Action of $O(n+1)$ on $S^n$, i.e., $\Phi:O(n) ~ \times ~S^n \to S^n$ defined as $~\Phi(A,x) := Ax$, i.e., action by matrix multiplication is continuous.
How do I show that this map is continuous? Since $O(n+1) \times S^n \subset \mathbb{R}^{(n+1)^2} \times \mathbb{R}^{n+1}$, can I think of applying the sequential criteria for continuity? If yes, then for any sequence $(A_n,x_n)_{n \in \mathbb{N}}$ converging to $(A,x)$, we have $(A_nx_n) \to Ax$. How do I show this?
This approach works fine. To show $A_mx_m\to Ax$, you can simply note that the components of the vector $A_mx_m$ are finite(!) sums of products of entries of $A_m$ and components of $x_m$. Concretely, if $A_m = (a_{ij}^{(m)})_{ij}$ and $x_m = (x_i^{(m)})_i$, then $A_mx_m$ is a vector with components $(\sum_{j=1}^{n+1} a_{ij}^{(m)}x_j^{(m)})_i$. Now take the limit as $m\to\infty$, and apply the assumption $A_m\to A$ and $x_m\to x$.