1. **Problem Statement:** If a subset $Y$ of a topological space $X$ is dense in itself, prove that its closure $\overline{Y}$ is also dense in itself. 2. **Definitions:** - A set $Y$ is **dense in itself** if it contains no isolated points, i.e., every point of $Y$ is a limit point of $Y$. - The **closure** $\overline{Y}$ of $Y$ is the union of $Y$ and its limit points. 3. **Goal:** Show that $\overline{Y}$ has no isolated points. 4. **Proof:** - Take any point $x \in \overline{Y}$. - If $x \in Y$, since $Y$ is dense in itself, $x$ is a limit point of $Y$. - If $x \in \overline{Y} \setminus Y$, then $x$ is a limit point of $Y$ by definition of closure. - In either case, every neighborhood of $x$ contains points of $Y$ different from $x$. - Since $Y \subseteq \overline{Y}$, every neighborhood of $x$ contains points of $\overline{Y}$ different from $x$. - Therefore, $x$ is a limit point of $\overline{Y}$. 5. **Conclusion:** Since every point of $\overline{Y}$ is a limit point of $\overline{Y}$, $\overline{Y}$ is dense in itself. **Final answer:** The closure $\overline{Y}$ of a set $Y$ dense in itself is also dense in itself.