[proofplan] The proof shows that $\prod_{\alpha \in A} X_\alpha$ is not first-[countable](/page/Countable%20Set), and since every [metrizable space](/page/Metrizable%20Space) is first-countable, this implies non-metrizability. Fix a point $p$ in the product. Each basic open neighbourhood of $p$ constrains only finitely many coordinates. A countable family of such neighbourhoods therefore constrains at most countably many coordinates — but $A$ is uncountable, so some coordinate $\beta$ is entirely unconstrained. We use this unconstrained coordinate to exhibit an open neighbourhood of $p$ that cannot contain any member of the proposed countable basis, yielding a contradiction. [/proofplan]

[step:Record the finite-support structure of basic open sets in the product topology]Let $X := \prod_{\alpha \in A} X_\alpha$ carry the [product topology](/page/Product%20Topology). A basis for this topology consists of sets of the form \begin{align*} \prod_{\alpha \in A} U_\alpha, \qquad \text{where each } U_\alpha \subset X_\alpha \text{ is open and } U_\alpha = X_\alpha \text{ for all but finitely many } \alpha. \end{align*} For any such basic open set $B = \prod_{\alpha \in A} U_\alpha$, define its **support** as \begin{align*} \operatorname{supp}(B) := \{\alpha \in A : U_\alpha \ne X_\alpha\}. \end{align*} By definition of the product topology, $\operatorname{supp}(B)$ is a finite subset of $A$.[/step]

[guided]The product topology (also called the [Tychonoff topology](/page/Product%20Topology)) is the coarsest topology making each projection $\pi_\alpha: X \to X_\alpha$ [continuous](/page/Continuity). Its subbasis consists of sets $\pi_\alpha^{-1}(U_\alpha)$ for $U_\alpha \subset X_\alpha$ open, and its basis consists of finite intersections of such subbasis elements. Equivalently, a basic open set is a product $\prod_{\alpha \in A} U_\alpha$ where $U_\alpha = X_\alpha$ for all but finitely many indices $\alpha$. The support records which coordinates are genuinely constrained. If $\alpha \notin \operatorname{supp}(B)$, then $B$ imposes no restriction on the $\alpha$-th coordinate — any value in $X_\alpha$ is permitted. This finite-support property is the structural feature we will exploit.[/guided]

[step:Show that a countable family of neighbourhoods leaves uncountably many coordinates unconstrained]Fix a point $p = (p_\alpha)_{\alpha \in A} \in X$. Suppose for contradiction that there exists a countable neighbourhood basis $\{W_n\}_{n \in \mathbb{N}}$ at $p$, meaning that for every open set $V$ containing $p$, there exists $n$ with $W_n \subset V$. For each $n \in \mathbb{N}$, since $W_n$ is open and contains $p$, there exists a basic open set $B_n = \prod_{\alpha \in A} U_\alpha^{(n)}$ with $p \in B_n \subset W_n$. Define \begin{align*} S := \bigcup_{n=1}^{\infty} \operatorname{supp}(B_n). \end{align*} Each $\operatorname{supp}(B_n)$ is finite, so $S$ is a countable union of finite sets, hence countable. Since $A$ is uncountable, $A \setminus S \ne \varnothing$. For every $\beta \in A \setminus S$ and every $n \in \mathbb{N}$, we have $\beta \notin \operatorname{supp}(B_n)$, so $U_\beta^{(n)} = X_\beta$. In other words, no basic open set $B_n$ constrains the $\beta$-th coordinate.[/step]

[guided]Each basic open set $B_n$ constrains only the finitely many coordinates in $\operatorname{supp}(B_n)$. Collecting all constrained coordinates across the entire countable family gives the set $S = \bigcup_{n=1}^{\infty} \operatorname{supp}(B_n)$. A countable union of finite sets is countable, so $|S| \le \aleph_0$. Since $|A| > \aleph_0$, the complement $A \setminus S$ is nonempty — in fact uncountable. Any index $\beta \in A \setminus S$ satisfies $\beta \notin \operatorname{supp}(B_n)$ for every $n$, so $U_\beta^{(n)} = X_\beta$ for all $n$: the $\beta$-th coordinate is completely free in every $B_n$, and therefore in every $W_n \supset B_n$.[/guided]

[step:Exploit an unconstrained coordinate to contradict the neighbourhood basis property]If every factor $X_\alpha$ for $\alpha \in A \setminus S$ has the indiscrete topology, then the product topology is determined entirely by the countably many factors indexed by $S$. But each indiscrete $X_\alpha$ with $|X_\alpha| \ge 2$ is not $T_0$ (distinct points share all open neighbourhoods), so the product is not $T_0$ and hence not metrizable. Non-metrizability follows without the first-countability argument. We may therefore assume there exists $\beta \in A \setminus S$ and a proper [open subset](/page/Open%20Set) $V_\beta \subsetneq X_\beta$ with $p_\beta \in V_\beta$. (If every proper nonempty open set in $X_\beta$ misses $p_\beta$, then $p_\beta$ cannot be separated from any point in such an open set, making the product non-$T_1$ and hence non-metrizable. So we may assume such $V_\beta$ exists.) Define $V := \pi_\beta^{-1}(V_\beta) = \{x \in X : x_\beta \in V_\beta\}$. This is open in $X$ and $p \in V$. If $\{W_n\}$ were a neighbourhood basis at $p$, there would exist $n_0 \in \mathbb{N}$ with $B_{n_0} \subset W_{n_0} \subset V$. But $\beta \notin \operatorname{supp}(B_{n_0})$, so $U_\beta^{(n_0)} = X_\beta$. Since $V_\beta \subsetneq X_\beta$, choose $q_\beta \in X_\beta \setminus V_\beta$ and define $z \in X$ by \begin{align*} z_\alpha := \begin{cases} p_\alpha & \text{if } \alpha \ne \beta, \\ q_\beta & \text{if } \alpha = \beta. \end{cases} \end{align*} For $\alpha \in \operatorname{supp}(B_{n_0})$: since $\beta \notin \operatorname{supp}(B_{n_0})$, we have $\alpha \ne \beta$, so $z_\alpha = p_\alpha \in U_\alpha^{(n_0)}$ (because $p \in B_{n_0}$). For $\alpha \notin \operatorname{supp}(B_{n_0})$: $U_\alpha^{(n_0)} = X_\alpha$, so $z_\alpha \in U_\alpha^{(n_0)}$ automatically. Therefore $z \in B_{n_0} \subset V$. But $z_\beta = q_\beta \notin V_\beta$, so $z \notin V$ — a contradiction.[/step]

