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[parent] derivation of properties on interior operation (Derivation)

Let $ X$ be a topological space and $ A$ a subset of $ X$. Then

  1. $ \operatorname{int}(A)\subseteq A$.
    Proof. If $ a\in \operatorname{int}(A)$, then $ a\in U$ for some open set $ U\subseteq A$. So $ a\in A$. $ \qedsymbol$
  2. $ \operatorname{int}(A)$ is open.
    Proof. Since $ \operatorname{int}(A)$ is a union of open sets, $ \operatorname{int}(A)$ is open. $ \qedsymbol$
  3. $ \operatorname{int}(A)$ is the largest open set contained in $ A$.
    Proof. If $ U$ is open set with $ \operatorname{int}(A)\subseteq U\subseteq A$, then $ U\subseteq\bigcup \lbrace V\subseteq A\mid V$ open $ \rbrace = \operatorname{int}(A)$, so $ U=\operatorname{int}(A)$. $ \qedsymbol$
  4. $ A$ is open if and only if $ A=\operatorname{int}(A)$.
    Proof. If $ A$ is open, then $ A$ is the largest open set contained in $ A$, and so $ \operatorname{int}(A)=A$ by property 3 above. On the other hand, if $ \operatorname{int}(A)=A$, then $ A$ is open, since $ \operatorname{int}(A)$ is, by property 2 above. $ \qedsymbol$
  5. $ \operatorname{int}(\operatorname{int}(A))=\operatorname{int}(A)$.
    Proof. Since $ \operatorname{int}(A)$ is open by property 2, $ \operatorname{int}(A)=\operatorname{int}(\operatorname{int}(A))$ by property 4. $ \qedsymbol$
  6. $ \operatorname{int}(X)=X$ and $ \operatorname{int}(\varnothing )=\varnothing $.
    Proof. This is so because both $ X$ and $ \varnothing $ are open sets. $ \qedsymbol$
  7. $ \overline{A^\complement}=(\operatorname{int}(A))^\complement$.
    Proof. (LHS $ \subseteq$ RHS). If $ a\in \overline{A^\complement}$, then $ a\in B$ for every closed set $ B$ such that $ A^\complement \subseteq B$. In particular, $ a\in (\operatorname{int}(A))^\complement$, for $ (\operatorname{int}(A))^\complement$ is the complement of an open set by property 2, and $ A^\complement \subseteq (\operatorname{int}(A))^\complement$ by taking the complement of property 1.

    (RHS $ \subseteq$ LHS). If $ a\in (\operatorname{int}(A))^\complement$, then $ a\notin \operatorname{int}(A)$. If $ B$ is a closed set such that $ A^\complement \subseteq B$, then $ B^\complement \subseteq A$. Since $ B^\complement$ is open, $ B^\complement \subseteq \operatorname{int}(A)$ by property 3, so $ a\notin B^\complement$, and thus $ a\in B$. Since $ B$ is arbitrary, $ a\in \overline{A^\complement}$ as desired. $ \qedsymbol$

  8. $ \overline{A}^\complement = \operatorname{int}(A^\complement)$.
    Proof. Set $ B=A^\complement$, and apply property 7. So $ \overline{A}^\complement = \overline{B^\complement}^\complement = (\operatorna... ...\complement\complement}=\operatorname{int}(B)=\operatorname{int}(A^\complement)$. $ \qedsymbol$
  9. $ A\subseteq B$ implies that $ \operatorname{int}(A)\subseteq \operatorname{int}(B)$.
    Proof. This is so because $ \operatorname{int}(A)$ is open (property 2), contained in $ A$ (and therefore contained in $ B$), so contained in $ \operatorname{int}(B)$, as $ \operatorname{int}(B)$ is the largest open set contained in $ B$ (property 3). $ \qedsymbol$
  10. $ \operatorname{int}(A)=A\setminus \partial A$, where $ \partial A$ is the boundary of $ A$.
    Proof. Recall that $ \partial A=\overline{A}\cap \overline{A^\complement}$. So $ \partial A = \overline{A}\cap (\operatorname{int}(A))^\complement$ by property 7. By direct computation, we have $ A\setminus \partial A = A \setminus (\overline{A}\cap (\operatorname{int}(A))^... ... (A\setminus \overline{A})\cup (A\setminus (\operatorname{int}(A))^\complement)$. Since $ A\setminus \overline{A}=\varnothing$ and $ A\setminus (\operatorname{int}(A))^\complement = A\cap (\operatorname{int}(A))^{\complement\complement}= A\cap \operatorname{int}(A)$, which is $ \operatorname{int}(A)$ by property 2. $ \qedsymbol$
  11. $ \overline{A} = \operatorname{int}(A)\cup \partial A$.
    Proof. Again, by direct computation:
    $\displaystyle \operatorname{int}(A)\cup \partial A$ $\displaystyle = \operatorname{int}(A)\cup (\overline{A}\cap (\operatorname{int}(A))^\complement)$    because $\displaystyle \partial A = \overline{A}\cap (\operatorname{int}(A))^\complement$    
      $\displaystyle = (\operatorname{int}(A)\cup \overline{A})\cap (\operatorname{int}(A)\cup (\operatorname{int}(A))^\complement)$ $\displaystyle \qquad \cap$ distributes over $\displaystyle \cup$    
      $\displaystyle =\overline{A}\cap X=\overline{A}.$ $\displaystyle \qquad \operatorname{int}(A)\subseteq A\subseteq \overline{A}$    

    $ \qedsymbol$
  12. $ X=\operatorname{int}(A)\cup \partial A \cup \operatorname{int}(A^\complement)$.
    Proof. By property 11, $ \operatorname{int}(A)\cup \partial A \cup \operatorname{int}(A^\complement) = \overline{A} \cup \operatorname{int}(A^\complement)$, which, by property 8, is $ \overline{A} \cup \overline{A}^\complement$, and the last expression is just $ X$. $ \qedsymbol$
  13. $ \operatorname{int}(A\cap B)=\operatorname{int}(A)\cap \operatorname{int}(B)$.
    Proof. (LHS $ \subseteq$ RHS). Let $ C=\operatorname{int}(A\cap B)$. Since $ C$ is open and contained in both $ A$ and $ B$, $ C$ is contained in both $ \operatorname{int}(A)$ and $ \operatorname{int}(B)$, since $ \operatorname{int}(A)$ and $ \operatorname{int}(B)$ are the largest open sets in $ A$ and $ B$ respectively. (RHS $ \subseteq$ LHS). Let $ D=\operatorname{int}(A)\cap \operatorname{int}(B)$. So $ D$ is open and is a subset of both $ A$ and $ B$, hence a subset of $ A\cap B$, and therefore a subset of $ \operatorname{int}(A\cap B)$, since it is the largest open set contained in $ A\cap B$. $ \qedsymbol$

Remark. Using property 7, we see that an alternative definition of interior can be given:

$\displaystyle \operatorname{int}(A)=\overline{A^\complement}^\complement.$



"derivation of properties on interior operation" is owned by CWoo. [ full author list (2) ]
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Cross-references: interior, expression, boundary, implies, complement, closed set, property, contained, union, open, open set, subset, topological space
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This is version 6 of derivation of properties on interior operation, born on 2008-03-19, modified 2008-03-20.
Object id is 10418, canonical name is DerivationOfPropertiesOnInteriorOperation.
Accessed 127 times total.

Classification:
AMS MSC: 54-00 (General topology :: General reference works )
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