In the last few days I've been thinking on and off about these two problems and I can't get my head around them:

Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a continous open map.

1. If $f$ is surjective and $\varnothing \neq A \subseteq \mathbb{R}$ is an open set, can $f^{-1}(A)$ be bounded?

2. Can $f$ be also closed?

The only progress I managed to make is saying that in 2. since $\mathbb{R}^2$ is both open and closed, then its image should be both open and closed, thus it must be that $f$ is surjective, since $\mathbb{R}$ is connected.

In the last few days I've been thinking on and off about these two problems and I can't get my head around them:

Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a continuous open map.

1. If $f$ is surjective and $\varnothing \neq A \subseteq \mathbb{R}$ is an open set, can $f^{-1}(A)$ be bounded?

2. Can $f$ be also closed?

The only progress I managed to make is saying that in 2. since $\mathbb{R}^2$ is both open and closed, then its image should be both open and closed, thus it must be that $f$ is surjective, since $\mathbb{R}$ is connected.

In the last few days I've been thinking on and off about these two problems and I can get my head around them:

Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a continous open map.

1. If $f$ is surjective and $\varnothing \neq A \subseteq \mathbb{R}$ is an open set, can $f^{-1}(A)$ be bounded?

2. Can $f$ be also closed?

The only progress I managed to make is saying that in 2. since $\mathbb{R}^2$ is both open and closed, then its image should be both open and closed, thus it must be that $f$ is surjective, since $\mathbb{R}$ is connected.

In the last few days I've been thinking on and off about these two problems and I can't get my head around them:

Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a continous open map.

1. If $f$ is surjective and $\varnothing \neq A \subseteq \mathbb{R}$ is an open set, can $f^{-1}(A)$ be bounded?

2. Can $f$ be also closed?

The only progress I managed to make is saying that in 2. since $\mathbb{R}^2$ is both open and closed, then its image should be both open and closed, thus it must be that $f$ is surjective, since $\mathbb{R}$ is connected.

In the last few days I've been thinking on and off about these two problems and I can get my head around them:

Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a continous open map.

1. If $f$ is surjective and $\varnothing \neq A \subseteq \mathbb{R}$ is an open set, can $f^{-1}(A)$ be bounded?

2. Can $f$ be also closed?

The only progress I managed to make is saying that in 2. since $\mathbb{R}^2$ is both open and closed, then its image should be both open and closed, thus it must be that $f$ is surjective, since $\mathbb{R}$ is connected. Now since $[0,1] \times [0,1]$ is compact and connected we have that its image is compact and connected, thus let's call it $[a, b]$, with $a \le b$. Now if 1. is false and $a \neq b$ then $f^{-1}((a, b))$ is unbounded, which is impossible since it should be contained in the unit square. So either 1. is actually true or $f$ must be constant on any compact set, or possibly both.

At this point I don't know how to articulate formally the reasoning but I really think that having a function constant on any compact implies that it's not surjective, so to at least prove that 1. implies 2. however I don't know how to prove it, or to prove 1.

In the last few days I've been thinking on and off about these two problems and I can get my head around them:

Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a continous open map.

1. If $f$ is surjective and $\varnothing \neq A \subseteq \mathbb{R}$ is an open set, can $f^{-1}(A)$ be bounded?

2. Can $f$ be also closed?

The only progress I managed to make is saying that in 2. since $\mathbb{R}^2$ is both open and closed, then its image should be both open and closed, thus it must be that $f$ is surjective, since $\mathbb{R}$ is connected.