $\overline{B}_{Y^*}$. By [Weak Metrizability on Bounded Sets](/theorems/989), the weak topology $\sigma(Y, Y^*)$ restricted to any norm-bounded subset of $Y$ is metrizable. [guided] We need $Y^*$ to have a countable norm-dense subset in its unit ball. This is a consequence of $Y$ being separable, though the argument is worth spelling out. Let $\{y_k\}_{k=1}^\infty$ be dense in the unit sphere $\{y \in Y : \|y\|_Y = 1\}$. For each $k$, define a linear functional on $\operatorname{span}\{y_k\}$ by $\lambda y_k \mapsto \lambda \|y_k\|_Y$. This functional is dominated by the sublinear functional $p(y) = \|y\|_Y$ on $Y$. By the [Hahn-Banach Theorem](/theorems/879), it extends to $g_k \in Y^*$ with $\|g_k\|_{Y^*} \le 1$ and $g_k(y_k) = 1$, so $\|g_k\|_{Y^*} = 1$. Now consider the countable collection $\mathcal{D} := \{\sum_{k=1}^N q_k g_k : N \in \mathbb{N}, q_k \in \mathbb{Q}\} \cap \overline{B}_{Y^*}$. We claim $\mathcal{D}$ is norm-dense in $\overline{B}_{Y^*}$. Let $g \in \overline{B}_{Y^*}$ and $\varepsilon > 0$. By a standard argument, $g$ is determined by its values on a dense set: if two functionals in $\overline{B}_{Y^*}$ agree on $\{y_k\}$, [continuity](/page/Continuity) forces them to agree on all of $Y$. Using the rational approximation of $g(y_k)$ by linear combinations of the $g_j(y_k)$ values, one can approximate $g$ in norm. With the dense sequence $\{g_m\}$ in $\overline{B}_{Y^*}$ in hand, [Weak Metrizability on Bounded Sets](/theorems/989) applies: Theorem 989 requires the [dual space](/page/Dual%20Space) $Y^*$ to be separable (which follows from density of the countable set $\mathcal{D}$ in $\overline{B}_{Y^*}$, and hence in all of $Y^*$ by scaling). The theorem then gives a metric $d$ on $Y$ whose topology agrees with $\sigma(Y, Y^*)$ on every norm-bounded subset. Note: we are applying Theorem 989 to the Banach space $Y$ (not $X$). We do NOT require $X^*$ to be separable — only $Y^*$, which is guaranteed because $Y$ is separable. [/guided] [/step] [step:Extract a weakly convergent subsequence using metrizability and compactness] The sequence $(x_n)$ lies in $A \cap Y$. Since $K = \overline{A}^{\,\sigma(X, X^*)}$ is weakly compact and hence norm-bounded (every weakly compact set in a [normed space](/page/Normed%20Vector%20Space) is norm-bounded, by the [Uniform Boundedness Principle](/theorems/549) applied to the family $\{\phi(x) : x \in K\} \subset X^{**}$), the set $A$ is norm-bounded. Let $M > 0$ satisfy $\|x\|_X \le M$ for all $x \in A$. Consider the norm-bounded set $B_M(Y) := \{y \in Y : \|y\|_Y \le M\}$. By the metrizability established above, $\sigma(Y, Y^*)|_{B_M(Y)}$ is metrizable. The sequence $(x_n)$ lies in $B_M(Y)$. We need to connect the weak topology of $Y$ with that of $X$. Every $f \in X^*$ restricts to $f|_Y \in Y^*$, and by Hahn-Banach, every $g \in Y^*$ extends to some $f \in X^*$. Therefore $\sigma(X, X^*)|_Y = \sigma(Y, Y^*)$: a net in $Y$ converges weakly in $X$ if and only if it converges weakly in $Y$. The set $K \cap Y$ is weakly closed in the weakly compact set $K$, hence weakly compact in $X$. By the identification $\sigma(X, X^*)|_Y = \sigma(Y, Y^*)$, $K \cap Y$ is weakly compact in $Y$, and in particular $K \cap Y \cap B_M(Y)$ is compact in the metrizable topology $\sigma(Y, Y^*)|_{B_M(Y)}$. Since $(x_n)$ is a sequence in $A \cap Y \subset K \cap Y \cap B_M(Y)$, and compact [metrizable spaces](/page/Metrizable%20Space) are sequentially compact, there exists a subsequence $x_{n_k}$ converging in $\sigma(Y, Y^*)|_{B_M(Y)}$ to some $x \in K \cap Y \subset X$. This means $x_{n_k} \rightharpoonup x$ weakly in $Y$, hence weakly in $X$. [guided] This step combines several ingredients. Let us trace through carefully. **Norm-boundedness of $A$.