Topologist's sine curve

Question

When we will spread the topologist's sine curve into components will receive part $\{(x,\sin \frac{1}{x}:x \in (0,1]\}$

How to argue that part is a borel set?

Answer 1

Since $\sin\frac1x$ is a continuous function $\Bbb R\setminus\{0\}\to \Bbb R$, its graph $\Gamma$ is closed in $(\Bbb R\setminus\{0\})\times\Bbb R$. Your set $A$ is precisely $\Gamma \cap ((0,1]\times\Bbb R)$, which is a closed subset of $(\Bbb R\setminus\{0\})\times\Bbb R$. Since $(\Bbb R\setminus\{0\})\times\Bbb R$ is open in $\Bbb R^2$, $A$ is the intersection of an open subset with a closed subsets. Hence Borel.