Prove that for any $S \subset \mathbb{R}^n$ and functions $f,g : S \rightarrow \mathbb{R}^m$ $f$ is homotopic to $g$. Prove that for any $S \subset \mathbb{R}^n$ and functions $f,g : S \rightarrow \mathbb{R}^m$ we have $f \sim g$. (Here by $f \sim g$ I mean f is homotopic to $g$).
Here is my attempt at a proof:
Let $h : S \times I \rightarrow \mathbb{R}^m$ be given by,
$$h(x,t)=(1-t)f(x)+tg(x)$$
Then since $h(x,0)=(1-0)f(x)+0\cdot g(x)=f(x)$ and $h(x,1)=(1-1)f(x)+1 \cdot g(x)=g(x)$ the function $h$ is a homotopy between $f$ and $g$ for any $f,g : S \rightarrow \mathbb{R}^m$.
I have two questions about this proof:

*

*Is the above function $h$ in fact a homotopy between $f$ and $g$? I thought that only two continuous functions can be homotopic to each other, but $f$ and $g$ are not defined to be continuous.

*Is there any necessity for this problem only to refer to functions between two Euclidean spaces?


$(1)$: In algebraic topology, maps are often assumed to be continuous, without explicitly mentioning so.
$(2)$: You are using the convexity of $\mathbb R^n$ for $n\ge 1$. So, you can use the same argument for any convex space.
