Let $\mathcal{J}$ be an uncoutable set and form the $\mathcal{J}$-indexed product $\prod_\mathcal{J}I$ of copies of the unit interval $I=[0,1]$. The claim is that the singleton containing the origin $$\{(0)_{\mathcal{J}}\}\subseteq \prod_\mathcal{J}I$$ is not a zero set (i.e. cannot be written $f^{-1}(\{0\})$ for some continuous function $f:\prod_\mathcal{J}I\rightarrow\mathbb{R}$).
How can I see that this claim is true?
Let $f: \prod_{\mathcal{J}} I \to \mathbb R$ be continuous with $f((0))=0$. Let $U_n = f^{-1}((-1/n, 1/n))$. These are open sets whose intersection is $f^{-1}(\{0\})$. Thus each $U_n$ contains a basic open neighbourhood $B_n$ of $(0)$. The basic open set $B_n$ is a cylinder set $\{p: p_j \in J_j\ \text{for}\ j \in \mathcal{F}_n\}$ where $\mathcal{F}_n \subset \mathcal{J}$ is finite and each $J_j$ is a neighbourhood of $0$. In particular, it contains $\{p: p_j = 0\ \text{for}\ j \in \mathcal{F}_n\}$. Now $\mathcal{F} = \bigcup_n \mathcal{F}_n$ is countable, and any $p$ with $p_j = 0$ for all $j \in \mathcal{F}$ is in all $B_n$, and therefore in $f^{-1}(\{0\})$. Thus $f^{-1}(\{0\})$ contains points other than $(0)$.
