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I am trying to characterize the continuum $\mathcal{C}$ using only the notion of midpoint, i.e. the operation $\mu : \mathcal{C}\times\mathcal{C} \to \mathcal{C}$ assigning to each pair of points the midpoint between them. I thought of defining “unbounded interval” as a set $I \subseteq \mathcal{C}$ such that $I$ as well as $\mathcal{C}\backslash I$ are closed under $\mu$. But is that even true?

My thoughts so far: If $I$ has the above property but is not an interval, then both $I$ and $\mathcal{C}\backslash I$ are dense in $\mathcal{C}$ and noncountable (it is easy to find an injection from one into the other)...

Note: The more general question whether it is possible to define "interval" or order in the above setting has now been answered here (in the comment by Matt F.) and the answer is no.

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  • $\begingroup$ Does your closedness of $I$ under $\mu$ mean: for all $x,y\in I,\mu(x,y)\in I$? $\endgroup$ – User Apr 15 '17 at 14:37
  • $\begingroup$ Yes, that's what I mean. $\endgroup$ – Larry Apr 15 '17 at 14:44
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Working with the usual definition of $\mathbb R$ (and assuming choice), we can define a set $I$ that's not an unbounded interval, but for which both $I$ and $\mathbb R \setminus I$ are closed under taking midpoints. (I use $\mathbb R$ to refer to the usual notion of realnumberness to distinguish from $\mathcal C$ which you want to build from the ground up.)

Choose a Hamel basis for $\mathbb R$ as a vector space over $\mathbb Q$, and let $b$ be an element of that basis. Then define $I$ to be the set of all elements of $\mathbb R$ so that, when represented in this basis, the coefficient of $b$ is nonnegative.

If we do this, then both $I$ and $\mathbb R\setminus I$ are closed under midpoint. If $x, y \in \mathbb R$ and have $b$-coefficients $x_b, y_b$, then $\frac{x+y}{2}$ has $b$-coefficient $\frac{x_b+y_b}{2}$. So if $x_b, y_b \ge 0$, then $\frac{x_b+y_b}{2} \ge 0$, and if $x_b, y_b < 0$, then $\frac{x_b + y_b}{2} <0$.

It's also worth noting that if $I$ and $\mathbb R\setminus I$ are closed under midpoints, and $I$ is bounded from above, then $I$ is an unbounded interval. Let $x = \sup I$. For any $\epsilon>0$, there is some $y \in I$ with $y > x-\epsilon$. If we now take $z < x-2\epsilon$, then $2y-z > x$, so $2y-z \notin I$. So we must have $z \in I$, or else $\frac{z + (2y-z)}{2} = y \notin I$ would be a contradiction. Therefore $(-\infty, x-2\epsilon) \subseteq I$ for all $\epsilon$, and therefore their union, $(-\infty, x)$, is contained in $I$. So $I$ is either $(-\infty,x]$ or $(-\infty,x)$ depending on whether $x \in I$ or not.

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  • $\begingroup$ That was quick! Unfortunately I don't know how to define “bounded (from above)“... $\endgroup$ – Larry Apr 15 '17 at 15:10
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Are you putting metric structure on these midpoints; such as as a weakened version of midpoint convexity.

Or just that the midpoint functions values exist in that appropriates, F(c/2+c1/2) exists in the structure somewhere between the c1 and c2 c1

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