Open maps which are not continuous What is an example of an open map $(0,1) \to \mathbb{R}$ which is not continuous? Is it even possible for one to exist? What about in higher dimensions? The simplest example I've been able to think of is the map $e^{1/z}$ from $\mathbb{C}$ to $\mathbb{C}$ (filled in to be $0$ at $0$). There must be a simpler example, using the usual Euclidean topology, right?
 A: Explicit examples are moderately difficult to construct, but it’s not too hard to come up with non-constructive examples; here’s one such.
For $x,y\in\mathbb{R}$ define $x\sim y$ iff $x-y\in \mathbb{Q}$; it’s easy to check that $\sim$ is an equivalence relation on $\mathbb{R}$. For any $x\in\mathbb{R}$, $[x] = \{x+q:q\in\mathbb{Q}\}$, where $[x]$ is the $\sim$-equivalence class of $x$. In particular, each equivalence class is countable. For any infinite cardinal $\kappa$, the union of $\kappa$ pairwise disjoint countably infinite sets has cardinality $\kappa$, so there must be exactly as many equivalence classes as there are real numbers. Let $h$ be a bijection from $\mathbb{R}/\sim$, the set of equivalence classes, to $\mathbb{R}$. Finally, define $$f:(0,1)\to\mathbb{R}:x\mapsto h([x])\;.$$
I claim that if $V$ is any non-empty open subset of $(0,1)$, $f[V]=\mathbb{R}$, which of course ensures that $f$ is open. To see this, just observe that every open interval in $(0,1)$ intersects every equivalence class. (It should be no trouble at all to see that $f$ is wildly discontinuous!)
A: There is in fact a rather easy example of a function $\mathbb R \to \mathbb R$ such that the image of every open set is $\mathbb R$: Let $(x_i)_{i\in\mathbb Z_+}$ be the binary expansion of $x$, so that each $x_i \in \{0,1\}$. Let then $f : \mathbb R \to \mathbb R$ be given by
$$f(x) = \sum_{k=1}^\infty\frac{(-1)^{x_k}}k\quad \textrm{if the series converges}$$
$$f(x) = 0\quad \textrm{otherwise.}$$
Since the harmonic series (or a tail of it) can be made to converge to any real number by changing signs in the appropriate way, this function has $f((a,b)) = \mathbb R$ for any real $a < b$. Hence this function is open, though clearly not continuous at any point.
The harmonic series can be substituted with any other absolutely unbounded series where the summand goes to zero.
A: Let me conceptualize around Brian's answer a bit.

Definition 0. If $X$ and $Y$ are topological spaces, a function $f:X→Y$ is said to be strongly Darboux iff for all non-empty open sets $A⊆X$, we have $f(A)=Y$.

Here's the basic facts:

Proposition.
  
  
*
  
*Every strongly Darboux function is an open function.
  
*If $X$ is non-empty, every Darboux function $X \rightarrow Y$ is surjective.
  
*If $X$ is non-empty and $f : X \rightarrow Y$ is a continuous Darboux mapping, then $Y$ carries the indiscrete topology.
  

Proofs.


*

*Trivial.

*Since $X$ is open and non-empty, hence $f(X)=Y.$ That is, $f$ is surjective.

*Let $B \subseteq Y$ denote a non-empty open set. Our goal is to show that $B=Y$. Since $f$ is surjective, $f^{-1}(B)$ is non-empty. Since $f$ is continuous, $f^{-1}(B)$ is open. Hence $f(f^{-1}(B))=Y$. But since $f$ is surjecive, hence $f(f^{-1}(B))=B.$ So $B=Y$.
Putting these together, we see that every strongly Darboux function $f:\mathbb{R} \rightarrow \mathbb{R}$ is a discontinuous open mapping.


*

*$f$ is an open mapping by (1).

*$f$ is discontinuous by (3), because it's domain is non-empty and it's codomain doesn't carry the indiscrete topology.
And, of course, Brian's answer guarantees the existence of a strongly Darboux function $\mathbb{R} \rightarrow \mathbb{R}$. This completes the proof.
