If {Ui | i ∈ I} is a collection of vector spaces over a field F then ⊕i∈IUi is a subspace of U i∈I (Ui). This is the external direct sum of the spaces {Ui|i ∈ I}
In the case that I is finite, can we define the external direct sum of Ui as a subspace of the direct product of Ui, as in that case they are identical? How can we prove this in the case that I is infinite? Sorry for bad formatting
if $U_i$ is a subspace of $U$, then it is a vector space over $\Bbb F$ in its own right. Then we can take the direct sum of $\{U_i\}_{i\in I}$ by taking
$$\bigoplus_{i\in I}U_i=\left\{\sum_{i=1}^n a_iu_i\mid u_i\in U_i,\, a_i\in \Bbb F,\, n\in\Bbb N_0\right\}$$$$=\{(u_i)_{i\in I}\mid u_i\in U_i \text{ and finitely many of the }u_i\ne0\}.$$
If one does not require these sums to be finite (i.e. require finite support), then we obtain the definition of the direct product:
$$\prod_{i\in I}U_i=\left\{(u_i)_{i\in I}\mid u_i\in U_i\right\}.$$
Note: The $a_i$ is superfluous in the direct sum written above, since $a_iu_i\in U_i$.