Three examples of possibly closed and bounded sets involving $\sin\big(\frac{1}{x}\big)$ Assume that we have the following sets of numbers in $\mathbb{R}$.
\begin{equation}
E_1 = \bigg\{ x \in \mathbb{R}:\sin^2\bigg(\frac{1}{x}\bigg) =0\bigg\}
\end{equation}
\begin{equation}
E_2 = \bigg\{ x \in \mathbb{R}:\sin^2\bigg(\frac{1}{x}\bigg) \geq 0.000001x^2\bigg\}
\end{equation}
\begin{equation}
E_3 = \bigg\{ x \in \mathbb{R}:\sin^2\bigg(\frac{1}{x}\bigg) =x^2\bigg\}
\end{equation}
The question is whether these sets are closed and bounded. 
I think the first set is closed and bounded since it can be seen that $E_1=\bigg\{x:x=\frac{1}{k\pi}\Big| k \in \mathbb{Z} -\{0\}\bigg\}$. I think that $E_2$ and $E_3$ are bounded since for large values of $x$ $x^2$ is going to be larger that the value on the left hand side. However I can't see if these sets are closed. My intuition is that $E_3$ is closed by looking at the behavior of $\sin^2\big(\frac{1}{x}\big)$ around zero. However I cannot prove this rigorously. Any help/ tips? Thanks!
 A: For $E_1$, as you noted we have $\sin(1/x) = 0$ if and only if $x = 1/(k\pi)$. Now we can see that $0$ is not in $E_1$, but $1/(k\pi) \to 0$. This means that $E_1$ is not closed. It is clear that $E_1$ is bounded.
For $E_2$ you can see that for large $x$ the inequality is false. Therefore you have $E_2$ bounded. The same question arises here: is it possible to attain $0$ as a limit point of a sequence in $E$? If this is true, then $E_2$ is not closed, since $0$ is not in $E_2$. Now, let's look at points where $\sin(1/x) = 1$. We need to have $1/x = \pi/2+2k\pi$ so $x_k = 1/(\pi/2+2k\pi)$. It is not difficult to see that $x_k \to 0$ as $k \to \infty$ and moreover, since $\sin(1/x_k) = 1$ we have $$\sin^2(1/x_k) = 1 \geq 0.000001 x_k^2$$
for $k$ large enough. Therefore $x_k \in E$(for large $k$) and $x_k \to 0$. Since $0$ is not a limit point of $E_2$ we conclude that $E_2$ is not closed.
For $E_3$ we see that $x^2$ needs to be between $0$ and $1$, therefore we have boundedness again. We must ask the same question. Can we have a sequence $x_k \in E$ with $x_k \to 0$? If yes, then there's a contradiction. If you imagine the graph of $\sin(1/x)$ you have something which oscillates a lot towards zero. It is not difficult to see that $y=\sin(1/x)$ intersects $y=x$ infinitely often and we can find $x_k \in E$ as close to zero as we want. Since $0$ is not in $E_3$ we conclude that $E_3$ is not closed.
To make this rigorous, you could study the function $\sin(1/x)-x$ around $0$.
In conclusion, you see that all problems occur when we are close to $0$, because $0$ is not in $E$. In fact we know that if $f$ is a continuous function then the sets $\{f(x) = 0\},\{f(x) = x\},\{f(x) = g(x)\}$ with $g$ continuous, are all closed. Since $\sin(1/x)$ is continuous everywhere but in $0$ it is natural to ask what happens in $0$.