** Since $K$ is weakly compact, $K$ is norm-bounded. To see this: for each $f \in X^*$, the map $x \mapsto f(x)$ is weakly continuous, so $f(K)$ is compact in $\mathbb{R}$, hence bounded. This means $\sup_{x \in K} |f(x)| < \infty$ for every $f \in X^*$. Identifying $x$ with $\phi(x) \in X^{**}$ via the [canonical embedding](/theorems/875), the family $\{\phi(x)\}_{x \in K}$ is pointwise bounded on $X^*$. By the [Uniform Boundedness Principle](/theorems/549) (applied with the Banach space $X^*$ and the normed space $\mathbb{R}$, noting that $X^*$ is a Banach space as the dual of a Banach space), $\sup_{x \in K} \|\phi(x)\|_{X^{**}} = \sup_{x \in K} \|x\|_X < \infty$. Since $A \subset K$, $A$ is norm-bounded. **Relating weak topologies of $Y$ and $X$.** We claim $\sigma(X, X^*)|_Y = \sigma(Y, Y^*)$. The restriction map $R: X^* \to Y^*$, $f \mapsto f|_Y$, is surjective by the [Hahn-Banach Theorem](/theorems/879): given $g \in Y^*$, the sublinear functional $p(x) = \|g\|_{Y^*} \|x\|_X$ dominates $g$ on $Y$, so Hahn-Banach provides $f \in X^*$ with $f|_Y = g$. Therefore $\{f|_Y : f \in X^*\} = Y^*$, which gives $\sigma(X, X^*)|_Y = \sigma(Y, Y^*)$. **Compactness in the metrizable topology.** The set $K \cap Y$ is weakly closed in the weakly compact $K$ (since $Y$ is norm-closed, hence weakly closed by [Mazur's Theorem](/theorems/985) applied to the convex set $Y$). Therefore $K \cap Y$ is weakly compact in $X$, hence in $Y$ by the identification above. Since all elements have norm $\le M$, $K \cap Y \subset B_M(Y)$, and the weak topology on $B_M(Y)$ is metrizable. [Compact subsets](/page/Compact%20Space) of metrizable spaces are sequentially compact. So the sequence $(x_n)$ in $K \cap Y \cap B_M(Y)$ has a convergent subsequence $x_{n_k} \to x$ in the metrizable weak topology, i.e. $x_{n_k} \rightharpoonup x$ in $Y$, hence in $X$. This completes the proof that $(1) \Rightarrow (2)$. [/guided] [/step] [step:Begin $(3) \Rightarrow (1)$ — set up the Banach-Alaoglu framework in $X^{**}$] Assume (3): every sequence in $A$ has a weak cluster point in $X$. We must show that $\overline{A}^{\,\sigma(X,X^*)}$ is weakly compact. By the [Uniform Boundedness Principle](/theorems/549), condition (3) implies that $A$ is norm-bounded: if $A$ were unbounded, we could pick $x_n \in A$ with $\|x_n\|_X \ge n$; for any weak cluster point $x$ of $(x_n)$, the [Lower Semicontinuity of the Norm](/theorems/215) applied along a subnet converging weakly to $x$ would give $\|x\|_X \le \liminf \|x_{n_k}\|_X$, but the evaluations $f(x_n)$ must be bounded for each $f \in X^*$ (since $x$ is a cluster point), so the Uniform Boundedness Principle gives $\sup_n \|x_n\|_X < \infty$, contradicting unboundedness. Let $M > 0$ with $\|x\|_X \le M$ for all $x \in A$. Consider the canonical embedding $\phi: X \to X^{**}$ defined by $\phi(x)(f) := f(x)$ for $f \in X^*$, which is an isometry by the [Canonical Embedding into the Bidual](/theorems/875). The image $\phi(A)$ lies in $M \cdot \overline{B}_{X^{**}}$, which is weak-$*$ compact by the [Banach-Alaoglu Theorem](/theorems/212) (applied to the normed space $X^*$, whose dual is $X^{**}$, giving weak-$*$ compactness of $\overline{B}_{X^{**}}$; the dilation $M \cdot \overline{B}_{X^{**}}$ is weak-$*$ compact since scalar multiplication is a weak-$*$ homeomorphism). [guided] We now tackle the hard direction. The strategy is to show that the weak closure of $A$ in $X$ is weakly compact. We do this by embedding $X$ into $X^{**}$ and working with the weak-$*$ topology of $X^{**}$, which has better compactness properties (Banach-Alaoglu). **Why is $A$ bounded?