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I've been working on answering this question on and off for a while (months). I can't seem to solve it, and I've presented it to a few people who also cannot solve it. I will present a special case of the problem here, as I intuitively suspect that if a counterexample can be found, it will be found in $ \mathbb{R}$.

$S \subseteq \mathbb{R}$ is said to have "Property 1" if and only if:

$\forall m \in \mathbb{R}$, $\exists$ $k \in \mathbb{R}$ such that $m+k \in S$ and $m-k \in S$.

Prove or disprove the following:

If $S$ has "Property 1", then $\exists$ $S' \subseteq S$ such that $S'$ has "Property 1" and $\forall$ $C \subseteq S'$, $C \neq \emptyset$, $S'-C$ does not have "Property 1."

I attempted a proof using Zorn's Lemma and quickly discovered that in order to even satisfy the hypothesis of said lemma, I would pretty much have to assume what we're trying to prove here (unless there's some clever way around this I'm not seeing). I didn't want to post this anywhere, I really wanted to prove it myself, but I'm having quite a bit of trouble. Any insight would be appreciated.

M10687
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1 Answers1

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Yes I think this is possible. Forgetting for the moment the relative version you have stated, just try to get even one minimal symmetric set.

Do this by a diagonalization of length the continuum. Enumerate all the real numbers $\{m_{\alpha}\}$ and successively chose $p_{\alpha},q_{\alpha}$ symmetric about $m_{\alpha}$. The set $S_{\alpha}$ is thus constructed in stages. At the same time we build sets $T_{\alpha}$ which are elements that it is forbidden to use.

Lets say $m_{\alpha}$ is the next in the list. If it already has a symmetric pair from the elements already chosen, just move on to the next element.

Note: I use $\beta <\alpha$ in all notation.

If it does not have a pair chose a pair $p_{\alpha},q_{\alpha}$ such that it is symmetric for $ m_{\alpha}$ and not in the list of forbidden elements. Now we add new forbidden elements, first for any element $x\in S_{\alpha}$ already chosen look at the element $y$ symmetric to $x$ about $m_{\alpha}$, so $x$ and $y$ are symmetric about $m_{\alpha}$. $y$ is not in $S_{\alpha}$ because we assumed that $m_{\alpha}$ did not already have a symmetric pair. Put $y\in T_{\alpha+1}$ so $y$ is on the forbidden list. This means that the "honor" of $p_{\alpha},q_{\alpha}$ is preserved. They are symmetric about $m_{\alpha}$, and no further point will be able to destroy that honor, since all possible pairs are, at this stage, being destroyed by putting one side on the forbidden list. Of course we must continue to protect the honor of all previous pairs $p_{\beta},q_{\beta}$, so if $x$ is one of the new elements $p_{\alpha}$ or $q_{\alpha}$ and $m_{\beta}$ is old center and if $y$ is symmetric to $x$ about
$m_{\beta}$ the add $y$ to the forbidden list. (Note that $y$ cannot already be in the set since then $x$ ($=p_{\alpha}$ or $q_{\alpha}$) would be in the forbidden list and we chose them not on that list.)

So to sum up pairs are added because they are needed, and they remain needed throughout the entire construction. Thus no point can be removed to still retain symmetry.

Kind of complicated but in fact is has become a standard type of set theory argument.

Rene Schipperus
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  • First, why do we put $y \in T_{\alpha+1}$? Does this not assume the index set $I$, $\alpha \in I$ is countable? Secondly, how can you guarantee that for each $\alpha$ $\exists$ $p_\alpha, q_\alpha$ symmetric about $m_\alpha$ and NOT on any forbidden list $T_\beta$ ($\beta< \alpha$)? Third, if indeed one assumes countable $I$, are we not then assuming CH? Forgive me, I'm not a set theorist and this is the first time I've seen such an argument, but a little bit of research has revealed that these arguments often require CH. – M10687 Jan 03 '16 at 01:22
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    Ok, so the index set is $2^{\aleph_0}=|\mathbb{R}|$. The continuum hypothesis is not relevant here. Second all the sets $S_{\alpha}$, $T_{\alpha}$ have size less than $2^{\aleph_0}$, so that is why the next p,q can always be found there are continuum many choices and less than continuum many forbidden elements. – Rene Schipperus Jan 03 '16 at 01:29
  • Ok. I'll have to read more about this particular kind of argument to fully understand its inner workings but thanks so much for taking the time to answer. – M10687 Jan 03 '16 at 01:44
  • The book Linear Orderings by Rosenstein has some nice examples of diagonalizations of length the continuum. – Rene Schipperus Jan 03 '16 at 01:46