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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.

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    $\begingroup$ Okay, I think I got it now. Let $A = f^{-1}((-\infty, a])$, $B = f^{-1}((a,b))$, and $C = f^{-1}([b,\infty))$. Assume $B$ is bounded. Then $A$ and $C$ are both closed, one is unbounded, and one is not. So one is compact, contradiction. $\endgroup$
    – Nik Weaver
    Commented May 31 at 11:08
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    $\begingroup$ (This doesn't use openness, only continuity and surjectivity.) $\endgroup$
    – Nik Weaver
    Commented May 31 at 11:13
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    $\begingroup$ You can also consider sequences of points $(p_n)$ and $(q_n)$ such that $f(p_n)\to\infty$ and $f(q_n)\to-\infty$ and consider paths between $p_n$ and $q_n$ to see that $f^{-1}(\{a\})$ is unbounded for all $a$ (again, using only continuity and surjectivity). This also shows that if $f:\mathbb{R}^2\to\mathbb{R}$ is continuous and unbounded then it cannot be closed (as there is some sequence of points $(p_n)$ in $\mathbb{R}^2$ going to infinity with $f(p_n)=\frac{1}{n}$) $\endgroup$
    – Saúl RM
    Commented May 31 at 11:18
  • $\begingroup$ @NikWeaver sorry could you explain in more detail, eventually in an answer? It's not clear to me why it must follow that A or C is bounded. $\endgroup$
    – Ismo
    Commented May 31 at 16:20
  • $\begingroup$ @Ismo if $B$ is bounded then it is contained in a disk. The complement of that disk is connected, so it is either contained in $A$ or $C$, but not both. The other one is contained in the original disk. $\endgroup$
    – Nik Weaver
    Commented May 31 at 16:59

1 Answer 1

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Assume $f:\mathbb{R}^2\to\mathbb{R}$ is continuous, closed, and open. It is indeed surjective, because the range has to be closed and open. Since the map is open the set $f^\gets(0)$ is nowhere dense: the image of its interior would be $\{0\}$, which is not an open set. Also: if $f(x,y)=0$ and $r>0$ then there are $(u,v)$ and $(w,z)$ in thball $B((x,y),r)$ with $f(u,v)<0<f(w,z)$, because the image of the ball contains an interval around $0$. Hence $f^\gets(0)$ is the common boundary of $f^\gets[(-\infty,0)]$ and $f^\gets[(0,\infty)]$. Neither of these two sets is bounded because the respective images of their closures, $(-\infty,0]$ and $[0,\infty)$, are not compact. As in the comments this shows that $f^\gets(0)$ is unbounded: given $M>0$ take $(x,y)$ and $(u,v)$ outside the circle around $(0,0)$ of radius $M$ with $f(x,y)<0<f(u,v)$ and connect them by an arc outside the circle. That arc will intersect $f^\gets(0)$.

Now take a sequence $\bigl<(x_n,y_n):n\in\omega\bigr>$ in $f^\gets(0)$ that diverges to infinity. For each $n$ take $(u_n,v_n)\in B((x_n,y_n),2^{-n})$ such that $0<f(u_n,v_n)<2^{-n}$. The set $\{(u_n,v_n:n\in\omega\}$ is closed but its image is not.

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    $\begingroup$ Not to be a jerk, but I think there may be an etiquette issue about posting an answer a couple of hours after the question was answered in the comments ... $\endgroup$
    – Nik Weaver
    Commented May 31 at 16:05
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    $\begingroup$ @NikWeaver Is there a reason you didn't post an answer? $\endgroup$
    – Timothy Chow
    Commented May 31 at 21:53
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    $\begingroup$ @TimothyChow If a question is too basic I will usually advise the OP to ask on math.stackexchange. For a question like this one (undergraduate level with a two-line solution) it feels appropriate to put the answer in a comment rather than try to karma farm. $\endgroup$
    – Nik Weaver
    Commented May 31 at 23:17
  • $\begingroup$ I appreciate that others may consider a full answer to be appropriate, but I'm still a little ruffled that the posted answer didn't mention that the problem was already solved in the comments. $\endgroup$
    – Nik Weaver
    Commented May 31 at 23:18
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    $\begingroup$ @NikWeaver If that sort of thing bothers you, then I recommend that (as a general principle) you post your answers as answers rather than in the comments. It's my understanding of MO etiquette that answers should normally be posted as answers. I'm not sure what you mean by "karma farming," but if that's a philosophical objection to the reputation system then you can always make your answer community wiki. Otherwise, if you choose to buck convention by answering in the comments then it seems philosophically inconsistent to me to then turn around and complain about someone else gaining credit. $\endgroup$
    – Timothy Chow
    Commented Jun 1 at 1:45

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