Presumably you mean that you're considering embedded into in the natural way (all coordinates zero, except the ) and by the 'algebraic direct sum' you mean . If so, let's show that . To see that let . Then, for any we have, since there exists some such that . Define then by

Evidently then and moreover . Since was arbitrary the conclusion follows (technically you have to adjust the above argument so that but that's easy enough).

Conversely, suppose that . Then, there exists some such that for every there exists some such that . Evidently this implies that for every one has that . So, let be arbitrary. Then, by definition there exists some such that for every one has that . Consequently

Thus, . Since was arbitrary it follows that .

That should give you the basic idea...I might have skipped a minor detail in the case but it's easily fixable.