[Math] Can a subspace S containing vectors with a finite number of nonzero components contain the zero vector

linear algebravector-spaces

A set S consists of those vectors with a finite number of nonzero components.

Given a vector space V with infinite dimensions, or $V = R^\infty$, I am trying to prove that the set S is a subspace of it.

This is Problem 18 in Section 7.1 (Vector Spaces and Subspaces), page 285, from Linear Algebra by Jeffrey Holt.

By definition, to be a subspace, S must:

  1. Contain the zero vector
  2. Be closed under vector addition
  3. Be closed under scalar multiplication

I immediately want to say that this proof is false because of the first point above with the zero vector. Since S consists of those vectors with a finite number of nonzero components, it makes me think that it cannot contain the zero vector which would have the same number of components as S, with all of them being 0 (e.g. The zero vector contains a finite number of zero components!)

Can someone help me understand this better? Is my logic correct?

Best Answer

Your headline question is poorly phrased. Your longer description is better, though please note the comments about notation.

$S$ is a subspace. Let's look at the 3 rules.

  1. $0 \in S$. Of course it is, it has a finite number of nonzero components, i.e. $0$ of them!
  2. $s,t \in S \implies s+t \in S$. The indices of the nonzero components of $s+t$ is at most the union of the indices for $s$ and $t$, hence finite.
  3. $s \in S \implies \lambda s \in S$ for any scalar $\lambda$. Becuase we have already covered off the zero vector, we may assume $\lambda \neq 0$. Then the zero components of $s$ and $\lambda s$ are identical.
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