[guided]We need to produce an open neighbourhood of $p$ that constrains the $\beta$-th coordinate, which none of the $B_n$ do. **Degenerate case.** If every $X_\alpha$ for $\alpha \in A \setminus S$ is indiscrete (only open sets are $\varnothing$ and $X_\alpha$), then the subbasis elements $\pi_\alpha^{-1}(U_\alpha)$ for $\alpha \in A \setminus S$ are either $\varnothing$ or $X$, contributing nothing to the topology. The product topology depends only on the countably many factors indexed by $S$. But each indiscrete $X_\alpha$ with $|X_\alpha| \ge 2$ is not $T_0$: if $a, b \in X_\alpha$ with $a \ne b$, every open set containing $a$ also contains $b$ and vice versa. The product inherits this: the two points in $X$ that agree with $p$ on all coordinates except $\alpha_0 \in A \setminus S$ (where they take distinct values) cannot be topologically distinguished. Since every [metric space](/page/Metric%20Space) is $T_0$ (the open ball $B(x, d(x,y)/2)$ contains $x$ but not $y$), the product is not metrizable. **Main case.** We may assume there exists $\beta \in A \setminus S$ and a proper open subset $V_\beta \subsetneq X_\beta$ with $p_\beta \in V_\beta$. (Why can we find $V_\beta$ containing $p_\beta$? If $X_\beta$ has a non-indiscrete topology but every proper nonempty open set misses $p_\beta$, then $p_\beta$ is not contained in any proper open set. But then for any $q_\beta \ne p_\beta$ belonging to a proper open set $W$, the points $p$ and $q$ (agreeing on all coordinates except $\beta$) satisfy: $q$ has the open neighbourhood $\pi_\beta^{-1}(W)$ that does not contain $p$, while $p$ has no open neighbourhood excluding $q$. This makes the product non-$T_1$, hence non-metrizable.) Define \begin{align*} V := \pi_\beta^{-1}(V_\beta) = \{x \in X : x_\beta \in V_\beta\}. \end{align*} This is a subbasis element of the product topology, hence open, and $p \in V$ since $p_\beta \in V_\beta$. If the neighbourhood basis property holds, there exists $n_0 \in \mathbb{N}$ with $W_{n_0} \subset V$, hence $B_{n_0} \subset V$. Every element of $B_{n_0}$ must then have its $\beta$-th coordinate in $V_\beta$. We show this is impossible. Since $\beta \notin \operatorname{supp}(B_{n_0})$, the set $B_{n_0}$ imposes no restriction on the $\beta$-th coordinate: $U_\beta^{(n_0)} = X_\beta$. Since $V_\beta \subsetneq X_\beta$, there exists $q_\beta \in X_\beta \setminus V_\beta$. Construct the point $z \in X$ by \begin{align*} z_\alpha := \begin{cases} p_\alpha & \text{if } \alpha \ne \beta, \\ q_\beta & \text{if } \alpha = \beta. \end{cases} \end{align*} We verify $z \in B_{n_0}$ coordinate by coordinate: - For $\alpha \in \operatorname{supp}(B_{n_0})$: since $\beta \notin \operatorname{supp}(B_{n_0})$, we have $\alpha \ne \beta$, so $z_\alpha = p_\alpha$. Since $p \in B_{n_0}$, we know $p_\alpha \in U_\alpha^{(n_0)}$, hence $z_\alpha \in U_\alpha^{(n_0)}$. - For $\alpha \notin \operatorname{supp}(B_{n_0})$: $U_\alpha^{(n_0)} = X_\alpha$, so $z_\alpha \in U_\alpha^{(n_0)}$ regardless. Therefore $z \in B_{n_0} \subset V$. But $z_\beta = q_\beta \notin V_\beta$, so $z \notin V$. This contradicts $B_{n_0} \subset V$.[/guided]

[step:Conclude non-metrizability from the failure of first countability] The contradiction shows that no countable neighbourhood basis at $p$ exists, so $X$ is not first-countable. Every metrizable space is first-countable: if $(Y, d)$ is a metric space and $y \in Y$, then $\{B(y, 1/n)\}_{n \in \mathbb{N}}$ is a countable neighbourhood basis at $y$, since for every open $U \ni y$ there exists $\varepsilon > 0$ with $B(y, \varepsilon) \subset U$ and hence $B(y, 1/n) \subset U$ for $n > 1/\varepsilon$. Since $X$ is not first-countable, $X$ is not metrizable. [/step]