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In order to give an example of a function from $\mathbb{R}$ to itself whose graph is connected in $\mathbb{R} \times \mathbb{R}$, yet is not continuous, the book Berkley Problems on Mathematics refers to a function: $f(x)=\sin(\frac{1}{x})$.

Let $S_1=\left\{(x,\sin\left(\frac{1}{x}\right))\mid x <0\right\}, \ S_2=\left\{(x,\sin\left(\frac{1}{x}\right)) \mid x >0\right\}.$ $S_1$ and $S_2$ are both connected subsets of $\mathbb{R} \times \mathbb{R}$. Since $(0,0)$ belongs to the closure in $\mathbb{R} \times \mathbb{R}$ of both $S_1$ and $S_2$, the sets $S_1 \cup \{(0,0)\}$ and $S_2 \cup \{(0,0)\}$ are connected. Since these sets have a point in common, their union $(S_1 \cup \{(0,0)\}) \cup (S_2 \cup \{(0,0)\})$ is connected. This union is the graph of the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x)=\sin(\frac{1}{x})$, $x \neq 0$, $f(0)=0$, which is not continuous at $x=0$.

In this example, what I don't understand is why $(0,0)$ belongs to the closure of $S_1$ and $S_2$. As I know, the function $f(x)$ vibrates strongly near $0$, so how can the graph get near $(0,0)$? How can $(0,0)$ be in the closure of the two sets?

Many thanks!

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For every natural $n$, $S_2$ contains the point $(1/ (n \pi),0)$. As $n$ tends to infinity, these points tend to $(0,0)$, so $(0,0)$ is in the closure of $S_2$. Similarly, $S_1$ contains $(-1/ (n \pi),0$).

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    Alternatively, any neighborhood of$(0,0)$intersects both $S_1$ and $S_2$, making it a limit point of each of the sets, so that (0,0) must belong to the closure of both.2011-04-20