Consider the sequence of vector spaces $i_n : \mathbb R^n\to \mathbb R^{n+1}$ given by the inclusions $x\mapsto (x,0)$.
We can consider the direct limit of this sequence, call it $\mathbb R^\infty$.
This object is defined by the universal property that any collection of homomorphisms $\phi_n:\mathbb R^n\to Y$ such that $\phi_{n+1}\circ i_n = \phi_n $ induces a map $u:\mathbb R^\infty\to Y$ such that the obvious diagram commutes.
It seems to me that this definition is satisfied by both $\bigoplus_{i\geq 0} \mathbb R$ and $\prod_{i\geq 0} \mathbb R$ because for each $n$ they split as $\mathbb R^n \oplus \bigoplus_{i> n} \mathbb R$ and $\mathbb R^n \oplus \prod_{i> n} \mathbb R$. However the two are not isomorphic.
What am I missing here ?
It is not satisfied by the infinite product. Consider the natural sequence of inclusions $\phi_n : \mathbb{R}^n \to \bigoplus_{i \ge 0} \mathbb{R}$; by the universal property we get a map $\mathbb{R}^{\infty} \to \bigoplus_{i \ge 0} \mathbb{R}$ which of course is an isomorphism. We do not get a map $\prod_{i \ge 0} \mathbb{R} \to \bigoplus_{i \ge 0} \mathbb{R}$ because there is nowhere for the sequences which are not finitely supported to go. This map would have to be a retract of the inclusion $\bigoplus_{i \ge 0} \mathbb{R} \to \prod_{i \ge 0} \mathbb{R}$ and I believe it is consistent with ZF + the negation of the axiom of choice that no such retract exists; even assuming AC, such a retract is not unique, which contradicts the uniqueness part of the universal property.