I want to show that if an open set $G$ dense in $\mathbb{R}$ then $G-{x}$ is dense in $\mathbb{R}$ for all $x\in \mathbb{R}$

So lets assume that $G$ is dense in $\mathbb{R}$. Then, given $x \in \mathbb{R}$, $N_{\epsilon}(x) \cap G$ $\neq \emptyset$ for all $ x \in \mathbb{R}$

Let $y \in N_{\epsilon}(x) \cap G$. Then, since $G$ is an open set, $\exists \epsilon_2>0$ such that $N_{\epsilon_2}(y) \subset G$.

Therefore, $N_{\epsilon}(x) \cap N_{\epsilon_2}(y)$ is not empty, and since this is the nonempty intersection of two open subsets of $\mathbb{R}$, it must contain infinitely many points.

Thus, given any $x\in\mathbb{R}$, $N_{\epsilon}(x) \cap G$ contains infinitely many points $\forall \epsilon>0$ and therefore removing one of them won't be a problem, and thus $G-x$ is dense in $\mathbb{R}$.

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    $\begingroup$ @bof I think the exercise is about set-theoretic difference rather than translation. $\endgroup$ – Andrés E. Caicedo Jun 16 '18 at 2:44
  • $\begingroup$ It is simpler to verify that $y \to y-x$ is a homeomorphism.and hence preserves denseness. $\endgroup$ – Kavi Rama Murthy Jun 16 '18 at 4:58
  • $\begingroup$ Please clarify that this is about the set-theoretic difference and (if so) write it $G-\{x\}.$ $\endgroup$ – spaceisdarkgreen Jun 16 '18 at 8:39

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