A function $f:\mathbb{R}^2\to\mathbb{R}^2$ that is open and closed, but not continuous.

Question

Does there exist a function $f:\mathbb{R}^2\to\mathbb{R}^2$ that is open and closed, but not continuous?

Note that I require $f$ to be defined on the entirety of $\mathbb{R}^2$.

There are a few examples of open functions that are not continuous. Most examples I found were functions $f$ with the property that $f(U)=\mathbb{R}^2$ for all open $U\subseteq\mathbb{R}^2$. The extra requirement that $f$ needs to be closed makes things a lot harder, though.

I believe the answer is no, such a function does not exist. My main motivation is the following observation.

Proposition: Let $f:\mathbb{R}^2\to\mathbb{R}^2$ be open and closed. If $x_n\to x$ and $f(x_n)$ is bounded, then $f(x_n)\to f(x)$.

Proof of 1: Let $f(x_{n_i})$ be a convergent subsequence of $f(x_n)$ with limit $p$. For all $i$ let $y_i=x_{n_i}$ if $f(x_{n_i})\neq p$. Otherwise, let $r=d(x_{n_i},x)$. Then $B_r(x_{n_i})$ is open, so $f(B_r(x_{n_i}))$ is open as well. So $f(B_r(x_{n_i}))\cap B_r(p)\setminus\{p\}$ is not empty. So we can choose $y_i\in B_r(x_{n_i})$ such that $f(y_i)\in B_r(p)\setminus\{p\}$.

We find $y_i\to x$ and $f(y_i)\to p$ and $f(y_i)\neq p$ for all $i$. Let $S=\{y_i:i\in\mathbb{N}\}$. We find that $p$ is a limit point of $f(S)$ not contained in $f(S)$. However, $C=S\cup\{x\}$ is closed, so $f(C)$ is also closed, and hence contains $p$. We conclude $f(x)=p$, so $f(x_{n_i})\to f(x)$.

Assume for the contrary that $f(x_n)\not\to f(x)$. Then there is a subsequence of $f(x_n)$ that always stays a certain distance from $f(x)$. By the Bolzano Weierstrass theorem, this subsequence itself has a convergent subsequence. By the previous observation, this subsequence converges to $f(x)$. This contradicts the fact that it always stays a certain distance from $f(x)$. $\square$

If you manage to prove that such a function does not exist, it might be neat to also look at how general the domain and codomain of $f$ can be made. For example, all arguments in the proposition still work for $f:X\to Y$ with any metric space $X$, and any finite dimensional vector space $Y$. However, with tweaking the arguments only a bit you find the following.

Let $X$ be a first countable Hausdorff topological space, and let $Y$ be a first countable topological space with no isolated points. Let $f:X\to Y$ be open and closed. If $x_n\to x$ and $f(x_n)$ is contained in some sequentially compact set, then $f(x_n)\to f(x)$.

Anyways, please let me know your thoughts.

Answer 1

I am not sure what happens with the maps of the plane but here is the situation with higher-dimensional domains:

Theorem 1. For every $n\ge 3$ there exists a closed and open function $f: R^n\to R^n$ which is not continuous.

Proof. While this sounds like a real analysis theorem, the proof uses some topology.

The key result is a rather nontrivial theorem due to John Walsh (he proved something stronger, I am stating a special case):

Theorem 2. Fix $n, m\ge 3$. Then for any pair of compact connected triangulated manifolds (possibly with boundary) $M, N$ of dimensions $m, n$ respectively, every continuous map $g: M\to N$ inducing the surjective map of fundamental groups $\pi_1(M)\to \pi_1(N)$ is homotopic to a (necessarily surjective) open continuous map $h: M\to N$.

See corollary 3.7.2 of

J. Walsh, Monotone and open mappings on manifolds. I. Trans. Amer. Math. Soc. 209 (1975), 419-432.

This deep theorem is a generalization of earlier results on existence of open continuous dimension-raising maps from $m$-cubes to compact triangulated manifolds due to Keldysh and Wilson.

Now, take $n=m$.

