On the bottom of page 38 of Roman's Advanced Linear Algebra is written the following (here $V$ is a vector space over the field $F$ and $\mathcal{S}(V)$ is the set of linear subspaces of $V$):

"...if $S$, $T\in \mathcal{S}(V)$ (and $F$ is infinite), then $S \cup T\in \mathcal{S}(V)$ iff $S \subseteq T$ or $T \subseteq S$."

I cannot for the life of me figure out why the finiteness of $F$ is mentioned; it seems irrelevant. The proof of $\Leftarrow$ is obvious in any case, and the proof of $\Rightarrow$ can be done by contrapositive as follows: take $s \in S \setminus T$ and $t \in T \setminus S$ and note $s+t \in T \Rightarrow s\in T$ (contradiction) and similarly for $S$, so $s+t \notin S \cup T$.

How could this argument break down for finite $F$?

  • 3
    $\begingroup$ Looks OK. What does break down is that if $F$ is infinite, then $V$ js not a finite union of proper subspaces, while if $F$ is finite, $V$ can be such a finite union. $\endgroup$ – André Nicolas Apr 6 '13 at 20:15
  • $\begingroup$ Confirming André's comment. The assertion Roman makes here is independent of the size of $F$, and holds whenever $V$ is a module over any ring, not necessarily a vector space. (On the other hand, his Theorem 1.2 on the next page does require infiniteness of $F$.) $\endgroup$ – darij grinberg Apr 6 '13 at 21:48
  • $\begingroup$ OK. Suppose I attempt to prove that $V$ is not a finite union of proper subspaces $V_j \subset V$ as follows: The argument that I gave above shows that $V$ cannot be a union of two proper subspaces. Now suppose $V$ cannot be a union of $n-1$ proper subspaces. Take $n\in \mathbb{Z}^+$, and suppose for contradiction that $V_1 \cup \dotsb \cup V_n = V$ for $V_i \subset V$. Then $V_1 \cup \dotsb \cup V_{n-1}$ is a proper subspace of $V$ by the induction hypothesis, and so $V$ is a union of $(V_1 \cup \dotsb \cup V_{n-1}) \cup V_n$, i.e. a union of two proper subspaces, a contradiction. $\endgroup$ – Eric Auld Apr 6 '13 at 22:31
  • $\begingroup$ This proof seems not to require $F$ to be infinite. Where does it break down? $\endgroup$ – Eric Auld Apr 6 '13 at 22:32

As André Nicolas mentioned in the comments the argument you have also holds for finite $F$.

What does not hold anymore is that a vector space can be a finite union of proper subspaces, e.g. $$\mathbb{F}_2^2=\langle (1,0)\rangle \cup \langle (1,1)\rangle \cup \langle (0,1)\rangle.$$

What does break down in your induction argument in the comments is that the union of two of these subspaces is not a subspace anymore. Thus you can not apply induction hypothesis as you have done for $(V_1\cup \dots\cup V_{n-1})$ and $V_n$ (here $V_1\cup\dots\cup V_{n-1}$ is not a subspace).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.