# 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 $\mathbb{R}$

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).

Thanks.

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.

• is the last property that you state the Archimedean property of $\mathbb{R}$? – Iconoclast Sep 24 '18 at 17:21
• 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 $0 < ny - x < y < \epsilon$. – Umberto P. Sep 24 '18 at 17:40
• 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 $\Bbb R$. – DanielWainfleet Sep 25 '18 at 1:55