Projection operator in Hilbert space

Let $H$ be a Hilbert space, can we find an increasing net of finite rank projections which converge to the identity in the strong operator topology?

And I think if $H$ is separable, we can find an orthonormal basis $\{v_{n}\}_{n=1}^{\infty}$ and let $p_{n}$ be the projection (from $H$) onto span$\{v_{1}, ..., v_{n}\}$. Then, we can verify $||p_{n}x-x||\rightarrow 0$ for all $x\in H$. But what about the case that $H$ is not separable?

You always can, even if $H$ is not separable. Let $\{e_j\}_{j\in J}$ be an orthonormal basis of $H$. Now let $\mathcal F=\{F\subset J:\ |F|<\infty\}$, ordered by inclusion. This order is not total, but that is not relevant.
Now consider the net $\{P_F\}_{F\in\mathcal F}$, where $P_F$ is the orthogonal projection onto the span of $\{e_j:\ j\in F\}$.
Convergence in the strong operator topology is pointwise convergence. So fix $h\in H$ and $\varepsilon> 0$. Then $h=\sum_jc_je_j$, where the limit is taken along the net $\mathcal F$ and there exists $F_0\in\mathcal F$ such that $\|h-\sum_{j\in F_0}c_je_j\|<\varepsilon$. But this is $\|h-P_{F_0}h\|<\varepsilon$. For any $F\in\mathcal F$ such that $F\geq F_0$, we have $\|h-P_Fh\|\leq\|h-P_{F_0}h\|<\varepsilon$. So $$\lim_{\mathcal F}P_Fh=h,$$ which implies that $$\lim_{\mathcal F}P_F=I$$ in the strong operator topology.
• Is there anything wrong in my viewpoint if $H$ is a separable Hilbert space? And in your answer, what do you mean by $F \geq \{h\}$ and then how can you get $P_{F}h=h$? – Rancho Mar 18 '14 at 11:36
• No, nothing wrong. The set $\mathcal F$ is ordered by inclusion, so $F\geq \{h\}$ means $h\in F$. The equality is not such, I'm editing the answer right now. – Martin Argerami Mar 18 '14 at 11:47
• Why does $||\sum_{j\notin F_{0}}c_{j}e_{j}||<\varepsilon$? – Rancho Mar 19 '14 at 1:51
• It's the definition of convergence. In the separable case you would write $\|\sum_{j\geq j_0}c_je_j\|<\varepsilon$. – Martin Argerami Mar 19 '14 at 1:54