# Union of uncountably many subspaces

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.

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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

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 $V_{n_k}=\mathbb{R}^n$.

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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 $$V_t=\{x\in\mathbb{R}^n:x_1\cos\frac{2\pi}{t+1}+x_2\sin\frac{2\pi}{t+1}=0\}$$

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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. – Norbert 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 $\phi(t)$.

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