If $S \subseteq \mathbb{R}$ has limit point and $x+y, x-y \in S \forall x,y \in S$ then $S$ is dense in $\mathbb{R}$

My attempt (By counterpositive)

If $S$ is not dense in $\mathbb{R}$, then there is a non-empty $O \subseteq_{op} \mathbb{R}$ such that $O \cap S = \emptyset$, then I am supposed to construct, given a point $x \in S$, a neighborhood $V_x$ of $x$ that has infinitely many points of $S$, or $x+y \notin S$ or $x-y \notin S$, given $x,y \in S$. Can I say $V_x = (\mathbb{R} - O) \cap S$ has infinitely many points of $S$ (I know that $S$ is at least countable, because has limit point)? But then I don't know how to conclude about $x+y$ and $x-y$... Any help would be appreciated (even if you give help for a direct or absurd proof).



Let $x$ be a limit point of $S$. Then there exists a sequence of distinct values $\{y_k\} \subset S$ with the property that $y_k \to x$. By hypothesis each difference $y_k - y_j$ belongs to $S$ too.

Fix $\epsilon > 0$. Since $\{y_k\}$ is convergent it is Cauchy, so there exist distinct indices $j,k$ with the property that $|y_j - y_k| < \epsilon$.

Let $y = |y_j - y_k|$. Then $0 < y < \epsilon$ and both $y$ and $-y$ belong to $S$.

By hypothesis $2y = y+y \in S$, and thus so is $3y = y + 2y$, and in general $$\{ny \mid n \in \mathbb Z\} \subset S.$$

However, every $x \in \mathbb R$ has the property that $|x - ny| < \epsilon$ for some $n \in \mathbb Z$. Since $\epsilon > 0$ is arbitrary you can conclude $S$ is dense.

  • $\begingroup$ is the last property that you state the Archimedean property of $\mathbb{R}$? $\endgroup$ – Iconoclast Sep 24 '18 at 17:21
  • 1
    $\begingroup$ Yes. If $x > 0$ there exists $n$ with the property that $ny > x$. The least such $n$ has the property that $ny > x \ge (n-1)y$, implying $0 < ny - x < y < \epsilon$. $\endgroup$ – Umberto P. Sep 24 '18 at 17:40
  • $\begingroup$ To the proposer: $S$ is an additive sub-group of the additive group $\Bbb R$. An additive sub-group of $\Bbb R$ that has a non-zero member but no least positive member is dense in $\Bbb R$. $\endgroup$ – DanielWainfleet Sep 25 '18 at 1:55

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