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Let $V$ be an infinite dimensional vector space $V$ with a basis $\{e_\alpha\}_{\alpha \in A}$ and $\{E_\alpha\}_{\alpha \in A}$ a corresponding set of subspaces (where $E_\alpha = span(e_\alpha)$).
Then V has a representation as the direct sum, $V = \oplus_{\alpha \in A} E_\alpha$.
I.e., each $v \in V$ is a finite sum of vectors $v_\alpha \in V_\alpha$
I have seen this also expressed as each $v \in V$ = $\Sigma_{\alpha \in A} v_\alpha$ where only finitely many of the $v_\alpha \ne 0$.

Lets assume that V has no norm or inner product so there is no possibility of defining an infinite sum by convergence. Is it valid to say that the infinite number of zero vectors sum to zero ?

Besides idle curiosity, the question gains relevance when considering the inclusions $\{i_\alpha\}_{\alpha \in A}$ and projections $\{p_\alpha\}_{\alpha \in A}$ associated with the direct sum. I would like to say that $v = \Sigma_{\alpha \in A} i_\alpha \circ p_\alpha (v)$.

There is a well-answered question Why is $\infty \cdot 0$ not clearly equal to $0$? which looks at summing infinitely many zeroes. This generally seems to to be answered as it depending on context: I don't see this context mentioned, hence this post.

Tom Collinge
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    It is harmless to talk about sums over an infinite index set, subject to the proviso that only finitely many summands are non-zero. – Rob Arthan Feb 13 '21 at 16:06
  • Yes, $0 \cdot \infty$ is an indeterminate form. But nevertheless it is still true that if $f(x) = 0$ identically and $g(x) \to \infty$ we may conclude $f(x)g(x) \to 0$. Similarly, $0^0$ is an indeterminate form, but if $f(x) \searrow 0$ and $g(x) = 0$ identically, then $f(x)^{g(x)} \to 1$. – GEdgar Feb 13 '21 at 16:47

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It is valid to say whatever you want, as long as you define what you mean by it. In this case, there is a common convention to define an infinite sum $\sum_{\alpha\in A} x_\alpha$ as long as $x_\alpha=0$ for all but finitely many $\alpha$, even in the absence of anything like a topology as would usually be needed to define infinite sums. This convention is convenient so that you can refer to such sums without constantly needing to restrict the index set to different finite subsets.

(If you like, you could also consider this as a limit of finite partial sums with respect to the discrete topology--the finite partial sums are eventually constant since only finitely many of the terms are nonzero.)

Eric Wofsey
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