Is it true that for any metrics, $f(\lambda)=\lambda x+(1-\lambda) y$ is continuous where $\lambda \in \mathbb{R}, x,y \in \mathbb{R}^n$?

Question

Let's say we only know $\mathbb{R},\mathbb{R}^n$ are metric space but not sure what the metrics is. Intuitively I can see how this may be true. But I find it very frustrating to write it down rigorously so I doubt some further information is required . Specifically, this step is where I'm having trouble with: Take a point $a:=\lambda_a x+(1-\lambda_a) y$ on the image. Then for all points $b:=\lambda_b x+(1-\lambda_b)y $ such that $d(a,b)<\epsilon \iff d(\lambda_a x+(1-\lambda_a) y,\lambda_b x+(1-\lambda_b) y) <\epsilon$, I'm now supposed to show that it is possible to find some $w>0$ such that $(\lambda_a-w,\lambda_a+w)$'s image is contained in $d(a,b)<\epsilon$. But now even though I feel this is true for $w$ very small but I can't write anything without further specification of the metrics involved.

Answer 1

Here's a partial answer which is too big to fit in the comments:

Let's consider $f_{(x,y)}(\lambda): \mathbb{R} \to \mathbb{R}^n$ for a fixed pair $(x,y)\in\mathbb{R}^n\times\mathbb{R}^n$.

If the metric $d$ on $\mathbb{R}^n$ comes from a norm and $\mathbb{R}$ is equipped with the Euclidean metric, then we have

$$d(\lambda_a x+(1-\lambda_a)y, \lambda_b x+(1-\lambda_b) y)=\| (\lambda_a-\lambda_b)x+(\lambda_b-\lambda_a)y\|$$ By the triangle inequality and the fact that we're dealing with a norm now

$$\| (\lambda_a-\lambda_b)x+(\lambda_b-\lambda_a)y\|\leq |\lambda_a-\lambda_b|(\|x\|+\|y\|)$$

So, obviously in this case $\delta < \frac{\epsilon}{\|x\|+\|y\|}$ works.

Now consider $\mathbb{R}^n$ with the discrete topology but keep $\mathbb{R}$ as it is. If $x$ and $y$ are linearly independent, then $\lambda_a x+(1-\lambda_a)y=\lambda_b x+(1-\lambda_b)y$ if and only if $\lambda_a = \lambda_b$. Now consider $\lambda_b = \lambda_a + \frac{\delta}{2}$.

Then we get two different points on the line joining $x$ and $y$. So, the definition of continuity fails for any $\epsilon<1$. Hence, $f$ is discontinuous.