Let $T = (X, \mathcal{T})$ be a D1106: Topological space such that

(i) | $U \in \mathcal{T}$ is an D97: Open set in $T$ |

Then
\begin{equation}
\text{int} \langle U \rangle = U
\end{equation}

Result R4541
on D519: Set interior

Open set is its own interior

Formulation 0

Let $T = (X, \mathcal{T})$ be a D1106: Topological space such that

(i) | $U \in \mathcal{T}$ is an D97: Open set in $T$ |

Then
\begin{equation}
\text{int} \langle U \rangle = U
\end{equation}

Subresults

▶ | R1149: Every point in open set is an interior point |

Proofs

Let $T = (X, \mathcal{T})$ be a D1106: Topological space such that

(i) | $U \in \mathcal{T}$ is an D97: Open set in $T$ |

By definition
\begin{equation}
\text{int} \langle U \rangle : = \bigcup \{ V \in \mathcal{T} : V \subseteq U \}
\end{equation}
Result R125: Subset relation is reflexive shows that $U$ itself satisfies $U \subseteq U$. Since $U$ is also an open set, then $U$ is a set in the union $\text{int} \langle U \rangle$ and result R4142: Union is an upper bound to each set in the union implies that $U \subseteq \text{int} \langle U \rangle$.

Conversely, result R3945: Set is a superset to its interior guarantees that $\text{int} \langle U \rangle \subseteq U$. Since we have inclusions in both directions, the claim then follows as a consequence of R2741: Set equality iff inclusion in both directions. $\square$

Conversely, result R3945: Set is a superset to its interior guarantees that $\text{int} \langle U \rangle \subseteq U$. Since we have inclusions in both directions, the claim then follows as a consequence of R2741: Set equality iff inclusion in both directions. $\square$