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The corollary below is from Hoffman and Kunze's book, Linear Algebra.

Corollary. If $W$ is a $k$-dimensional subspace of an $n$-dimensional vector space $V$, then $W$ is the intersection of $(n-k)$ hyperspaces in $V$.

In the proof, they find $n-k$ linear functionals $f_{i}$ such that $W=\cap_{i=1}^{n-k}\ker (f_{i})$.

I want to know if the following is true: For all proper subspace $W$ of a infinite-dimensional vector space $V$ there are a set of linear functionals $\{f_{i}|i\in I\}$, where $I$ is a set of index such that $W=\cap_{i\in I}\ker (f_{i})$.

Thanks for your kindly help.

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hyperspaces should be hyperplanes all over the place, no? –  Mariano Suárez-Alvarez Feb 9 '12 at 20:38
    
@MarianoSuárez-Alvarez: Yes! –  spohreis Feb 10 '12 at 23:45

3 Answers 3

up vote 3 down vote accepted

Let $(w_i)_{i\in I}$ be a Hamel basis of $W$. Extended this basis to a basis of $V$, let us call it $(w_i)_{i\in I}\cup (v_j)_{j\in J}$. Now for each $j\in J$ defined $f_j(v_j)=1$ and zero elsewhere. It follows that $W=\bigcap_{j\in J} ker(f_j)$.

If you are interested in normed vector space then $W$ need to be a closed subspace and you can mimic the proof using Hahn-Banach.

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You mean, if $W$ is a closed subspace of a normed vector space, the functionals can be chosen to be continuous. –  Robert Israel Feb 9 '12 at 20:31

Yes. For every $x \notin W$, there is a linear functional $f_x$ on $V$ such that $f_x = 0$ on $W$ and $f_x(x) \ne 0$. Then $W = \bigcap_{x \notin W} \text{ker}(f_x)$.

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The linear functionals on $V$ which are zero on every element of $W$ are precisely the linear functionals on the quotient $V/W$. So the question reduces to the case that $W = 0$, which follows by the axiom of choice because $V/W$ has a basis.

Without the axiom of choice, the result can be false. For example, it's consistent with ZF that the vector space $\prod_{i=1}^{\infty} \mathbb{R} / \bigoplus_{i=1}^{\infty} \mathbb{R}$ has no nonzero linear functionals on it (see this MO question).

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