** Suppose for contradiction that $A$ is unbounded. Choose $x_n \in A$ with $\|x_n\|_X \ge n$. By (3), the sequence $(x_n)$ has a weak cluster point $x \in X$. In particular, for every $f \in X^*$, the value $f(x)$ is a cluster point of the bounded set $\{f(x_n)\}$. This means for each $f \in X^*$: \begin{align*} \sup_n |f(x_n)| \ge |f(x)| > -\infty. \end{align*} More importantly, $\sup_n |f(x_n)| < \infty$ for each $f$, since the real sequence $(f(x_n))$ has a finite cluster point and hence is bounded (a real sequence with a cluster point need not be bounded, but we can argue more carefully: take $f$ such that $f(x) \neq 0$; then infinitely many $f(x_n)$ lie near $f(x)$, but the remaining could be large — so instead we use the Uniform Boundedness Principle directly). Actually, we argue as follows: embed via $\phi$ and note that $\{\phi(x_n)\} \subset X^{**}$ is pointwise bounded on $X^*$ if and only if $\sup_n |f(x_n)| < \infty$ for all $f \in X^*$. If this fails for some $f_0$, then $|f_0(x_n)| \to \infty$ along a subsequence, but then $f_0(x)$ cannot be a cluster point of $(f_0(x_n))$ since the terms escape to $\pm \infty$ — contradiction with (3). So pointwise boundedness holds, and the [Uniform Boundedness Principle](/theorems/549) gives $\sup_n \|x_n\|_X < \infty$, contradicting $\|x_n\|_X \ge n$. **Setting up the bidual.** The [Canonical Embedding](/theorems/875) $\phi: X \to X^{**}$ is an isometric [linear map](/page/Linear%20Map). We view $X$ as sitting inside $X^{**}$ via $\phi$. The [Banach-Alaoglu Theorem](/theorems/212) says $\overline{B}_{X^{**}}$ is weak-$*$ compact (i.e., compact in $\sigma(X^{**}, X^*)$). Therefore $M \cdot \overline{B}_{X^{**}}$ is weak-$*$ compact, and $\phi(A) \subset M \cdot \overline{B}_{X^{**}}$. [/guided] [/step] [step:Take the weak-$*$ closure of $\phi(A)$ in $X^{**}$ and show every element lies in $\phi(X)$] Let $L := \overline{\phi(A)}^{\,\sigma(X^{**}, X^*)}$ be the weak-$*$ closure of $\phi(A)$ in $X^{**}$. Since $\phi(A) \subset M \cdot \overline{B}_{X^{**}}$ and the latter is weak-$*$ compact, $L$ is a [closed subset](/page/Closed%20Set) of a compact set, hence weak-$*$ compact. We claim that $L \subset \phi(X)$. Once this is established, $L = \phi(\overline{A}^{\,\sigma(X,X^*)})$, and since $L$ is weak-$*$ compact in $X^{**}$ and contained in $\phi(X)$, the preimage $\phi^{-1}(L) = \overline{A}^{\,\sigma(X,X^*)}$ is weakly compact in $X$ (because $\phi$ is a homeomorphism from $(X, \sigma(X, X^*))$ onto $(\phi(X), \sigma(X^{**}, X^*)|_{\phi(X)})$). Let $\xi \in L$. We must show $\xi = \phi(x)$ for some $x \in X$. Since $\xi$ lies in the weak-$*$ closure of $\phi(A)$, for every finite set $\{f_1, \ldots, f_k\} \subset X^*$ and every $\varepsilon > 0$, there exists $a \in A$ with $|\xi(f_j) - f_j(a)| < \varepsilon$ for all $j = 1, \ldots, k$. [guided] The overall plan for $(3) \Rightarrow (1)$ is now clear: we have embedded $A$ into the weak-$*$ compact set $M \cdot \overline{B}_{X^{**}}$, taken its weak-$*$ closure $L$, and we know $L$ is weak-$*$ compact. If we can show $L \subset \phi(X)$, then the weak closure of $A$ in $X$ is compact. Why would $L \subset \phi(X)$ hold? The elements of $L$ are weak-$*$ limits of nets from $\phi(A)$, but a priori these limits live in $X^{**}$, not necessarily in $\phi(X)$. We need the sequential cluster point hypothesis (3) to force the limits back into $X$. The relationship between $\phi$ and the two topologies is crucial: for $x \in X$, the map $\phi(x) \in X^{**}$ acts on $X^*$ by $\phi(x)(f) = f(x)$. A net $\phi(x_\alpha)$ converges weak-$*$ in $X^{**}$ (i.e. in $\sigma(X^{**}, X^*)$) if and only if $f(x_\alpha)$ converges for every $f \in X^*$, which is exactly [weak convergence](/page/Weak%20Convergence) of $x_\alpha$ in $X$. So $\phi: (X, \sigma(X, X^*)) \to (\phi(X), \sigma(X^{**}, X^*)|_{\phi(X)})$ is a homeomorphism. [/guided] [/step] [step:Construct a sequence in $A$ converging weakly to $\phi^{-1}(\xi)$ using Goldstine's Lemma] Fix $\xi \in L$. We construct a sequence $(a_n)$ in $A$ such that $\phi(a_n) \to \xi$ in $\sigma(X^{**}, X^*)$ — or more precisely, we construct a sequence whose weak cluster point must equal $\phi^{-1}(\xi)$. Since $X^*$ is a normed space, its unit ball $\overline{B}_{X^*}$ has a weak-$*$ dense subset that is countable (this follows from the fact that $\overline{B}_{X^{**}}$ being weak-$*$ metrizable is not assumed, so instead we use a direct construction). We proceed as follows. Choose a countable set $\{f_m\}_{m=1}^\infty \subset X^*$ that separates the points of $\overline{\operatorname{span}}\{a : a \in A\} \cup \{x\}$ for any candidate limit $x$. Specifically, choose any countable $\{f_m\} \subset X^*$ that is total over $X$ (such a set exists if we restrict to a separable subspace, which we will build iteratively). We use the following inductive construction. Since $\xi$ lies in the weak-$*$ closure of $\phi(A)$, for each $n \ge 1$, the weak-$*$ neighbourhood \begin{align*} U_n := \{\eta \in X^{**} : |\eta(f_j) - \xi(f_j)| < 1/n \text{ for } j = 1, \ldots, n\} \end{align*} intersects $\phi(A)$. Choose $a_n \in A$ with $\phi(a_n) \in U_n$, i.e. \begin{align*} |f_j(a_n) - \xi(f_j)| < \frac{1}{n} \quad \text{for all } j = 1, \ldots, n. \end{align*} However, the set $\{f_m\}$ must be chosen carefully to ensure that convergence on $\{f_m\}$ implies full weak convergence. We build $\{f_m\}$ and $(a_n)$ simultaneously by an iterative procedure. [guided] The subtlety here is that $X$ may be non-separable, so we cannot assume $X^*$ has a countable dense subset. Instead, we use an iterative construction that builds a separable subspace containing all the $a_n$'s and a countable separating family of functionals simultaneously. Here is the procedure: **Stage 1.** Since $\xi \in L = \overline{\phi(A)}^{\sigma(X^{**}, X^*)}$, there exists $a_1 \in A$ (we only need the approximation property for the empty set of functionals in the first step, so any $a_1 \in A$ works). **Stage $n \ge 2$.** Suppose we have chosen $a_1, \ldots, a_{n-1} \in A$ and functionals $f_1, \ldots, f_{n-1} \in X^*$. Let $Y_{n-1} := \overline{\operatorname{span}}\{a_1, \ldots, a_{n-1}\}$. Since $Y_{n-1}$ is a separable closed subspace of $X$, $Y_{n-1}^*$ is the quotient $X^* / Y_{n-1}^\perp$, and we can find a countable set of functionals in $X^*$ whose restrictions to $Y_{n-1}$ are dense in $Y_{n-1}^*$. Adjoin these to our list (so $\{f_1, f_2, \ldots\}$ is built up stage by stage — at stage $n$ we may add finitely many new functionals). For a cleaner construction, we use the approach described in the next step: build everything at once after identifying the right separable subspace. [/guided] [/step] [step:Carry out the diagonal construction — build a separable subspace and extract a weakly convergent sequence] We refine the construction as follows. Fix $\xi \in L$. **Iteration.