Given a set $S$ in $\mathbb{R}^n$ and a point $x$ with the property that every ball $B(x;r)$ contains both interior and exterior points of $S$. Prove that $x$ is a boundary point of S.
My attempt
Proof. Suppose $x$ is not a boundary point, then $x$ is either an interior point or an exterior point. Assume $x$ is an interior point for this case, the case where $x$ is an exterior point follows the same argument. Since $x$ is an interior point, there is an open ball $B(x;a)$ that is a subset of $B(x;r)$. But this implies that $x$ is also an exterior point, which is a contradiction. Hence $x$ cannot be an interior point, by the same argument $x$ cannot be an exterior point. Thus x is a boundary point. QED
Any suggestions or comments?
Answer
Yes your proof certainly is correct. But that is the definition of boundary point, proof wasn't needed.
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