If $X$ is a vector space endowed with a metric that is NOT norm induced and If $x_n\to x$ and $y_n\to y$, then $x_n+y_n\to x+y$


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). $ $