Take $N=T^n$, the $n$-dimensional torus (the $n$-fold product of circles). Fix a triangulation of $N$. Let $M\subset M$ be the complement to the interior of one of the $n$-dimensional simplices in the triangulation of $N$. Take $g: M\to N$ be the identity embedding. Then $g$ induces an isomorphic of fundamental groups (since $n=m\ge 3$). Therefore, by Theorem 2, $g$ is homotopic to a continuous open map $h: M\to N$. The universal covering of $N$ is $p: R^n\to N$; take $X:= p^{-1}(M)$; the restriction of $p$ to $X$ is the universal covering $X\to M$. By the basic covering theory, $h$ lifts to a continuous map $\tilde{h}: X\to R^n$. Since $h$ is homotopic to $g$, the mapping $\tilde{h}$ is properly homotopic to the identity embedding $X\to R^n$, in particular, $\tilde{h}$ is a proper map, hence, a closed map. Since the map $h$ is open, so is the map $\tilde{h}$, where we equip $X$ with the subspace topology induced from $R^n$.

So far, all our maps were continuous, I will now introduce a discontinuity (a very mild one). Each component $C_i$ of $R^n-X$ is an open $n$-dimensional simplex. Therefore, for each $i$ there exists a homeomorphism $c_i: C_i\to R^n$. Now, define the function $f: R^n\to R^n$ whose restriction to $X$ equals $\tilde{h}$ and whose restriction to each $C_i$ equals $c_i$. This function is clearly discontinuous (with discontinuities at the boundaries of the simplices $C_i$).

a. The function $f$ is closed. It suffices to prove that every convergent sequence $x_i\in R^n, x_i\to x$, whose accumulation set in $R^n$ is $\{f(x)\}$. After extraction, we can assume that $(x_i)$ either lies in one of the components $C_k$ as above, or it lies in $X$. In the former case, $$ \lim_{i\to\infty} f(x_i)= \lim_{i\to\infty} c_k(x_i)=\infty, $$ so the sequence has no accumulation points in $R^n$. In the latter case, since $X$ is closed, $x\in X$ and since $\tilde{h}$ is continuous, $$ \lim_{i\to\infty} f(x_i)= \lim_{i\to\infty} \tilde{h}(x_i)= f(x). $$ Thus, $f$ is closed.

b. The function $f$ is open. It suffices to show that every point $x\in R^n$ has a basis of neighborhoods $U$ whose images are open in $R^n$.

b1. The restriction of $f$ to the complement $C= R^n-X$ is an open map (since each component is mapped homeomorphically to $R^n$). Thus, for $x$ in $C$, the claim is clear.

b2. Suppose that $x$ lies in of $X$. Since the mapping $\tilde{h}: X\to R^n$ is open, there is a neighborhood basis $\{U_\alpha\}$ of $x$ in $R^n$ such that $\tilde{h}(U_\alpha)= f(U_\alpha\cap X)$ is open in $R^n$. But $U_\alpha\cap C$ is open, hence, $f(U_\alpha\cap C)$ is open as well (see part b1). Therefore, $f(U_\alpha)$ is open for every $\alpha$. This completes the proof. qed

Edit. A better proof is to appeal to the work of David Wilson directly, but you have to dig through his propositions and notation, specifically,

Proposition 3 of

D. Wilson, Open mappings of the universal curve onto continuous curves. Trans. Amer. Math. Soc. 168 (1972), 497–515.

(this is the real power tool behind all the results)

and

Propositions 1 and 3 of

D. Wilson, Open mappings of the universal curve onto continuous curves. Trans. Amer. Math. Soc. 168 (1972), 497–515.

Here is what you then get:

Proposition. Let $I^n$ be the closed $n$-dimensional cube, $n\ge 3$, and $J^n\subset int(I^n)$ is a closed subcube. Let $Q$ denote the interior of $J^n$. Then, there exists an open continuous map $g: I^n - Q\to I^n$ which equals the identity on the boundary of $I^n$ ends sends $\partial Q$ to the interior of $I^n$.

Given this, proceed as follows. First, extend the map $g$ by identity to the complement of $I^n$ in $R^n$. The result is a proper continuous open map $g: R^n - Q\to R^n$. Extend $g$ to $Q$ by a homeomorphism $Q\to R^n$ (this exists since $Q$ is an open cube). The result is an open and closed discontinuous function $f: R^n\to R^n$, $n\ge 3$. The discontinuity set of $f$ is the boundary of $Q$.