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?
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3$\begingroup$ Since $(0,1)$ and $\mathbb R$ are homeomorphic via a linear map composed with $\arctan$, it suffices to find a map $\mathbb R \to \mathbb R$ that is open but not continuous. Googling that gives you mathforum.org/library/drmath/view/62395.html $\endgroup$– lhfOct 25, 2011 at 0:55
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3$\begingroup$ this is obviously not much help, but if you can find a continuous bijection $f$ with discontinuous inverse, then $f^{-1}$ will do. $\endgroup$– user12014Oct 25, 2011 at 1:13
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9$\begingroup$ One can build such a function from a Cantor set $C$ (the usual "middle thirds" set will do). Send each point in $C$ to $0$, and map each connected component of the complement of $C$ homeomorphically to the interval $(-1,1)$. Then the image of any open set intersecting $C$ will be $(-1,1)$ (thus open), and the image of any open set not meeting $C$ will also be open, since it's a union of homeomorphic images of open sets. Of course, each point of $C$ will be a discontinuity. $\endgroup$– user83827Oct 25, 2011 at 1:16
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3$\begingroup$ @PZZ for instance the map wrapping [0,1) around the unit circle. $\endgroup$– JSchlatherOct 25, 2011 at 1:37
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2$\begingroup$ @PZZ: In fact there are no counterexamples of the type you're suggesting: if $I$ and $J$ are intervals in $\mathbb{R}$ and $f: I \rightarrow J$ is a continuous bijection, then $f^{-1}$ is necessarily continuous. By coincidence this is exactly where I am in my Spivak calculus course, so see e.g. Theorem 37 in $\S 6.4$ of math.uga.edu/~pete/2400calc2.pdf. (Or see Spivak's text!) $\endgroup$– Pete L. ClarkOct 25, 2011 at 3:18
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4 Answers
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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!)
answered Oct 25, 2011 at 1:41
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1$\begingroup$ Just curious: Is the axiom of choice used anywhere in your proof? $\endgroup$ Jan 17, 2017 at 8:14
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8$\begingroup$ I think I'm going to start calling $\sim$ the "Vitali equivalence relation"... $x$ and $y$ are Vitali equivalent iff $x-y \in \mathbb{Q}$, etc. Honestly, this thing is useful enough to deserve a name. $\endgroup$ Mar 2, 2017 at 13:54
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1$\begingroup$ @DavidFeng: No. All $x$ from the same equivalence class give the same value. For example, $f(\frac12)=f(\frac13)$ since $\frac12-\frac13\in\mathbb Q$ $\endgroup$– celtschkMar 2, 2019 at 21:55
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2$\begingroup$ @SanajitPatra: For any $x\in(0,1)$, $[x]$ is a dense subset of $\Bbb R$ (since it’s just a translate of $\Bbb Q$), so $[x]\cap(0,1)$ is a dense subset of $(0,1)$. Fix $x\in(0,1)\setminus\Bbb Q$, and let $y=h(x)$; $x$ is irrational, so $x\not\sim\frac12$. Now show that there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $[x]$ converging to $\frac12$, and observe that $h(x_n)=y$ for each $n\in\Bbb N$, but $h\left(\frac12\right)\ne y$, so $h$ is discontinuous at $\frac12$. The argument can be generalized to show that $h$ is nowhere continuous. $\endgroup$ Apr 30, 2022 at 20:33
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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.
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1$\begingroup$ Thank you! I made up the function for a topology assignment some years ago, and figured it was significantly easier than the other answers here. $\endgroup$ Aug 7, 2022 at 10:30
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3$\begingroup$ This "everywhere surjective" property also appears in the Conway Base 13 function en.wikipedia.org/wiki/Conway_base_13_function. In fact your function satisfies I think all the properties of the Conway base 13 function listed in the Wikipedia! $\endgroup$– D.R.Jul 6, 2023 at 23:42
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$\begingroup$ I am not familiar with Conway's construction (it seems awfully complicated) but my function certainly satisfies that every open interval is mapped onto ℝ, and therefore it trivially has the intermediate value property. $\endgroup$ Jul 8, 2023 at 8:39
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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.
answered Nov 17, 2016 at 13:29
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This was supposed to be a comment on Brian's answer, but it is just too long:
To emphasize how weird the world of this example is: Let $q\colon \mathbb R\to \mathbb R/\mathbb Q$ be the quotient map of $\mathbb Q$-vector spaces. Now assuming the axiom of choice, any vector space has a basis, so we may choose a basis $B$ of $\mathbb R/\mathbb Q$. But then $\dim_{\mathbb Q}(\mathbb R) = \mathfrak c = |\mathbb R|$, by standard facts about cardinality. Thus we may pick a bijection between $B$ and $\mathbb R$ which extends to give a $\mathbb Q$-linear surjection $c\colon \mathbb R/\mathbb Q \to \mathbb R$. It follows (in much the same way as it does for Brian's map) that $h = c\circ q\colon \mathbb R\to \mathbb R$ is a $\mathbb Q$-linear map (equivalently, a homomorphism of the abelian group $(\mathbb R,+)$) for which $h(I)=\mathbb R$ for any interval $I$.
