# If V is an infinite-dimensional vector space, and S is an infinite-dimensional subspace of V, must the dimension of V/S be finite? Explain

I'm having some issues thinking about this one. I'm going to think aloud here. If $V$ is infinite dimensional, this means that there infinitely many basis vectors. I know the choice of the representative of the space doesn't matter. Likewise for $S$, only any vector we choose from $S$ will also already be in $V$.

Now, $\frac{V}{S}$ is a space of vectors in V modulo the vectors in $S$. This only means that the difference of say $v-s \in \frac{V}{S}$, right? I could really use some help here. Thanks.

• Consider the space of infinite real-valued sequences and the subspace thereof that consists of those sequences that are zero in every even coordinate. – David Mitra Feb 4 '14 at 14:40
• Okay, I think I see what you're saying here. So this means that V/S isn't necessarily finite. "modding out" the subspace of sequences that consist of zero in every coordinate do not change anything. Right? – Calculus08 Feb 4 '14 at 14:48

## 2 Answers

Let $a_1,\cdots, a_n, \cdots$ is a basis for $V$. Let $S$ be the vector subspace spanned by $a_2, \cdots, a_{2k}, \cdots$, then $V/S$ is spanned by $a_1+S, \cdots, a_{2k+1}+S. \cdots,$, which is infinite dimension.

Consider the space $V$ of all functions $\Bbb R\to\Bbb R$ as a vector space over $\Bbb R.$ The set $P$ of all polynomial functions is an infinite-dimensional subspace, but (for example) consider the set $F$ of functions $f:\Bbb R\to\Bbb R$ such that $\{x\in\Bbb R:f(x)\ne 0\}$ is finite. Then $F$ is again infinite-dimensional, and can readily be shown to be isomorphic to its image modulo $P,$ so $V/P$ is again infinite-dimensional.