Uniform limit of finite-rank operators with the same rank. Let $\{T_n\in\mathcal{B}(X)\,|\,\text{rank}(T_n)=R\,\}^{\infty}_{n=1}$ is a sequence of linear bounded finite-rank operators on a Banach space with the same rank $R$. Let it converge uniformly to an operator $T\in\mathcal{B}(X)$, that is $\Vert T_n - T\Vert_{\mathcal{B}(X)}\longrightarrow 0\;(n \rightarrow\infty)$, that is $\forall\epsilon>0\;\exists N\in\mathbb N\;\forall n\ge N\;\forall x\in X\;|\;\Vert x\Vert_{X}\le 1$ holds $\Vert T_n x - T x\Vert_X\le\epsilon$.
Prove that $\text{rank}(T)\le R$.
 A: I'll assume we are working over a real Banach space; the complex case is similar.
Suppose $x_1, \dots, x_{R+1} \in X$.  We will show that $Tx_1, \dots, Tx_{R+1}$ are linearly dependent.
For each $n$, the rank of $T_n$ is $R$, and so $T_n x_1, \dots, T_n x_{R+1}$ are linearly dependent.  Thus there are coefficients $a_{n,1}, \dots, a_{n,R+1} \in \mathbb{R}$, not all zero, such that $\sum_{i=1}^{R+1} a_{n,i} T_n x_i = 0$.  By rescaling, we may assume that $\sum_{i=1}^{R+1} |a_{n,i}|^2 = 1$, i.e. $(a_{n,1}, \dots, a_{n,R+1})$ lies in the unit sphere $S^R$ of $\mathbb{R}^{R+1}$.  Now $S^R$ is compact, so we can extract a subsequence $(a_{n_k, 1}, \dots, a_{n_k, R+1})$ converging to some $(a_1, \dots, a_{R+1}) \in S$.  In particular, for each $i$ we have $a_{n_k, i} \to a_i$, and the $a_i$ are not all zero.  But since $T_{n_k} \to T$ uniformly, we have $$\sum_{i=1}^{R+1} a_i T x_i = \lim_{k \to \infty} \sum_{i=1}^{R+1} a_{n_k, i} T_{n_k} x_i = 0. \quad (\ast)$$
Note it was sufficient that $\operatorname{rank}(T_n) \le R$ for all but finitely many $n$, so we have in fact shown that rank is lower semicontinuous with respect to the uniform topology on $\mathcal{B}(X)$.  
The same argument goes through if $T_n \to T$ in the strong operator topology (SOT), or even the weak operator topology (WOT, intepreting the limit in (*) as a weak limit in $X$).  We could also replace the sequences with nets to get the strongest statement: rank is WOT-lower semicontinuous on $\mathcal{B}(X)$.
