Alternative characterisation of supremum

Question

Given a sequence $\{b_j\}_{j\in\mathbb{N}}$, can we define $\alpha_k=\sup_{j\geq k} b_j$ for any $k\in\mathbb{N}$ recursively through $\alpha_j=\max\{b_j,\alpha_{j+1}\}$ for all $j\geq k$? I am able to show that $\alpha_k\geq \sup_{j\geq k} b_j$, but due to the backward structure of the definition I am unable to show the other direction, as my attempt by contradiction failed. Could it be that my intuition about the equivalence of this characterisation of the supremum is wrong? Thanks.