# Can a subspace S containing vectors with a finite number of nonzero components contain the zero vector?

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?

• $R^\infty$ is not a good notation, nor a legit one, AFAIK. – user228113 Nov 2 '16 at 22:00
• And, what do you mean by component? – user228113 Nov 2 '16 at 22:04
• What does it mean for the zero vector to "have the same number of components as $S$?" Can you give some examples of elements in $S$ and explain why the zero vector does not fit this description? – Matt Nov 2 '16 at 22:05
• Sorry, that is how my textbook wrote the question. – zhughes3 Nov 2 '16 at 22:05
• @G. Sassatelli Matt I am trying to show that S is a subspace, which means it must contain the zero vector. For it to be a member of S, it must have the same number of components as a vector in S. So if S is in $R^3$, a valid vector would <1, 2, 3> – zhughes3 Nov 2 '16 at 22:07

$S$ 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.
• Not really. I should have been more precise. Let $s_i$ be the $i$-th component of s. $i$ is the index. For $(s+t)_i$ to be nonzero, at least one of $s_i$ and $t_i$ must be nonzero. This can only occur a finite number of times. – Scott Burns Nov 5 '16 at 2:09