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Short version of question. Is there a set $S\subseteq [0,1]$ with $|S|=2^{\aleph_0}$ such that all points of $S$ have distinct pairwise distances?

Formal version of question. If $X$ is a set, let $[X]^2=\big\{\{x,y\}:x\neq y\in X\big\}$. Let $(X,d)$ be a metric space. Let $d_{\text{set}}:[X]^2\to \mathbb{R}$ be defined by $$\{x,y\}\in [X]^2\mapsto d(x,y)=d(y,x).$$ We say $S\subseteq X$ has the distinct pairwise distance property (dpdp) if the restriction of $d_{\text{set}}$ to $S$ is injective. Is there a set $S\subseteq [0,1]$ with (dpdp) and $|S|=2^{\aleph_0}$, where $[0,1]$ is endowed with the Euclidean metric?

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    $\begingroup$ yes, use Zorn to find a maximal such set and observe it is uncountable. $\endgroup$
    – Uri Bader
    Commented Oct 12, 2018 at 9:39
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    $\begingroup$ but my answer remains correct: replacing "uncountable" with $2^{\aleph_0}$... $\endgroup$
    – Uri Bader
    Commented Oct 12, 2018 at 9:45
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    $\begingroup$ We will catch it when it sneaks! If $S_\alpha$ is a chain with union $S$ and $x_1,x_2,y_1,y_2\in S$ having $d(x_1,y_1)=d(x_2,y_2)$ then they all should appear in some $S_\alpha$, don't they? $\endgroup$
    – Uri Bader
    Commented Oct 12, 2018 at 9:58
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    $\begingroup$ Got it! I am convinced now every maximal element with dpdp is uncountable, but I'm not sure about $2^{\aleph_0}$ yet. As soon as I understand, I'll delete the question $\endgroup$
    – Dominic van der Zypen
    Commented Oct 12, 2018 at 10:00
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    $\begingroup$ Don't delete the question please! As it turns out, it's a nice (routine?) application of Zorn's lemma: it has some didactical value. Maybe, just move it to MSE! $\endgroup$
    – Qfwfq
    Commented Oct 12, 2018 at 11:55

2 Answers 2

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Consider $\Bbb R$ as a vector space over $\Bbb Q$, choose (Zorn) a basis $S$ consisting (after rescaling with rational scalars) of elements in $(0,1)$. Then we have no relation of the shape $\pm(s_1-s_2)=(s_3-s_4)$ for $s_1,s_2,s_3,s_4\in S$. The cardinality of the basis is also the required one.

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    $\begingroup$ By the way, one does not need a basis, a linearly independent subset is enough, and such subsets of cardinal $2^{\aleph_0}$ can be explicitly constructed. $\endgroup$
    – YCor
    Commented Oct 12, 2018 at 15:46
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You can actually make $S$ a (copy of the) Cantor set.

First a claim: if $[a_1,b_1]$, $[a_2,b_2]$, ..., $[a_n,b_n]$ is a finite set of intervals, ordered left-to-right ($b_1<a_2$, $b_2<a_3$, etc) then we can choose points $p_{2i-1}<q_{2i-1}<p_{2i}<q_{2i}$ in $(a_i,b_i)$ with the following property: if $i$, $j$, $k$ and $l$ are distinct and $x\in[p_i,q_i]$, $y\in[p_j,q_j]$, $z\in[p_k,q_k]$, and $w\in[p_l,q_l]$ then six possible distances are distinct. To prove this draw inspiration from the first answer and choose $x_{2i-1}<x_{2i}$ in $(a_i,b_i)$, such that $X=\{x_j:j\le 2n\}$ is linearly independent over $\mathbb{Q}$. Then $d$ is injective on $[X]^2$ and by continuity we can choose the intervals $[p_j,q_j]$ to preserve this for points taken from distinct intervals.

Given this claim follow the common construction of the Cantor set where at stage $n$, instead of deleting the middle thirds of all $2^n$ intervals, you take the next $2^{n+1}$ intervals using the claim.

The resulting Cantor set is as required: given two distinct unordered pairs $\{x,y\}$ and $\{z,w\}$ find a stage where distinct points are in distinct intervals; it follows that $|x-y|\neq|z-w|$.

The existence of such a Cantor set also follows from Mycielski's theorem on Independent sets in topological algebras.

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