Let $a,b\in\mathbb{R}$ and $a<b$. Which of the following statement(s) is/are true?

Question

Let $a,b\in\mathbb{R}$ and $a<b$. Which of the following statement(s) is/are true?

1. There exists a continuous function $f:[a, b]\rightarrow (a,b)$ such that $f$ is one-one.
2. There exists a continuous function $f:[a,b]\rightarrow(a,b)$ such that $f$ is onto.
3. There exists a continuous function $f:(a, b)\rightarrow[a,b]$ such that $f$ is one-one.
4. There exists a continuous function $f:(a, b)\rightarrow[a,b]$ such that $f$ is onto.

I have found an example for option 4: take a compact subset $[c,d]$ of $[a,b]$, find a bijection $f : [c,d] \to [a,b]$, and then extend $f$ to have domain $(a, b)$.

Also 2 can not be correct, because a continuous function on a compact set has compact image.

How to eliminate the others?

Answer 1

1. Yes, can shrink the closed interval to fit in $(a,b)$.
2. As $f$ is continuous we must have $f([a,b])$ closed and bounded i.e compact.
3. Yes, the identity.
4. Yes, we can use the following continuous surjection (there are plenty of others, too) $(0,1) \to [0,1]$, $x\mapsto \frac{1}{2}\sin (4\pi x) + \frac{1}{2}$. Then a continuous surjection $$(a,b)\to (0,1) \to [0,1] \to [a,b] $$ is straightforward to find.

Answer 2

For 1) take any closed interval $[c,d]$ contained in $(a,b)$. Can you find a homeomorphism from $[a,b]$ onto $[c,d]$? (There is one of the type $f(x)=\alpha x+\beta$).

For 3) the identity function is an example.

Answer 3

Not if $f$ is not one-to-one then $f$ can "double up". If $f$ is not onto it doesn't have to "hit them all". Your argument that the image of a compact set must be compact is good... if the function isn't onto the codomain need not be the image!

1)$f:[a,b]\to (a,b)$ is continuous and one to one. It seems we want be able to get close to $\lim_{x\to w}f(w) =a$ without some issue. But we don't have to!

If it's not onto it needn't hit all points.

Let $a < c < d < e$. Let $f(a) = c$ and $f(d) = b$ and all the other points just do a linear stretch. That is for $a\le x \le b$ let $f(x) = (x-a)\frac {d-c}{b-a} + c$. That's one-to-one.

2) If $f$ is onto then the image is the codomain and your argument that a continuous image of a compact set is compact is valid. This is in essence the Extreme value system. This can not be done.

3) $f:(a,b)\to [a,b]$ is one-to-one. But it doesn't have to be onto. So it doesn't have to hit them all.

Let $a< c< d< b$ and let those $x$ that are "close" to $a$ have $f(x)$ "close" to $c$ and those "close" to $b$ have $f(x)$ "close" to $c$ so the image of $f$ is $(c,d)$ and $f$ just stretches between the points. Like the function in 1).

That is for $a< x < b$ let $f(x) = (x-a)\frac {d-c}{b-a} + c$. That's one-to-one.

(Hmmm.... even easier.... $f(x) = x$....The image is $(a,b)$.)

4) is different from $2$ A continuous function from a compact set must have a compact image, but a compact image needn't come from a compact set. The function is onto but not one to one. So for $f(w)=a$ we must have some $a< x< w$ where $f(x)$ is equal to some $f(y)$ for some $w< y$ but that's okay.

Let $a<c<d<b$ an let $f(c) =a$ and $f(d)=b$ and just stretch it out for the points between. That is, for $c\le x\le d$ let $f(x) = (x-c)\frac {b-a}{d-c}$. And for $x < c$... welll do whatever you want. Let $f(x)=a$ and for $x > d$ let $f(x)=b$. Not one-to-one but onto.