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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 $F$?

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    $\begingroup$ 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
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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).

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