Proof that the graph of $\sin(\frac{1}{x + 1})$ is not a closed set A set is closed if a point being not a limit point implies it is an exterior point. 
Using this definition, I am trying to understand that the graph of $\sin(\frac{1}{x + 1})$ is not a closed set. On other questions, this set is given as an example for a bounded but non-closed set. I can see that on a compact domain that includes the point $-1$, say $[-2, 0]$, grahp being also a compact set implies this function is continuous. This function is not continuous at $x = -1$. Thus the graph is not compact. Then it is not closed.
This type of argumentation I can follow. But I am looking for an explanation that goes to the definition. Because I want to understand the nature of the point $(-1, f(-1))$. If I redefined the function piece-wise as follows; 
$$f(x) =
  \begin{cases}
    \sin(\frac{1}{x + 1}) & \text{if }  x \neq -1 \\
    0 & \text{otherwise} \\
  \end{cases}
$$
This is still not continuous at $x = -1$. Since this function oscillates bizarrely around that point, I think I can't complete this function at that point to be continuous. 
All the other points in the graph being a limit point, this $(-1, f(-1))$ is the only point I am having problem. For this graph to be not closed, this point should be a limit point but not an exterior point. How is this so? How can we construct a sequence in this graph that converges to a point outside the graph like $(-1, 1)$ or $(-1, -1)$?
 A: The sequence of points $(-1+1/(\pi/2 + 2n\pi),1), n=1,2,\dots$ all lie in the graph $f,$ and approach $(-1,1)$ as $n\to \infty.$ But $(-1,1)$ is not in the graph of $f.$ Therefore the graph of $f$ does not contain all its limit points, which implies it is not a closed subset of $\mathbb R^2.$
A: Fix some constant $c \in [-1,1]$. Since $y \mapsto \sin(y)$ is periodic there exists a sequence $(y_n)_{n \in \mathbb{N}}$ such that $y_n \to \infty$ and $\sin(y_n)=c$ for all $n \geq 1$. If we define $x_n \in (-1,\infty)$ by $$y_n = \frac{1}{1+x_n}$$ then $x_n \to -1$ and $$f(x_n)=\sin \left( \frac{1}{x_n+1} \right) = \sin(y_n)= c \qquad \text{for all $n \geq 1$.}$$ Thus, $$(x_n,f(x_n)) \xrightarrow[]{n \to \infty} (-1,c).$$ As $c \in [-1,1]$ was arbitrary, this shows that the line $\{-1\} \times [-1,1]$ is contained in the closure of the graph. In particular, the graph of $f$ is not closed (no matter how we choose $f(-1)$).
A: Note that if $x_n=-\frac{2}{(4n+1)\pi}-1$ then $x_n\in [-2,-1]$, $x_n\rightarrow -1$ but $\sin (\frac{1}{1+x_n})\rightarrow -1\neq 0$.