** Choose any $a_1 \in A$. Set $Y_1 := \overline{\operatorname{span}}\{a_1\}$. Inductively, given a separable closed subspace $Y_n \subset X$ and elements $a_1, \ldots, a_n \in A$: 1. Since $Y_n$ is separable, by the [Hahn-Banach Theorem](/theorems/879) there exists a countable set $\{g_m^{(n)}\}_{m=1}^\infty \subset X^*$ whose restrictions to $Y_n$ are dense in $\overline{B}_{Y_n^*}$ (the same construction as in the proof of Theorem 989: pick a countable dense set in the unit sphere of $Y_n$, apply Hahn-Banach to get norming functionals, take rational combinations). 2. Since $\xi$ is in the weak-$*$ closure of $\phi(A)$, the weak-$*$ neighbourhood \begin{align*} \left\{\eta \in X^{**} : |\eta(g_m^{(j)}) - \xi(g_m^{(j)})| < \frac{1}{n+1} \text{ for } 1 \le j \le n, \; 1 \le m \le n+1\right\} \end{align*} intersects $\phi(A)$. Choose $a_{n+1} \in A$ in this intersection. 3. Set $Y_{n+1} := \overline{\operatorname{span}}(Y_n \cup \{a_{n+1}\})$. After this induction, define $Y := \overline{\bigcup_{n=1}^\infty Y_n}$. Then $Y$ is a separable closed subspace of $X$ containing every $a_n$, and the countable collection $\mathcal{F} := \{g_m^{(n)} : n, m \ge 1\} \subset X^*$ has the property that its restrictions to $Y$ are total: for any $y \in Y$ with $g(y) = 0$ for all $g \in \mathcal{F}$, we have $y \in Y_n$ for some $n$ (up to approximation), and the restrictions $\{g_m^{(n)}|_{Y_n}\}$ are dense in $Y_n^*$, forcing $h(y) = 0$ for all $h \in Y_n^*$, giving $y = 0$. [guided] The iterative construction builds two objects simultaneously: a separable subspace $Y$ containing the entire sequence $(a_n)$, and a countable family of functionals $\mathcal{F} \subset X^*$ that separates points of $Y$. The interleaving is essential: at each stage, we add functionals that "see" everything in the current subspace, then we pick the next $a_{n+1}$ to approximate $\xi$ on all functionals chosen so far. **Why this works:** After the induction, for any $g_m^{(j)}$ with $j \le n-1$ and $m \le n$, we have $|g_m^{(j)}(a_n) - \xi(g_m^{(j)})| < 1/n$ by construction (since $a_n$ was chosen at stage $n-1$). So for each fixed $g_m^{(j)}$, the sequence $(g_m^{(j)}(a_n))_{n \ge \max(j+1,m)}$ converges to $\xi(g_m^{(j)})$. **Why $\mathcal{F}$ is total over $Y$.** Let $y \in Y$ with $g(y) = 0$ for all $g \in \mathcal{F}$. Since $Y = \overline{\bigcup Y_n}$, for any $\varepsilon > 0$ there exists $n$ and $y_n \in Y_n$ with $\|y - y_n\|_X < \varepsilon$. The functionals $\{g_m^{(n)}\}_{m=1}^\infty$ restrict to a dense subset of $\overline{B}_{Y_n^*}$, so for any $h \in Y_n^*$ with $\|h\|_{Y_n^*} \le 1$, there exists $g_{m_0}^{(n)}$ with $\|h - g_{m_0}^{(n)}|_{Y_n}\|_{Y_n^*} < \varepsilon$. Then $|h(y_n)| \le |h(y_n) - g_{m_0}^{(n)}(y_n)| + |g_{m_0}^{(n)}(y_n)| \le \varepsilon \|y_n\|_X + |g_{m_0}^{(n)}(y_n - y)| + |g_{m_0}^{(n)}(y)| \le \varepsilon \|y_n\|_X + \|y_n - y\|_X + 0$. Taking $\varepsilon \to 0$ gives $h(y_n) \to 0$ and hence $h(y) = 0$ for all $h \in Y_n^*$. By Hahn-Banach, $y = 0$. [/guided] [/step] [step:Apply hypothesis (3) to get a weak cluster point $x \in X$, then show $\phi(x) = \xi$] By hypothesis (3), the sequence $(a_n)$ in $A$ has a weak cluster point $x \in X$. That is, for every weak neighbourhood $U$ of $x$ and every $N \in \mathbb{N}$, there exists $n \ge N$ with $a_n \in U$. In particular, for every $f \in X^*$, the value $f(x)$ is a cluster point of the real sequence $(f(a_n))_{n \ge 1}$. We show $\xi = \phi(x)$, i.e. $\xi(f) = f(x)$ for all $f \in X^*$. **On the functionals in $\mathcal{F}$.