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The definition of limit point z for a set S in complex analysis states that there exists at least one point of the set inside the deleted neighbourhood of z.Does this imply that all interior points of the set are limit points? Edit: I had written the question other way round in the title.

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A point $z \in S$ is called an interior point of $S$ if there exists some $\epsilon_0 > 0$ such that $\{x \in \mathbb{C} \colon |z-y| < \epsilon_0 \} \subset S$.

One can show that each deleted neighborhood of $z$ has non-empty intersection with $\{x \in \mathbb{C} \colon |z-y| < \epsilon_0 \}$, and hence with $S$ in case $z$ is an interior point.

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