Consider $A = \{(x, \sin\frac{1}{x}) \mid 0< x \leq 1 \}$, a subset of $\mathbb R^2$. Find int($A$).
We can see graphically that the interior of $A$ is definitely empty, but I want to check by the definition of interior point. Let if possible there exist a point $(a,b) \in \operatorname{int}(A)$. Then there exists an open ball $(p,q) \times (r,s)$ such that $(a,b) \in (p,q)\times (r,s) \subseteq A$ .
Please give me a hint, How to get a contradiction ?
Best Answer
Suppose $r < y < s$. The basic open set $(p,q) \times (r,s)$ contains the line $(p,q) \times \{y\}$. Is it possible that this line is contained in that subset of $\mathbb{R}^2$?