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Let $V$ to be an infinite dimensional linear space over some field $k$.

(you can take $k=\mathbb{C}$, or further assume $V$ is a complex Hilbert space).

And assume $W$ is a finite dimensional subspace of $V$, 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?


Remarks:

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.

3, If $V, V_i's$ are all Hilbert spaces, we think $P_W$ as the orthogonal projection.

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Thanks, I fix the typo. You are right, the hamel basis in uncountable for a countable Hilbert space. –  ougao Sep 4 '13 at 3:18
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up vote 2 down vote accepted

Let $(e_1,e_2,\ldots)$ be an orthonormal basis for a Hilbert space $V$, let $V_k=\{e_1,e_2,\ldots,e_k\}^\perp$, and let $W$ be the span of $\sum\limits_{n=1}^\infty\dfrac{1}{n}e_n$. If $P_W$ is orthogonal projection onto $W$, then $P_W(V_k)=W$ for all $k$, but $V_\infty=\{0\}$.

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You mean $W$ is a one dimensional subspace? But why $P_W(V_k)=W$? –  ougao Sep 4 '13 at 3:32
    
Oh, I see, thanks! –  ougao Sep 4 '13 at 3:34
    
Do you have any counterexample when $V_{\infty}$ is not zero? –  ougao Sep 4 '13 at 3:43
    
@ougao: Sure, let $W$ be the span of $\left\{e_1,\sum\limits_{n=1}^\infty\dfrac{1}{n}e_n\right\}$, and $V_k=\{e_2,e_3,\ldots,e_k\}^\perp$. –  Jonas Meyer Sep 4 '13 at 3:55
    
You are right, nice, I fail to realize we can add $e_1$ to $W$ if we add $e_1$ to all $V_i's$. –  ougao Sep 4 '13 at 4:00
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