## 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 $$d(x_{n},y_{n}) \to d(x,y).$$