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