# Is there a set $S\subseteq [0,1]$ with $|S|=2^{\aleph_0}$ and distinct pairwise distances?

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?

• yes, use Zorn to find a maximal such set and observe it is uncountable. – Uri Bader Oct 12 '18 at 9:39
• but my answer remains correct: replacing "uncountable" with $2^{\aleph_0}$... – Uri Bader Oct 12 '18 at 9:45
• 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? – Uri Bader Oct 12 '18 at 9:58
• 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 – Dominic van der Zypen Oct 12 '18 at 10:00
• 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! – Qfwfq Oct 12 '18 at 11:55

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.

• 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. – YCor Oct 12 '18 at 15:46

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 ($$b_1, $$b_2, etc) then we can choose points $$p_{2i-1} in $$(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} 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.