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I was wondering if you could help me with the following problem. I am trying to construct a set $A \subset [0,1]$ such that

(i) $x \in A$ implies $1 - x \in A$

(ii) $A$ is orderable with respect to $<$ in the sense that there exists a bijection $f: \mathbb{N} \to A$ such that $f(1) < f(2) < f(3) ...$

Any ideas? Or maybe this is not possible? Thanks in advance.

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Well, actually is seems that this is impossible since condition ii obviously implies that $A$ has a least element. But then condition (i) implies that $A$ has a greatest element. And then anon's comment applies. Sorry! – user12014 Jul 17 '11 at 4:38
up vote 10 down vote accepted

If $1\in A$ and $A$ is countably infinite then that implies there is a number $n\in\mathbb{N}$ such that $f(n)=1$. However, this would mean $f(n+1) > f(n) = 1$ and hence $f(n+1) \not\in [0,1]$, which is a contradiction. Your desired setup is therefore impossible.

Addendum: Even if you take out the condition $0,1 \in A$ the the setup is still impossible. Let $m$ be the number such that $f(m) = 1-f(1)$ - note that this is guaranteed to exist by your reflection-inclusion property. Then $f(m+1)>f(m)$ implies that $f(k) = 1-f(m+1) < 1-f(m) = f(1)$ for some likewise guaranteed-to-exist $k \in \mathbb{N}$, which contradicts $f(k)>f(1)$ (a result of the ordering property). No cake for you OP. :)

[Stated more succinctly by PZZ: the ordering implies a minimum in the set, which by reflection-inclusion implies there is a greatest element in the set, say $f(j)$, hence you get a contradiction with $f(j+1) > f(j)$.]

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Thank you, I should have seen that. I can live without the condition that $0,1 \in A$. Are the other 2 still possible? I will edit the question to take your answer into account. – user12014 Jul 17 '11 at 4:32

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