I undestand that the EVT states that if we have an interval like $[a, b]$ then if there is a continuous function $f$ defined over the interval then there are such $x_1, x_2\in [a, b]$ that $f(x_1) = \sup_{[a, b]} f$ and $f(x_2) = \inf_{[a, b]} f$ (in other words – the functions attains its extremes). So as I understand this fact implies that there can be no continuous (bijective?) function that'd transform eg. the interval [0, 1] into the $\mathbb R$ number line.
So my question is about sets like $[a, b)$. The set is not compact therefore the EVT does not apply yet it is somehow similar... Can we say that if we have a continuous $f:[a, b) \to \mathbb R$ then for some value(s) $f$ attains its maximum/minimum/both (but not none of those as with $(a, b)$)?
Hence there is no continuous bijection from eg. $[0, 1)$ to $\mathbb R$? If there is then how come? To me it seems that either we have the function bounded from either above or below by the $f(0)$ value and approaching $\pm\infty$ on the second end, or the function cannot be defined for every point on the interval if we want it to ,,produce'' both $+\infty$ and $-\infty$ in a continuous way (easier to draw than describe).
To generalise the examples: one can easily create a continuous bijection form $(a, b)$ to $\mathbb R$ whereas it is impossible with $[a, b]$ but how about those half-closed intervals? Does the EVT extend to those somehow?
Could someone clarify whether I understand this correctly?
If $f:[a,b)\longrightarrow\mathbb{R}$ is continuous and injective then it's either monotonic increasing or decreasing (otherwise there would exist $x,y,z\in[a,b)$ such that $x<y<z$ and $f(x)<f(y)$ and $f(z)<f(y)$, which would lead by mean value theorem to $c\in(x,z)$ with $f(c)=\max\{f(x),f(z)\}\in(\min\{f(x),f(z)\},f(y))$).
If $f$ is monotonic increasing then $f(a)\leq f(x)$ for all $x\in[a,b)$, so $f([a,b))\neq \mathbb{R}$ and $f$ is not surjective, hence not bijective. Same for $f$ decreasing.
If you don task $f$ to be injective you can actually do that kind of map, for example, with the function: $$x\in(0,1]\longmapsto f(x):=\dfrac{\sin\left(\dfrac{1}{x}\right)}{x}$$