Let $X$ be a real or complex vector space endowed with a metric $d$ which is not induced by a norm, that is there exist no norm $\| \cdot \|$ on $X$ such that $d(x,y)=\| x-y \|$. Prove or disprove:
(a) If $x_n\to x$ and $\lambda_n\to \lambda$ ($\lambda_n,\lambda$ are scalars), then $\lambda_n x_n \to \lambda x$.
(b) If $x_n\to x$ and $y_n\to y$, then $x_n+y_n\to x+y$.
This is throwing me off just because the vector space is endowed with a metric that is not norm induced, and I am not sure what all that implies. With that said, I think I may be doing this problem naively. Anyway, here are my solutions:
(a) Let $y\in X$ be fixed. Then, by the triangle inequality, $$d(\lambda_nx_n,y) \leq d(\lambda_nx_n,\lambda x) + d(\lambda x, y),$$ similarly, $$d(\lambda x,y)\leq d(\lambda x,\lambda_n x_n) + d(\lambda_n x_n , y),$$ which implies that $$d(\lambda x,y) - d(\lambda_n x_n , y) \leq d(\lambda x,\lambda_n x_n),$$ so that as $n\to\infty$ we have $\lambda_n x_n \to \lambda x$.
(b) We may establish an upper bound via the triangle inequality, so that $$ d(x_{n},y_{n}) \leq d(x_{n},x) + d(x,y) + d(y,y_{n}). $$ Again for a lower bound, we use the triangle inequality: $$ d(x,y) \leq d(x_{n},x) + d(x_{n},y_{n}) + d(y_{n},y); $$ hence we have that, $$ d(x,y) - d(x_{n},x) - d(y_{n},y) \leq d(x_{n},y_{n}). $$ Now as $n \to \infty$ we have $$ d(x_{n},y_{n}) \to d(x,y). $$