Let $2^{<\mathbb{N}}$ be the full binary tree, and let $B\subseteq 2^{<\mathbb{N}}$ be decidable. Suppose $B$ is a bar, meaning that for every infinite binary sequence $\alpha\in 2^{\mathbb{N}}$ there exists $n\in\mathbb{N}$ such that the length-$n$ prefix of $\alpha$ lies in $B$. Then $B$ is uniform: there exists $N\in\mathbb{N}$ such that every $\alpha\in 2^{\mathbb{N}}$ has a prefix of length at most $N$ in $B$.