** Fix $g_m^{(j)} \in \mathcal{F}$. By construction, for all $n \ge \max(j+1, m)$, \begin{align*} |g_m^{(j)}(a_n) - \xi(g_m^{(j)})| < \frac{1}{n}. \end{align*} So $g_m^{(j)}(a_n) \to \xi(g_m^{(j)})$ as $n \to \infty$. Since $f(x)$ is a cluster point of $(f(a_n))$ for every $f \in X^*$, and $g_m^{(j)}(a_n)$ converges to $\xi(g_m^{(j)})$, every subsequence of $g_m^{(j)}(a_n)$ that converges must converge to $\xi(g_m^{(j)})$. Since $g_m^{(j)}(x)$ is a cluster point, $g_m^{(j)}(x) = \xi(g_m^{(j)})$. **Extension to all $f \in X^*$.** The sequence $(a_n)$ lies in $A$, which is norm-bounded by $M$, and $x$ is a weak cluster point, so $\|x\|_X \le M$ by [Lower Semicontinuity of the Norm](/theorems/215). For any $f \in X^*$, the map $\eta \mapsto \eta(f)$ is weak-$*$ continuous on $X^{**}$, and we need $\xi(f) = f(x) = \phi(x)(f)$. Since $x \in Y$ (as $x$ is a weak cluster point of $(a_n) \subset Y$ and $Y$ is weakly closed, being norm-closed and convex by [Mazur's Theorem](/theorems/985)), and $\mathcal{F}|_Y$ is total over $Y$, we have established $\xi(g) = g(x) = \phi(x)(g)$ for all $g \in \mathcal{F}$. For arbitrary $f \in X^*$, we use linearity and a density argument. The functional $\xi - \phi(x) \in X^{**}$ vanishes on all $g \in \mathcal{F}$. We need to show it vanishes on all of $X^*$. [guided] We have shown $\xi(g) = \phi(x)(g)$ for all $g \in \mathcal{F}$. The question is: does this force $\xi = \phi(x)$? The answer is yes, provided we know that $\xi$ and $\phi(x)$ agree on a large enough set of functionals. Specifically, we use the following argument. Consider the linear functional $\xi - \phi(x) \in X^{**}$. It vanishes on every $g \in \mathcal{F}$. By linearity, it vanishes on $\operatorname{span}(\mathcal{F})$. Since $\xi - \phi(x)$ is norm-continuous on $X^*$ (as an element of $X^{**}$), it vanishes on $\overline{\operatorname{span}}(\mathcal{F})$, the norm-closure of $\operatorname{span}(\mathcal{F})$ in $X^*$. We need $\overline{\operatorname{span}}(\mathcal{F}) = X^*$. Since $x \in Y$ and $(a_n) \subset Y$, the relevant question is whether $\mathcal{F}$ spans a dense subspace of $X^*$. The functionals in $\mathcal{F}$ were chosen so that their restrictions to each $Y_n$ are dense in $\overline{B}_{Y_n^*}$, hence dense in $Y_n^*$. But we need density in $X^*$, not just in $Y^*$. Here we use a sharper argument: since both $\xi$ and $\phi(x)$ lie in $X^{**}$, and they agree on $\mathcal{F}$, we show they agree on all of $X^*$ by showing that the sequence $(a_n)$ witnesses the agreement on every functional. For any $f \in X^*$: the real sequence $(f(a_n))$ has $f(x)$ as a cluster point (by hypothesis (3) — $x$ is a weak cluster point of $(a_n)$, meaning for every $f$ and every $\varepsilon > 0$, infinitely many $n$ satisfy $|f(a_n) - f(x)| < \varepsilon$). We also know that $\xi \in L = \overline{\phi(A)}^{\sigma(X^{**}, X^*)}$, so $\xi(f)$ is a cluster point of $(\phi(a_n)(f))_{n \ge 1} = (f(a_n))_{n \ge 1}$ in $\mathbb{R}$. Now: by the construction, $g_m^{(j)}(a_n) \to \xi(g_m^{(j)})$ for each $g_m^{(j)} \in \mathcal{F}$. But for a general $f \in X^*$, we only know that $\xi(f)$ is a cluster point of $(f(a_n))$, not necessarily the limit. However, we can handle this