I am trying to show that the singular value decomposition is continuous. For a matrix $A$ of size $m\times n$, let $s = \min\{m,n\} =n$, i.e, $(m \ge n)$. Let $A = U_A\Sigma_A V_A^T$ be a SVD of $A$ and it is unique up to a sign change of both columns of $U_A$ and $V_A$.
First, let us choose any fixed $U_A$ and $V_A$. Let $(U_A^{(1)},V_A^{(1)}) = (U_A,V_A)$. Then, let us define $f_k:\mathbb{R}^{m\times n} \mapsto \mathbb{R}^{m\times n} \times \mathbb{R}^{n \times n} \times \mathbb{R}^{n \times n}$ such that $$ f_{k}(A) = [U_A^{(k)},\Sigma_A,V_A^{(k)}], $$ where $A = U_A^{(k)}\Sigma_A(V_A^{(k)})^T$ is a singular value decomposition of $A$ and $(U_A^{(k)},V_A^{(k)})$ represents the $k$-th sign change of $(U_A,V_A)$, assuming a certain order.
I would like to show the (upto sign) continuity of SVD. To do this, I want to show that for any $\epsilon >0$, there exists $\delta > 0$ such that for any $B$ satisfying $\|A-B\|_F < \delta$, we have $$ \min_{k,j}\|f_k(A) - f_j(B)\|_F = \sqrt{\|U_A^{(k)}-U_B^{(j)}\|_F^2 + \|\Sigma_A-\Sigma_B\|_F^2 + \|V_A^{(k)}-V_B^{(j)}\|_F^2} < \epsilon. $$
What I can show is that if $A$ and $B$ have the same singular vectors, i.e., $U_A = U_B$ and $V_A = V_B$, the continuity follows immediately, as $\min_{k,j}\|f_k(A) - f_j(B)\|_F = \|A-B\|_F$. However, not sure how to proceed when $U_A\ne U_B$ and $V_A\ne V_B$.
Any comments/suggestions/answers would be very appreciated.