Convergence of idempotents preserve dimension of range Suppose that $H$ is a Hilbert space and that $\{P_n\}_n$ is a sequence of bounded idempotents such that $\|P_n-P\|\rightarrow 0$ (where $P$ is another bounded idempotent, however I guess this follows from convergence). Does it then hold that $\dim\mathrm{ran}~P_n\rightarrow \dim\mathrm{ran}~P$? With this I mean: does there exist some $N\in \mathbb{N}$ such that $n\geq N$ implies that $\dim \mathrm{ran}~P_n =\dim \mathrm{ran}~P$ in the sense of cardinals?
Here $\mathrm{ran}$ denotes the range of the operator.
For orthogonal projections this is true (see for instance Theorem 4.35 in Linear Operators in Hilbert Spaces by J. Weidmann) so adding the condition that all operators are self-adjoint would imply the convergence however what I wonder is if this holds also for bounded idempotents.
 A: One can prove the following: if $P,Q\in B(H)$ are idempotents such that $\|P-Q\|<1$, then their ranges have the same dimension. Since from $P_n\to P$ we get that there exists an $n_0$ such that $\|P_n-P\|<1$ for all $n\geq n_0$, it follows that  $\dim\operatorname{ran}(P_n)=\dim\operatorname{ran}(P)$ for all $n\geq n_0$. 
To prove it, let $x_1,\ldots,x_m\in PH$ be linearly independent. Suppose that $\sum_j\alpha_jQx_j=0$. Then 
$$
\|\sum_j\alpha_jx_j\|=\|\sum_j\alpha_jPx_j-\sum_j\alpha_jQx_j\|\leq\|P-Q\|\,\|\sum_j\alpha_jx_j\|. 
$$
Since $\|P-Q\|<1$, the strict inequality is impossible, so we conclude that $\sum_j\alpha x_j=0$, and then by linearly independence $\alpha_1=\ldots=\alpha_m=0$. Thus $Qx_1,\ldots,Qx_m$ are linearly independent. 
The roles can be reversed, so for any linearly independent subset of $PH$ there is a linearly independent subset of $QH$ with the same cardinality. Thus $\dim PH=\dim QH$. 
And, as a consequence, it follows that if $\|P-Q\|<1$ then $P$ and $Q$ are unitarily equivalent. 