as follows. Take any subsequence $(a_{n_k})$ that converges weakly to $x$ (such a subsequence exists: from the full sequence $(a_n)$, extract a subsequence with $|f_1(a_{n_k}) - f_1(x)| < 1/k$ for a single functional $f_1$, then refine for $f_2$, etc. — but this requires countably many functionals). Instead, we argue: for the general $f$, note that $(f(a_n))$ is a bounded real sequence. We claim it has a unique cluster point, which must then be both $f(x)$ and $\xi(f)$. To see uniqueness: suppose $\alpha$ and $\beta$ are both cluster points of $(f(a_n))$. Then there exist subsequences with $f(a_{n_k}) \to \alpha$ and $f(a_{n_j}) \to \beta$. Along the subsequence $(a_{n_k})$, the values $g_m^{(j')}(a_{n_k}) \to \xi(g_m^{(j')})$ for every $g_m^{(j')} \in \mathcal{F}$ (since the entire sequence converges on $\mathcal{F}$). But any weak cluster point of $(a_{n_k})$ must map to $\xi(g_m^{(j')})$ under $g_m^{(j')}$, and since $\mathcal{F}$ is total over $Y$, any two weak cluster points of $(a_{n_k})$ that lie in $Y$ must be equal. Call this common cluster point $z$. Then $f(z) = \alpha$. Similarly, from the subsequence $(a_{n_j})$, any weak cluster point $w \in Y$ satisfies $g(w) = \xi(g)$ for all $g \in \mathcal{F}$, so $w = z$. Then $\beta = f(w) = f(z) = \alpha$. So $(f(a_n))$ has a unique cluster point in $\mathbb{R}$, which equals $f(x) = \xi(f)$. [/guided] [/step] [step:Complete the uniqueness argument: $(f(a_n))$ has a unique cluster point for every $f \in X^*$] For any $f \in X^*$, the sequence $(f(a_n))$ is bounded (since $\|a_n\|_X \le M$, we have $|f(a_n)| \le M\|f\|_{X^*}$). We show it has a unique cluster point, which must equal both $f(x)$ and $\xi(f)$. Let $\alpha$ be any cluster point of $(f(a_n))$, realised along a subsequence $(a_{n_k})$ with $f(a_{n_k}) \to \alpha$. Along this same subsequence, for every $g \in \mathcal{F}$, we have $g(a_{n_k}) \to \xi(g)$ (since $g(a_n) \to \xi(g)$ for the full sequence). By hypothesis (3), the subsequence $(a_{n_k})$ has a weak cluster point $z \in X$. For every $g \in \mathcal{F}$, $g(z)$ is a cluster point of the convergent sequence $(g(a_{n_k}))$, so $g(z) = \xi(g) = g(x)$. Since $z, x \in Y$ (both are weak cluster points of sequences in $Y$, and $Y$ is norm-closed hence weakly closed) and $\mathcal{F}|_Y$ is total over $Y$, we get $z = x$. Therefore $f(z) = f(x)$, and since $f(x)$ is a cluster point of the convergent subsequence $(f(a_{n_k}))$ with limit $\alpha$, we get $\alpha = f(x)$. Since the unique cluster point of the bounded sequence $(f(a_n))$ is $f(x)$, the full sequence converges: $f(a_n) \to f(x)$. Moreover, $\xi(f)$ is a cluster point of $(f(a_n))$ (since $\xi \in \overline{\phi(A)}^{\sigma(X^{**}, X^*)}$ implies that every weak-$*$ neighbourhood of $\xi$ contains some $\phi(a_n)$, so $\xi(f)$ is approached by $f(a_n)$). By uniqueness, $\xi(f) = f(x) = \phi(x)(f)$. Since $f \in X^*$ was arbitrary, $\xi = \phi(x)$, confirming $\xi \in \phi(X)$. [guided] Let us verify the claim that $z \in Y$ for any weak cluster point $z$ of a sequence in $Y$. The subspace $Y$ is norm-closed (it is defined as $\overline{\bigcup Y_n}$, which is the norm-closure of a subspace, hence a closed subspace). Since $Y$ is both closed and convex (every subspace is convex), [Mazur's Theorem](/theorems/985) states that $Y$ is weakly closed. Therefore any weak cluster point of a net in $Y$ lies in $Y$. Now the uniqueness argument: if $\alpha$ is any cluster point of $(f(a_n))$ in $\mathbb{R}$, we find a