Existence of Continuous bijective function

Question

I was looking for examples of bijective continuous functions infollowing cases

1. Does there exist Bijective continuous function $(0,1) \rightarrow \mathbb{R}$?

Yes, $f(x)=\frac{2x-1}{x-x^2}$

2. Does there exist Bijective continuous function from $\mathbb{R} \rightarrow (0,1)$ ?

Yes, as $(0,1) and \mathbb{R}$ are homeomorphic so inverse of $f(x)$ defined above will be example such a function.

3. Does there exist a bijective continuous function from $[a,b] \rightarrow \mathbb{R}$?

I think No, as continuous image of compact set is not compact.

4. Does there exist a bijective continuous function from $\mathbb{R} \rightarrow [a,b])$ ?

For this clearly it can be seen

a) continuous image of connected set is connected.

b) Inverse image of closed set under continuous map is closed.

So there exist such a function but i can not construct it.

5. Does Does there exist a bijective continuous function from $ (a,b) \rightarrow [a,b])$?

No. as inverse image of closed set is not closed so there can not exist any continuous bijective function.

But as $(a,b)$ is homeomorphic to $\mathbb{R}$So if we construct a function $\mathbb{R \rightarrow} [a,b]$ will that function work as an example of a function $ (a,b) \rightarrow [a,b])$?

6. Does Does there exist a bijective continuous function from $ [a,b] \rightarrow (a,b))$?

No. as inverse image of open set is not open so there can not exist any continuous bijective function.

Please correct me if I am wrong anywhere .And give examples of $(4)$ if any exists. Thanx very much for your time

Answer 1

4. No. Suppose there exists such a function $f$. Say $f(x_0) = a$, then $f(x) > a$ for both $x<x_0$ and $x> x_0$, in contradiction with $f$ being a bijection.

5 and 6. Your arguments are invalid. Given continuous function $f:X\to Y$, the pre-image of an open(closed) set in $Y$ under $f$ is open(closed) in $X$. Here $(a,b)$ is indeed closed in $(a,b)$, and $[a,b]$ likewise. That no continuous function maps $[a,b]$ onto $(a,b)$ can be easily seen by noticing that $[a,b]$ is compact while $(a,b)$ isn't. That no continuous bijection maps $(a,b)$ to $[a,b]$ can be reasoned similarly as in 4.

Also be aware that in general a continuous bijection doesn't have to be a homeomorphism, so you probably want to refine your arguments in 2. That said, we do have invariance of domain theorem in Euclidean spaces.

Answer 2

Your proposed $f$ for (1) is not a bijection. One that is:

$f(x)=\tan(\pi x-\frac{\pi}{2})$

Crucially, any such $f$ must be monotone (in-fact it cannot have derivative 0 over any interval, only at single points) (why?)

For (2), just invert (the corrected) function for #1.

(3) is no because $[a,b]$ is compact, but $\mathbb{R}$ is not (cannot satisfy surjectivity and continuity at the same time for reason you gave).

(4) is also no, but the reasoning is slightly different (see @QiyuWen's answer).

(5) You're reasoning is wrong, it's for the same reason as in (4) (this has to do w/ monotonicity again).

(6) Appeal to compactness again ($(a,b)$ is not compact; why?).