Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Supposing, $\{V_t\}, t > 0$ are an uncountable number of linear subspaces of $\mathbb{R}^n$. If $\bigcup_{t>0} V_t = \mathbb{R}^n$ is it true that $V_T = \mathbb{R}^n$ for some $T>0$?

Any help is appreciated. Thanks.

EDIT: I have forgot to add the condition that $V_t$ are increasing.

share|improve this question
Are your $V_t$ supposed to form a chain of linear subspaces? Otherwise $\bigcup_t V_t$ will not be a linear subspace in general... –  Zhen Lin Feb 3 '12 at 7:35
@Zhen: I am not sure what a chain of subspaces means, but I realize I completely forgot about the condition that $V_t$ have to be increasing. –  jpv Feb 3 '12 at 7:41
That is what is meant by (ascending) chain. This is very important as it changes the answer from no to yes! –  Zhen Lin Feb 3 '12 at 7:47
If $V_s$ and $V_t$ have the same dimension and $V_s \subseteq V_t$, then they are the same. There are only $n+1$ possible dimensions for a subspace of ${\mathbb R}^n$, so there are at most $n+1$ different subspaces in your chain, and the union is the one of largest dimension. If this is ${\mathbb R}^n$, it means one of the subspaces is ${\mathbb R}^n$. –  Robert Israel Feb 3 '12 at 8:35

3 Answers 3

Let $n=2$. Let $\{V_t\}$ be the set of all lines through the origin.

If you really want to index these lines with the positive reals, find a one-to-one correspondence between the set of all reals, plus the symbol $\infty$, and the set of all positive reals.

Clearly the union of the $V_t$ is $\mathbb{R}^2$, and none of the $V_t$ is $\mathbb{R^2}$.

Exactly the same example works for $\mathbb{R}^n$ for any $n\ge 2$.

The situation is very different if the $V_t$ are nested, that is, if $s<t$ implies that $V_s \subseteq V_t$. For then by some finite $t$, we will have $V_t=\mathbb{R}^n$. The argument is simple, and has nothing much to do with uncountability. There must be some integer $n_1$ such that $V_{n_1}$ has dimension $\ge 1$. But then there must be an $n_2>n_1$ such that $V_{n_2}$ has dimension $\ge 2$. And so on. Sooner or later, we must reach an integer $n_k$ such that $V_{n_k}=\mathbb{R}^n$.

share|improve this answer
Thanks for your answer. Yes, I forgot to add that $V_t$ are increasing. –  jpv Feb 3 '12 at 7:49
Thanks again! I have understood the argument. –  jpv Feb 3 '12 at 7:54
Alternatively and equivalently, if $\{e_1,\dots,e_n\}$ is a basis, there exist $t_1$, $\dots$, $t_n>0$ such that $e_i\in V_{t_i}$ for each $i$. But then $\mathbb R^n=V_{\max\{t_1,\dots,t_n\}}$. –  Mariano Suárez-Alvarez Feb 3 '12 at 7:56
@jpv: I added something to my post. We need only assume that if $s<t$ then $V_s\subseteq V_t$, sort of non-dcreasing rather than increasing, so a weaker condition than increasing. You certainly cannot have a strictly increasing sequence indexed by the positive reals, any strictly increasing sequence of subspaces is finite. –  André Nicolas Feb 3 '12 at 7:56

In general the answer is no. Consider family of subspaces of the form $$ V_t=\{x\in\mathbb{R}^n:x_1\cos\frac{2\pi}{t+1}+x_2\sin\frac{2\pi}{t+1}=0\} $$

share|improve this answer
Notice that your first example is the same as Nicolás's. Your second statement is strange, because the union of those spaces in not $\mathbb R^n$. –  Mariano Suárez-Alvarez Feb 3 '12 at 7:52
Oh, I forgot the second condition. –  userNaN Feb 3 '12 at 7:56

Consider a bijection $\phi:(0,+\infty)\to\mathbb R^n$ and for each $t>0$ let $V_t$ be the subspace of $\mathbb R^n$ spanned by $\phi(t)$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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