Intuition on why there can't be a continuous bijection between $(a,b)$ and $[c,d]$?

Question

Intuition on why there can't be a continuous bijection between $(a,b)$ and $[c,d]$?

I'm not (necessarily) looking for a proof for this, I want to understand why does this happen, intuitively: if I add uncountably many points to $(a,b)$ taking the set $(a,b+1)$ I can find a continuous bijection, but if I only add the extremes, this is impossible, why?

Answer 1

Suppose such a function exists with $x\in (a,b)$ such that $f(x)=d$. Pick points $y,z\in(a,b)$ with $y<x$ and $z>x$. We know such points exist because $x$ is an interior point of $(a,b)$. Without loss of generality, suppose $f(y)<f(z)$. We know that both $f(y)$ and $f(z)$ are less than $d$. By the intermediate value theorem, there exists $u\in[y,x]$ such that $f(u)=f(z)$. This is a contradiction to the hypothesis that $f$ is injective, so $f$ cannot exist.

Answer 2

Any continuous injective map on an interval must be strictly monotone. Let's assume that the map $f:(a, b) \to [a, b]$ is bijective and strictly increasing. Then there must be a point $c \in (a, b)$ that satisfies $f(c) = a$. But since $c > a$, all the points in the interval $(a, c)$ must be mapped to a value that is strictly less than $a$. This is impossible, since $a$ is already the smallest number in the range of $f$.

The case that $f$ is strictly decreasing works in a similar fashion by considering the interval $(c, b)$.