# Infinite dimension vector space decomposes into countable union of subspaces

The following problem is from Golan's linear algebra book. I have posted a solution in the answers.

Problem: Let $V$ be a vector space over a field $F$ having infinite dimension over $F$. Show there exists a countably-infinite collection of proper subspaces of V, the union of which equals $V$.

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By a standard theorem, there is a basis $B=\{b_\alpha\}$. Enumerate a countable subset of this and call it $\{b_i\}$. Let $F(D)$ denote all linear combinations of $D$ over the field $F$, for an arbitrary set of vectors $D$. Now consider the subspaces given by

$W_1=F(B-b_1)\\ W_2=F(B-b_2)\\ W_3=F(B-b_3)\\...$

There are a countable number of these, and it remains to show their union is all of $V$. But any $v\in V$ can be written as a finite linear combination of elements in $B$. Let the set of elements in $B$ used in this combination be $B_0$. Because $B_0$ is finite, there exists $b_i\in B$ with $b_i\notin B_0$. (If $B_0$ contained all of the $b_i$, it would be infinite.) Then $v\in W_i$, and we are done.

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Looks OK to me! –  rschwieb Jun 1 '12 at 0:14
What is $D{}{}$? –  Gerry Myerson Jun 1 '12 at 0:23
A typo. It is now fixed. –  Potato Jun 1 '12 at 0:24
That works. Alternatively, to express it as a countable ascending union, pick an embedding $f\colon \mathcal{N}\to B$, let $B'=B\setminus f(\mathbb{N})$, and then let $W_i = \mathrm{span}(B'\cup f(\{1,\ldots,n\}))$. –  Arturo Magidin Jun 1 '12 at 1:04