I was reading the proof of the following theorem in Hatcher's Algebraic Topology:
Theorem 2B.5. $\Bbb R$ and $\Bbb C$ are the only finite-dimensional divison algebra structure over $\Bbb R$ and have an identity.
In the proof, there is a statement that:
Suppose first that $\Bbb R^n$ has a commutative division algebra structure. Since the multiplication map $\Bbb R^n \times \Bbb R^n \to \Bbb R^n$ is bilinear, it is continuous.
I know that the multiplication map is by definition bilinear. However, I can't understand why this implies continuity (of course in the Euclidean topology), so I need some help. Thanks
Consider a bilinear map $\phi:\mathbb{R}^n\times \mathbb{R}^n\rightarrow \mathbb{R}^n$. Note that $\phi = (\phi_1,...,\phi_n)$, where each $\phi_i:\mathbb{R}^n\times \mathbb{R}^n\rightarrow \mathbb{R}$ is a bilinear form. Thus it suffices to show that every bilinear form $\phi:\mathbb{R}^n\times \mathbb{R}^n\rightarrow \mathbb{R}$ is continuous. For this note that $$\phi(v,w) = v^T\cdot A\cdot w$$ for some matrix $A = [a_{ij}]_{1\leq i,j\leq n}$. Hence $$\phi\left((v_i)_{1\leq i\leq n},(w_i)_{1\leq i\leq n}\right) = \sum_{i=1}^n\sum_{j=1}^nv_i\cdot a_{ij}\cdot w_j$$ and this is written in terms of the multiplication of arguments and addition of the results of this multiplications so it is clear that it is continuous.