# uncountably many numbers with lower bounded difference?

Does there exist an uncountable set U of reals s.t. $\forall a,b \in U\exists k>0$ giving $|a-b|>k$? This is impossible right?

Because if not, then $\{(a,b)_{\lambda}\}$ will be uncountably many disjoint intervals,which is a contradiction (eg. pick $x_{\lambda}\in (a,b)_{\lambda}\cap \mathbb{Q}$ then $f:\lambda\to x_{\lambda}$ is injective).

• What do you mean by $(a,b)$ in your second to last line? More precisely, given $a\in U$, how do you choose the $b$ that you are using as right end point of the interval? In order to do this you'd need to prove first that for all $a\in U$ there is a point of $U$ closest to $a$ from the right. – Andrés E. Caicedo Feb 22 '14 at 3:04
• Anyway, to fix this, check that there is a $k'>0$ such that for each $a\in U$, the interval centered at $a$ of radius $k'$ contains no other points of $U$. Once you have this, you can implement your argument. – Andrés E. Caicedo Feb 22 '14 at 3:06
• @TKM: just to make sure you understood me. The question as it is has obvious answer "no" because it does not specify $a\neq b$. But if you add $a\neq b$ to the question, then it has obvious answer "yes", because for any $a\neq b$ you can always find such $k$. What you need is to put $\exists k>0$ before $\forall a,b$ and also add $a\neq b$. – Vadim Feb 22 '14 at 3:32

Does there exist an uncountable set U of reals s.t. $\exists k>0$: $\forall a,b \in U$, $a\neq b$ implies $|a-b|>k$?
And your argument is correct: in this case for any $a\in U$: $(a-k/2,a+k/2)\cap U=\{a\}$, and if we peek a point $q_a\in(a-k/2,a+k/2)\cap\mathbb{Q}$, then for $a,b\in U$: $a\neq b$ implies $q_a\neq q_b$, therefore we have uncountable number of distinct rationals.