Continuous QR decomposition Let $F_n \rightarrow_{n\rightarrow \infty} 0$ and  $I+F_n=Q_nR_n$ be the QR decomposition with $r_{ii}>0$.
Then $Q_n \rightarrow I$. Why do we need the continuity of the mapping $A=QR\rightarrow Q$ for this conclusion?
 A: It is enough to show that any subsequence $(Q_{n_{k}})_{k}$ has a subsequence $(Q_{n_{k_{j}}})_{j}$ such that $Q_{n_{k_{j}}}\longrightarrow I$.
Since the columns of $Q_{n_{k}}$ are orthonormal for all $n$, then the sequence $(Q_{n_{k}})_{k\in\mathbb{N}}$ is bounded (you can consider any norm for our purposes, because all norms on a finite-dimensional space are equivalent). By compactness (of the closed balls in finite-dimensional normed spaces), there exists a convergent subsequence $(Q_{n_{k_{j}}})_{j}$, let's say, $Q_{n_{k_{j}}}\longrightarrow Q$. Therefore, $Q_{n_{k_{j}}}^{t}\longrightarrow Q^{t}$ (we are taking transposes)
Therefore $R_{n_{k_{j}}}=\underbrace{Q_{n_{k_{j}}}^{t}Q_{n_{k_{j}}}}_{=I}R_{n_{k_{j}}}=Q_{n_{k_{j}}}^{t}(I+F_{n_{k_{j}}})\longrightarrow Q^{t}$ as $j\rightarrow +\infty$
Hence $Q^{t}$ is upper triangular, and its diagonal consists of non-negative elements. But $Q^{t}$ is invertible (we are working with square matrices), because $I=Q_{n_{k_{j}}}^{t}Q_{n_{k_{j}}}\longrightarrow Q^{t}Q$, so $Q^{t}Q=I$. Hence, all the elements of the diagonal of $Q^{t}$ must be positive.
We have: $Q^{t}$ is upper triangular, all its diagonal consists of positive elements, and its rows are orthonormal. This implies that $Q^{t}=I$, and hence $Q=I$
