I'm trying to show that $\langle , \rangle$ is continuous on $V{\times}V$, ($V$ an inner product space). I've tried approaching it by showing $\langle\vec x,\vec y\rangle\rightarrow\langle\vec a,\vec b\rangle$ whenever $(\vec x,\vec y)\rightarrow(\vec a,\vec b)$ in $V{\times}V$. But the usual epsilon delta definition of the limit seems a little messy and I'm not sure it's going anywhere. Some input would be appreciated. Thanks.
Edit:
$\lim_{(x,y) \rightarrow (a,b)}\langle x,y\rangle = \lim_{(x,y) \rightarrow (a,b)} x^T y = a^T b = \langle a,b\rangle$
Where substitution in the third step comes from the combination theorem for limits. Is this sufficient or can I elaborate somewhere? Thanks again.
You need to do the exact same proof that you would do to show that the product of continuous functions is continuous.
Edit: in your edit, you are just writing the inner product in a different way, which even makes no sense in general (what's $x^T$ for $x\in V$?)
Here is what I meant: $$ |\langle x,y\rangle-\langle a,b\rangle|=\langle x,y\rangle-\langle x,b\rangle +\langle x,b\rangle -\langle a,b\rangle|=|\langle x,y-b\rangle| +|\langle x-a,b\rangle|\\ \leq \|x\|\,\|y-b\|+\|b\|\,\|x-a\|. $$ One would still need to bound $x$. But that is not an issue because for continuity we would consider $(x,y)$ in a ball around $(a,b)$.