Can we find a function $f:\mathbb{R}\rightarrow \mathbb{R}$ that is open, closed, but not continuous?

Question

The question is described in the title. To clarify, I'm looking for an example where the domain of such a function is $\mathbb{R}$, so the example here will not work.

The highest upvote answer here gives an example of discontinuous open map from $\mathbb{R}$ to $\mathbb{R}$, but that map can hardly be closed (closed set does not need to contain any interval!). I have difficulty generalizing the construction and know very little about any sufficient condition that leads to a closed map without continuity (piecewise constant will work, but that kind of map cannot be open!).

I'll appreciate any example, non-constructive prove/disprove, or any reference that covers this.

*In case I need to clarify this, we use std. metric topology on $\mathbb{R}$.

Answer 1

No such function exists.

Assume that $f((a,b))$ is a bounded open set. Then $f((a,b))= (c,d)$ for some $c,d\in \mathbb{R}$. Indeed, $f((a,b))$ is an open set hence of the form $\bigcup (a_n,b_n)$ however $f([a,b])$ has at most two more points than $f((a,b))$, so we conclude.

Claim: if $f((a,b))= (c,d)$ then $f$ is continuous and strictly monotonic on $(a,b)$.

Proof: Firstly, we will show that $f$ is injective. Indeed assume that $f(x_1)= f(x_2)$ for some $x_1,x_2\in (a,b)$. Then take the $f((x_1,x_2))=(c_1,d_1) $ for some $c_1,d_1$, a contradiction since $f( [x_1,x_2] )$ needs two points to be closed. Now, since $f$ has the IVT propoerty and it is injective it is strictly monotonic and continuous.

We will argue by contradiction. Assume that $f$ is not continuous at $x_0$, and set $V_n=( x_0-1/n,x_0+1/n )$. Now, the claim implies there are $z_n<w_n$ such that $f(V_n )=(-\infty ,z_n)\bigcup (w_n,\infty) $ where $z_n$ or $w_n$ can possibly take the value $\pm \infty$. Also, notice that $w_n$ is increasing and $z_n$ is decreasing.

First case, one of the $z_n,w_n$ converges (to a finite limit). WLOG assume that $w_n \rightarrow w$. Now, pick a number $\beta$ in $(w,\infty)$ that is different from $f(x_0)$. We can find a sequence $p_n\in V_{n-1}\setminus V_n$ such that $f(p_n)=\beta +1/n$. A contradiction since the closed set $\{x_0\}\bigcup \{p_n|n\in \mathbb{N}\}$ gets mapped via $f$ to $\{f(x_0)\}\bigcup \{\beta +1/n|n\in \mathbb{N}\}$ which is not closed.

In the second case, we have $z_n \rightarrow -\infty$ and $w_n \rightarrow\infty$. This is a contradiction since the previous implies $\bigcap_{n \in \mathbb{N}}f(V_n)= \emptyset$, however $\bigcap_{n \in \mathbb{N}}f(V_n)$ always contains $f(x_0)$.

Answer 2

This proof has a flaw in the second half:

Let $f:\mathbb R\rightarrow \mathbb R$ be a closed and open function.

First we prove that $f^{-1}(x)$ is finite for every $x\in \mathbb R$

suppose not, let $a_1>a_2>a_n\dots$ be a sequence of values whose image is all $x$, and such that for every $i$ there is an element in $f^{-1}(x)$ between $a_i$ and $a_{i+1}$.

Notice that $f((a_i,a_{i+1}))=f([a_i,a_{i+1}])$. Therefore $f((a_i,a_{i+1})=\mathbb R$, since it is a clopen set).

If the $a_i$'s converge then let $l$ be their limit.

We now take $b_1,b_2,\dots$ a sequence such that $a_i<b_i<a_{i+1}$ for all $i$, and such that $f(b_i)$ is a strictly increasing converging sequence ( if the $a_i$'s converge we also ask $f(a_i)>l$).

If $l$ does not exist then $\{b_1,b_2,\dots\}$ is closed while its image is not. And if $l$ exists then $\{l,b_1,b_2,\dots\}$ is closed while its image is not.

---

So now let $f^{-1}(0)=\{a_1,a_2,\dots, a_n\}$. Then $f$ is injective when restricted to $(-\infty,a_1),(a_i,a_{i+1})$ and $(a_n,\infty)$.

Proof: Let $U$ be a connected set which does not contain $0$ in the image, suppose $f(x)=f(y)$ for some $x\in U$. Notice that $f((x,y))$ and $f([x,y])$ differ by exactly one point. This means that $f(x,y)$ is an open set with at most $1$ limit point. Therefore $f(x,y)=\mathbb R$ or $\mathbb R \setminus f(x)$. Either case is a contradiction, since no point in $U$ maps to $0$.

We conclude that $f$ is injective when restricted to each of those open sets. Notice that the function $f$ is an open function with respect to the induced topology, since each of those sets is open. Moreover, each of these sets is isomorphic to $\mathbb R$. The following lemma finishes the proof.

Lemma: an open injection $g:\mathbb R\rightarrow \mathbb R$ is continuous.

proof: Let $h$ be the $g$ with a changed codomain, so that the function is surjective. Notice that $h^{-1}$ is a continuous bijection from a disjoint union of open sets to $\mathbb R$. (every open set is a disjoint union of open intervals).

It suffices to show that $h^{-1}$ is open on every interval, but this is clear, since a continuous injection $(a,b)\rightarrow \mathbb R$ is increasing, and hence a homeomorphism on its domain.

This shows that $f$ is continuous for every point $x$ with $f(x)\neq 0$. Using the same argument for $1$ instead of $0$ proves continuity everywhere.