Consider a topological space \(E\). For subsets \(A, B \subseteq E\) we have the equality \[
A^\circ \cap B^\circ = (A \cap B)^\circ\] and the inclusion \[
A^\circ \cup B^\circ \subseteq (A \cup B)^\circ\] where \(A^\circ\) and \(B^\circ\) denote the interiors of \(A\) and \(B\).

Let’s prove that \(A^\circ \cap B^\circ = (A \cap B)^\circ\).

We have \(A^\circ \subseteq A\) and \(B^\circ \subseteq B\) and therefore \(A^\circ \cap B^\circ \subseteq A \cap B\). As \(A^\circ \cap B^\circ\) is open we then have \(A^\circ \cap B^\circ \subseteq (A \cap B)^\circ\) because \(A^\circ \cap B^\circ\) is open and \((A \cap B)^\circ\) is the largest open subset of \(A \cap B\).

Conversely, \(A \cap B \subseteq A\) implies \((A \cap B)^\circ \subseteq A^\circ\) and similarly \((A \cap B)^\circ \subseteq B^\circ\). Therefore we have \((A \cap B)^\circ \subseteq A^\circ \cap B^\circ\) which concludes the proof of the equality \(A^\circ \cap B^\circ = (A \cap B)^\circ\).

One can also prove the inclusion \(A^\circ \cup B^\circ \subseteq (A \cup B)^\circ\). However, the equality \(A^\circ \cup B^\circ = (A \cup B)^\circ\) doesn’t always hold. Let’s provide a couple of counterexamples.

For the first one, let’s take for \(E\) the plane \(\mathbb R^2\) endowed with usual topology. For \(A\), we take the unit close disk and for \(B\) the plane minus the open unit disk. \(A^\circ\) is the unit open disk and \(B^\circ\) the plane minus the unit closed disk. Therefore \(A^\circ \cup B^\circ = \mathbb R^2 \setminus C\) is equal to the plane minus the unit circle \(C\). While we have \[A \cup B = (A \cup B)^\circ = \mathbb R^2.\]

For our second counterexample, we take \(E=\mathbb R\) endowed with usual topology and \(A = \mathbb R \setminus \mathbb Q\), \(B = \mathbb Q\). Here we have \(A^\circ = B^\circ = \emptyset\) thus \(A^\circ \cup B^\circ = \emptyset\) while \(A \cup B = (A \cup B)^\circ = \mathbb R\).

The union of the interiors of two subsets is not always equal to the interior of the union.

Let’s introduce and describe some properties of the line with two origins.

Let \(X\) be the union of the set \(\mathbb R \setminus \{0\}\) and the two-point set \(\{p,q\}\). The line with two origins is the set \(X\) topologized by taking as base the collection \(\mathcal B\) of all open intervals in \(\mathbb R\) that do not contain \(0\), along with all sets of the form \((-a,0) \cup \{p\} \cup (0,a)\) and all sets of the form \((-a,0) \cup \{q\} \cup (0,a)\), for \(a > 0\).

\(\mathcal B\) is a base for a topology \(\mathcal T\) of \(X\)

Indeed, one can verify that the elements of \(\mathcal B\) cover \(X\) as \[
X = \left( \bigcup_{a > 0} (-a,0) \cup \{p\} \cup (0,a) \right) \cup \left( \bigcup_{a > 0} (-a,0) \cup \{q\} \cup (0,a) \right)\] and that the intersection of two elements of \(\mathcal B\) is the union of elements of \(\mathcal B\) (verification left to the reader).

Each of the spaces \(X \setminus \{p\}\) and \(X \setminus \{q\}\) is homeomorphic to \(\mathbb R\)

Let’s prove it for \(X \setminus \{p\}\). The map \[
\begin{array}{l|rcll}
f : & X \setminus \{p\} & \longrightarrow & \mathbb R \\
& x & \longmapsto & x & \text{for } x \neq q\\
& q & \longmapsto & 0 \end{array}\] is a bijection. \(f\) is continuous as the inverse image of an open interval \(I\) of \(\mathbb R\) is an open subset of \(X\). For example taking \(I=(-b,c)\) with \(0 < b < c\), we have \begin{align*} f^{-1}[I] &= (-b,0) \cup \{q\} \cup (0,c)\\ &= \left( (-b,0) \cup \{q\} \cup (0,b) \right) \cup (b/2,c) \end{align*} One can also prove that \(f^{-1}\) is continuous. Continue reading The line with two origins →

Let’s play with the closure and the interior of sets.

To start the play, we consider a topological space \(E\) and denote for any subspace \(A \subset E\): \(\overline{A}\) the closure of \(A\) and \(\overset{\circ}{A}\) the interior of \(A\).

Warm up with the closure operator

For \(A,B\) subsets of \(E\), the following results hold: \(\overline{\overline{A}}=\overline{A}\), \(A \subset B \Rightarrow \overline{A} \subset \overline{B}\), \(\overline{A \cup B} = \overline{A} \cup \overline{B}\) and \(\overline{A \cap B} \subset \overline{A} \cap \overline{B}\).

Let’s prove it.
\(\overline{A}\) being closed, it is equal to its closure and \(\overline{\overline{A}}=\overline{A}\).

Suppose that \(A \subset B\). As \(B \subset \overline{B}\), we have \(A \subset \overline{B}\). Also, \(\overline{B}\) is closed so it contains \(\overline{A}\), which proves \(\overline{A} \subset \overline{B}\).

Let’s consider \(A,B \in E\) two subsets. As \(A \subset A \cup B\), we have \(\overline{A} \subset \overline{A \cup B}\) and similarly \(\overline{B} \subset \overline{A \cup B}\). Hence \(\overline{A} \cup \overline{B} \subset \overline{A \cup B}\). Conversely, \(A \cup B \subset \overline{A} \cup \overline{B}\) and \(\overline{A} \cup \overline{B}\) is closed. So \(\overline{A \cup B} \subset \overline{A} \cup \overline{B}\) and finally \(\overline{A \cup B} = \overline{A} \cup \overline{B}\).

Regarding the inclusion \(\overline{A \cap B} \subset \overline{A} \cap \overline{B}\), we notice that \(A \cap B \subset \overline{A} \cap \overline{B}\) and that \(\overline{A} \cap \overline{B}\) is closed to get the conclusion.

However, the implication \(\overline{A} \subset \overline{B} \Rightarrow A \subset B\) doesn’t hold. For a counterexample, consider the space \(E=\mathbb R\) equipped with the topology induced by the absolute value distance and take \(A=[0,1)\), \(B=(0,1]\). We have \(\overline{A}=\overline{B}=[0,1]\).

The equality \(\overline{A} \cap \overline{B} = \overline{A \cap B}\) doesn’t hold as well. For the proof, just consider \(A=[0,1)\) and \(B=(1,2]\). Continue reading Playing with interior and closure →