Is $\{x : \sin{\frac{1}{x}} = 0 \}$ open in $\mathbb{R}$? The set consists of elements that satisfy $\frac{1}{x} = n\pi$ (or $x = \frac{1}{n\pi}$), but I can't visualize any open balls around any points because this is a trigonometric function in $\mathbb{R^2}$ and we want to check if this is open in just $\mathbb{R}$.
 A: To check if this is open in $\Bbb R$, first look at the picture in $\Bbb R^{2}$, as you said, and then put a bold dot everywhere where this function crosses the $x$-axis.  Now let everything except the $x$-axis (along with these bold points) fade away.  For each bold point, can you find a ball around it that contains only other bold points and no non-bold points?  If so, the set is open.  If not, the set is not open.
A: Define $f:\Bbb R\setminus\{0\}\to \Bbb R$ by $f(x)=\sin\frac{1}{x}$. Show that $f$ is a continuous function. Then the required set is equal to the set $\{x\in\Bbb R\setminus\{0\}:f(x)=0\}=f^{-1}(\{0\})$ which the inverse image of a closed subset in $\Bbb R$, so is closed in $\Bbb R\setminus\{0\}$. 
Otherwise you can directly show that the given set is closed. As the given set is equal to $\{\frac{1}{n\pi}:n\in\Bbb Z\}\subseteq\Bbb R\setminus\{0\}$ which is a discrete set (i.e. without any limit points, the possible limit point is $0$ which is already being excluded by the definition of the function $f$) and hence closed in $\Bbb R\setminus\{0\}$.
Edit: Note that when we consider  the given set as a subset of $\Bbb R$ (w.r.t. the standard topology), the set fails to be closed in $\Bbb R$ as it misses the only possible limit point, which is in fact  $0$. 
