----- > [!proposition] Proposition. ([[open iff boundary is closure minus set]]) > Let $X$ be a [[topological space]] and $U \subset X$. Then, $U$ is [[open set|open in]] $X$ if and only if $\text{Bd }U= \overline{U}-U.$ > [!basicnonexample] > We really need that $U$ is open ($U=\text{Int }U$). To see why, let $\mathbb{I}$ denote the irrationals in $\mathbb{R}$. $\overline{ \mathbb{I}}=\mathbb{R}$ and $\overline{\mathbb{R}- \mathbb{I}}= \overline{\mathbb{Q}}=\mathbb{R}$, hence $\text{Bd }\mathbb{I}=\mathbb{R}$ while $\overline{\mathbb{I}}-\mathbb{I}=\mathbb{R} - \mathbb{I}=\mathbb{Q} \neq \mathbb{R}$. > [!proof]- Proof. ([[open iff boundary is closure minus set]]) > $\to$. Suppose $U$ is [[open set|open in]] $X$. Noting that this implies $X-U=\overline{X-U}$, we have$\begin{align} x \in \text{Bd }U \iff & x \in \overline{U} \cap \overline{X-U} \\ \iff & x \in \overline{U} \cap (X-U) \\ \iff & x \in X \cap (\overline{U}-U) \\ \iff & x \in \overline{U}-U. \end{align}$ $\leftarrow$. Suppose $\text{Bd }U= \overline{U} - U$, i.e., $\overline{U} \cap \overline{X-U} = \overline{U} - U$. Since $\overline{U} - U = X \cap (\overline{U} - U) = \overline{U} \cap (X-U) = \overline{U} \cap ({X-U}),$ we see that $\overline{U} \cap \overline{X-U} = \overline{U} \cap (X-U)$. Hence $X-U=\overline{X-U}$, implying that $X-U$ is [[clopen set|closed in]] $X$. Thus $U$ is [[open set|open in]] $X$. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```