Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that there is no function on open interval $(-1,1)$, which has only finite number of discontinuity point, such that its graph is invariant under rotation by the right angle around the origin.

share|cite|improve this question
nothing good really. – user64370 Feb 28 '13 at 13:13
What are some functions that are invariant under rotation by right angle about origin? except disc? – user45099 Feb 28 '13 at 13:47
Maybe it helps that such a function will satisfy $f(f(t))=-t$. – Berci Feb 28 '13 at 15:20
Any $f: \mathbb R \to \mathbb R$ that satisfies $f(f(t))=-t$ for all $t \in \mathbb R$ has infinitely many points of discontinuity. This is a problem from the 1985 Vietnam Team Selection Tests for the IMO (source:…). – marlu Feb 28 '13 at 20:55
can't figure out the proof, bu thanks – user64370 Feb 28 '13 at 20:55
up vote 2 down vote accepted

Let $f:(-1,1)\to\mathbb{R}$ be a function such that its graph is invariant under rotation by the right angle around the origin. It implies that if $(x,y)$ is the graph, then $(y,-x)$ is also in the graph, i.e. if $x\in(-1,1)$ and $y=f(x)$, then $y\in(-1,1)$ and $-x=f(y)$. It follows that $f$ maps $(-1,1)$ to itself, and $$f^{\circ 2}(x)=-x,\quad\forall x\in(-1,1).\tag{1}$$ From $(1)$ we know that $f$ must be bijective on $(-1,1)$, i.e. $f$ is $1$ to $1$ and onto. Therefore, $f$ cannot be continuous on $(-1,1)$, because if $f$ is continuous and injective on $(-1,1)$, $f$ must be monotone on $(-1,1)$, and hence $f^{\circ 2}$ must be increasing, contradicting to $(1)$.

Now suppose that $f$ has finitely many discontinuity points, which are $a_1<a_2<\dots<a_n$. Denote $I_i=(a_i,a_{i+1})$, $0\le i\le n$, where $a_0=-1$ and $a_{n+1}=1$. Moreover, denote $C=\cup_{i=0}^n I_i$, the collection of continuity points of $f$, and $D=\{a_i:1\le i\le n\}$, the collection of discontinuity points of $f$.

Since for each open interval $I_i$, $f$ is continuous and injective on $I_i$, $f(I_i)$ is also an open interval, and $f$ has a continuous inverse $g_i:f(I_i)\to I_i$. Since on $f(I_i)$, $f= f\circ f\circ g_i=-g_i$, $f$ is continuous on $f(I_i)$. That is to say, $f$ maps continuity points to continuity points. Combining this fact with $f$ being bijective, we have $$f(C)=C \quad\text{and}\quad f(D)=D.\tag{2}$$

In particular, for every $0\le i\le n$, there exists $0\le j\le n$, such that $f(I_i)\subset I_j$. Since $I_j\subset C$ and $C=\cup_kf(I_k)$, we know that $I_j=\cup_k(f(I_k)\cap I_j)$. Then by the connectedness of $I_j$, in fact $f(I_i)=I_j$. As a result, $f$ defines a permutation on $\mathcal{I}:=\{I_i:0\le i\le n\}$. Note that $f^{\circ 2 }(I_i)=-I_i$, so there are two cases. First, if $I_i\ne -I_i$, i.e. $0\notin I_i$, then $f^{\circ k}(I_i)$ are pairwise different, $k=0,1,2,3$. Second, if $I_i=-I_i$, i.e. $0\in I_i$, then $f(I_i)=I_i$, because $$f(I_i)=f(-I_i)=f^{\circ 3}(I_i)=f^{\circ 2}(f(I_i))=-f(I_i)\Rightarrow 0\in f(I_i)\Rightarrow I_i=f(I_i).$$ However, the second case cannot happen, and the reason is the same as in the first paragragh: if $f:I_i\to I_i$ is injective and continuous, then $f^{\circ 2}$ must be increasing. Finally, we can conclude that: (i) $0$ is a discontinuity point, and (ii), $\mathcal{I}$ is a disjoint union of the $f$-orbits, and each orbit is of length $4$, i.e.
$$ n+1=\#\mathcal{I}\equiv 0 \mod 4.\tag{3}$$

A similar argument can be applied to $f:D\to D$. For each $a_i\in D$, if $a_i\ne 0$, then $f^{\circ k}(a_i)$ are pairwise different, $k=0,1,2,3$; if $a_i=0$, then $f(a_i)=a_i$. Since $0\in D$, we can also conclude that $$n=\# D\equiv 1 \mod 4.\tag{4}$$

The contradiction between $(3)$ and $(4)$ completes the proof.

share|cite|improve this answer
Is it possible to explain some of the terms such as bijection, and to phrase this in a simpler way, for those who do not have much calculus experience? This is a good answer, but read the bounty conditions carefully. – cuabanana Apr 21 '13 at 1:46
@cuabanana: I edited my answer a little. Hope it looks clearer now. Since the answer is already very long in words, I don't want to expand it too much. Due to my poor English, this is the best I can do. – 23rd Apr 21 '13 at 8:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.