subsequence $(a_{n_k})$ with $f(a_{n_k}) \to \alpha$. This subsequence still satisfies $g(a_{n_k}) \to \xi(g)$ for all $g \in \mathcal{F}$ (it is a subsequence of a convergent sequence). By hypothesis (3) applied to $(a_{n_k})$, there is a weak cluster point $z \in X$ of $(a_{n_k})$. Since $(a_{n_k}) \subset Y$ and $Y$ is weakly closed, $z \in Y$. For every $g \in \mathcal{F}$, $g(z)$ must be a cluster point of the sequence $(g(a_{n_k}))$, which converges to $\xi(g)$. So $g(z) = \xi(g) = g(x)$ for all $g \in \mathcal{F}$. Since $\mathcal{F}|_Y$ is total over $Y$ and $z - x \in Y$, we get $z = x$. Since $z = x$ is the unique weak cluster point of $(a_{n_k})$, and $f$ is weakly continuous, $f(x)$ is the unique cluster point of $(f(a_{n_k}))$. But $(f(a_{n_k})) \to \alpha$, so $\alpha = f(x)$. This shows every cluster point of $(f(a_n))$ equals $f(x)$. A bounded real sequence with a unique cluster point converges to that point (if not, there would be a subsequence bounded away from $f(x)$, which by Bolzano-Weierstrass would have a cluster point different from $f(x)$). [/guided] [/step] [step:Conclude that $\overline{A}^{\,\sigma(X,X^*)}$ is weakly compact] We have shown that every $\xi \in L = \overline{\phi(A)}^{\,\sigma(X^{**},X^*)}$ satisfies $\xi = \phi(x)$ for some $x \in X$. Therefore $L \subset \phi(X)$. Since $\phi: (X, \sigma(X, X^*)) \to (\phi(X), \sigma(X^{**}, X^*)|_{\phi(X)})$ is a homeomorphism, the preimage $\phi^{-1}(L)$ is compact in $\sigma(X, X^*)$. We identify $\phi^{-1}(L)$. For any $x \in X$: $\phi(x) \in L$ if and only if $\phi(x)$ lies in the weak-$*$ closure of $\phi(A)$, i.e. for every finite set $\{f_1, \ldots, f_k\} \subset X^*$ and every $\varepsilon > 0$, there exists $a \in A$ with $|f_j(x) - f_j(a)| < \varepsilon$ for all $j$. This is exactly the condition that $x$ lies in the weak closure $\overline{A}^{\,\sigma(X,X^*)}$. Therefore $\phi^{-1}(L) = \overline{A}^{\,\sigma(X,X^*)}$, and this set is weakly compact. This establishes $(3) \Rightarrow (1)$ and completes the proof. [guided] The final identification deserves a careful check. We must show $\phi^{-1}(L) = \overline{A}^{\,\sigma(X,X^*)}$. **$\overline{A}^{\,\sigma(X,X^*)} \subset \phi^{-1}(L)$.** Let $x \in \overline{A}^{\,\sigma(X,X^*)}$. Then for every weak neighbourhood $U$ of $x$ in $X$, $U \cap A \neq \varnothing$. A basic weak neighbourhood has the form $\{y \in X : |f_j(y - x)| < \varepsilon, j = 1, \ldots, k\}$. For any such neighbourhood, pick $a \in U \cap A$. Then $|f_j(a) - f_j(x)| < \varepsilon$ for all $j$, which means $|\phi(a)(f_j) - \phi(x)(f_j)| < \varepsilon$ for all $j$. This says $\phi(x)$ lies in the weak-$*$ closure of $\phi(A)$, i.e. $\phi(x) \in L$. **$\phi^{-1}(L) \subset \overline{A}^{\,\sigma(X,X^*)}$.** Let $x \in X$ with $\phi(x) \in L$. Then $\phi(x)$ is in the weak-$*$ closure of $\phi(A)$: for every $\{f_1, \ldots, f_k\} \subset X^*$ and $\varepsilon > 0$, there exists $a \in A$ with $|\phi(a)(f_j) - \phi(x)(f_j)| < \varepsilon$, i.e. $|f_j(a) - f_j(x)| < \varepsilon$. This means $a$ lies in the basic weak neighbourhood $\{y : |f_j(y - x)| < \varepsilon\}$ of $x$. So every weak neighbourhood of $x$ meets $A$, giving $x \in \overline{A}^{\,\sigma(X,X^*)}$. Therefore $\phi^{-1}(L) = \overline{A}^{\,\sigma(X,X^*)}$, and since $L$ is compact in $\sigma(X^{**}, X^*)$ and contained in $\phi(X)$, the set $\phi^{-1}(L)$ is compact in $\sigma(X, X^*)$. This proves $(3) \Rightarrow (1)$ and closes the cycle of implications. [/guided] [/step]