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Let $V= \mathbb{Q}^{\mathbb{Z}}$, considered as an infinite dimensional linear space over $\mathbb{Q}$.

And assume $W=\mathbb{Q}^F$ is a finite dimensional subspace of $V$, so $F$ is just a finite set of the integers, and denote the projection from $V$ to $W$ by $P_W$.

Let $V_1\supseteq V_2\supseteq V_3\supseteq\cdots$ be a decreasing sequence of subspaces of $V$, denote $V_{\infty}=\cap_{i=1}^{\infty}V_i$.

Since $\{P_W(V_i)\}_{i=1}^{\infty}$ is a decreasing sequence of subspaces of the finite dimensional space $W$, then it would be stable after some sufficient large $j$, i.e, $P_W(V_{j})=P_W(V_{j+1})=\cdots :=\lim_iP_W(V_i)$.

My question is:

$$\lim_iP_W(V_i)=P_W(V_{\infty})?$$ Any counterexamples?


1, Note that $P_W(V_i)\neq V_i\cap W$, $W\cap (W_1+W_2)\neq W\cap W_1+W\cap W_2$ in general for linear subspaces $W, W_1, W_2$.

2, The nontrivial case is all the $V_i$ have infinite dimension.

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up vote 2 down vote accepted

My question is: $$\lim_iP_W(V_i)=P_W(V_\infty)?$$

Not necessarily.

Any counterexamples?

Let $F = \{0\}$, let $b_k = e_k + e_0$, where the $e_k$ are the tuples containing a $1$ in the $k$-th component and $0$ everywhere else, and $V_n = \operatorname{span} \{ b_k : k \geqslant n\}$.

Then $P_W(V_i) = \mathbb{Q}^F$ for all $i \geqslant 1$, but

$$V_\infty = \bigcap_{i=1}^\infty V_i = \{0\}.$$

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What is $b_k$ used for? Isn't that $P_W(V_i)=0$ in your example? – ougao Sep 4 '13 at 17:25
Typo. It was supposed to be $V_n = \operatorname{span} \{ b_k : k \geqslant n\}$. Thanks for spotting it. – Daniel Fischer Sep 4 '13 at 17:29
I guess you mean $b_k$ instead of $e_k$ in your definition of $V_n$. That would be correct. – ougao Sep 4 '13 at 17:30
Exactly. I've already corrected it. – Daniel Fischer Sep 4 '13 at 17:31

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