Consider $i:[a,b]\to \mathbb{R}$, defined in the simplest manner possible: $x\mapsto i(x)=x\in\mathbb{R}$. My first question would be: is this continuous? Studying topology, I realized that as it stands, this question makes no sense: in actual fact, I need to specify a topology on $[a,b]$ to be able to answer the continuity question.
Taking advantage of that, I induce a topology on $[a,b]$ such that this natural embedding is in fact continuous. Indeed $\mathcal{O}_{[a,b]}=\{[a,b]\cap O \ | \ O \text{open} \in \mathbb{R} \}$ will render $i$ continuous.
The real question I want to understand is this: can I add elements of $[a,b]$, which is evidently not a vector space? I would imagine that for $x,y\in [a,b]$, I could define $x+y:=i^{-1}(i(x)+i(y))$. Is this a feasible way of inducing a vector space structure on $[a,b]$? Of course it depends on the embedding and is not "natural". So is $[a,b]$ an embedding dependent vector space?
If someone had chosen a different embedding $j$, what would happen with the addition? Would I be correct in assuming that $$ j(x+y)=j\circ i^{-1}(i(x)+i(y)) = j(x)+j(y), $$ where I have used linearity (?) of $j\circ i^{-1}$, and hence $x+y=j^{-1}(j(x)+j(y))$, so the addition is embedding independent?
Am I right in assuming that $j\circ i^{-1}$ is linear?
Is there any contact with affine spaces and the above picture? Because for $f:(a,b)\to \mathbb{R}$, the derivative is defined via $\left.\dfrac{d}{d\epsilon}\right|_{\epsilon=0}f(x+\epsilon)$.
I want to make sense of $x+\epsilon$ for $(a,b)$ which is not a vector space.