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Let $R\subseteq\mathbb{R}^2$. Consider the set of all "horizontal sections" $H_R =${$Rb|b\in\mathbb{R}$}, where $Rb=${$a\in\mathbb{R} | (a,b)\in R$}. Similarly consider the set of "vertical sections" of $R$, $V_R =${$ aR|a\in\mathbb{R}$} where $aR=${$ b\in\mathbb{R} | (a,b)\in R$}. Now define the equivalence relation on $\wp (\mathbb{R^2})$ such that $R \sim S$ if, and only if, $H_R=H_S$ and $V_R=V_S$.

QUESTION: What is the equivalence class of a disk?

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I have some troubles understanding the definition of the sets $H_R$ and $V_R$. Are they the projection of $R$ on the $x$ and $y$ axis, respectively? If so, I think the equivalence class of a disk would be the set of all figures of constant width inscribed in the disk. – Manolito Pérez Feb 3 '12 at 17:41
Let $f:[0,1]\to[0,1]$ be any continuous, monotone decreasing function such that $f(0)=1$ and $f(1)=0$. Let $R_0$ be the region bounded by $y=f(x)$ and the coordinate axes, and let $R$ be the region, symmetric about the origin, obtained by reflecting $R_0$ in the axes. Every such $R$ is in the equivalence class of the unit disk, and my guess is that nothing else is. – Brian M. Scott Feb 3 '12 at 18:34
up vote 1 down vote accepted

Denote with $\pi_1,\pi_2: \Bbb R^2 \to \Bbb R$ the coordinate projections.

First, let me make explicit what exactly the elusive sets $H_R$ and $V_R$ are (since I tripped into a pitfall about that earlier on).

We have that $H_R = \{Rb: b \in \Bbb R\}$, where $Rb = \{a \in \Bbb R: (a,b)\in\Bbb R\}$. In terms of the projections, this comes down to $H_R = \{\pi_1[R \cap \pi_2^{-1}(b)]: b \in \Bbb R\}$. We can imagine this as the sets we see when moving a horizontal slit over the set $R$.

Similarly, $V_R = \{\pi_2[R \cap \pi_1^{-1}(a)]: a \in \Bbb R\}$, and it can be imagined as the subsets of $\Bbb R$ obtained from sliding over $R$ with a vertical slit.

For finding the equivalence class, let us reduce to the entirely generic case of the (closed) unit disk $B$.

Then we observe that $H_B = V_B = \{[-a,a]: a \in [0,1]\}$. The apparent symmetry allows us to restrict attention to the upper right quadrant, since $R$ will have to be extended in a unique way, by said symmetry. We will hereafter drop distinction between $R$ and the intersection of $R$ with the upper right quadrant.

The conditions $H_R = H_B$ and $V_R = V_B$ imply that for each $r,s \in [0,1]$, there are unique $a, b$ with $\pi_1[R \cap \pi_2^{-1}(s)] = [0,a]$ and $\pi_2[R \cap\pi_1^{-1}(r)] = [0,b]$. We define mappings $i_H, i_V$ by $i_H(s) =a$ and $i_V(r)=b$.

Now suppose that for given $a$, we find an $s \in [0,1]$ such that $i_H(s) = a$. If for some $s' > s$ we would have that $i_H(s') = a' > a$, then for $i_V(a')$ to be an interval, we would need $(a',s) \in R$ since $s \le s'$, contradicting that $a' > a$ and $i_H(s)=a$.

This contradiction provides us with the information that $s \le s'$ implies $i_H(s) \supseteq i_H(s')$. A similar reasoning can be applied to obtain that $r \le r'$ implies $i_V(r) \supseteq i_V(r')$.

Now what happens when $s < s'$ and $i_H(s) = i_H(s')$? Then $i_V(i_H(s)) = s'$, while for $a' > i_H(s)$ we have necessarily that $i_V(a') < s$; the above analysis hence renders it impossible that there exists an $a''$ such that $s < i_V(a'') < s'$, contradicting our assumptions.

Therefore, both $r \mapsto i_V(r)$ and $s \mapsto i_H(s)$ are strictly decreasing surjections, and by the above argument they are also injective; A cursory remark is that $i_H(i_V(r))=r$ and $i_V(i_H(s)) = s$ for all $r,s$.

In conclusion, $B \sim R$ iff $R$ is the area below a strictly decreasing bijection $f:[0,1]\to[0,1]$, reflected first the horizontal, then the vertical coordinate.

Real analysis tells us that on intervals, "strictly decreasing bijection" is synonymous with "continuous injection mapping endpoints to opposite endpoints". Thus this answer matches Brian M. Scott's comment.

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I think you got it wrong: $H_R$ and $V_R$ are not intervals, but sets of intervals. – Christian Blatter May 24 '13 at 13:53
@ChristianBlatter I'm confident that this new attempt, now with the correct definitions, settles the question. Thanks again. – Lord_Farin May 24 '13 at 18:51

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