Open Sets#

We need to prove three things:

• (I) The empty set \(\varnothing\) and \(X\) itself are open.
• (II) If \(\mathcal{S}\) is a collection of open sets, then its union is also open.
• (III) If \(\mathcal{S}\) is a finite collection of open sets, then its intersection is also open.

We assume that the topological space is characterized by a neighborhood system \(\mathcal{N}\).

Proof of (I):

To prove that \(\varnothing\) is open, we must show that \(\varnothing \in \mathcal{N}(x)\) for all \(x \in \varnothing\). This is vacuously true because \(\varnothing\) contains no elements. Thus, \(\varnothing\) is open.

To prove \(X\) is open, let \(x \in X\). By definition, the neighborhood system \(\mathcal{N}(x)\) has at least one neighborhood \(N \in \mathcal{N}(x)\). Since \(N\) is a subset of \(X\), one of the defining properties of a neighborhood system tells us that \(X\) is also a neighborhood of \(x\). Since this holds for all \(x \in X\), we know that \(X\) is open.

Proof of (II):

To show that \(\bigcup \mathcal{S}\) is open, we need to show that \(\bigcup \mathcal{S} \in \mathcal{N}(x)\) for each \(x \in \bigcup \mathcal{S}\).

Let \(x \in \bigcup \mathcal{S}\). By the definition of a union, there exists at least one \(V \in \mathcal{S}\) such that \(x \in V\). Since \(\mathcal{S}\) is a collection of open sets, then \(V\) must be open. Since \(x \in V\), the definition of an open set tells us that \(V\) is a neighborhood of \(x\). Since \(V \subseteq \bigcup \mathcal{S}\) and \(V \in \mathcal{N}(x)\), one of the defining properties of a neighborhood system tells us that \(\bigcup \mathcal{S}\) is also a neighborhood of \(x\).

This holds for all \(x \in \bigcup \mathcal{S}\), so \(\bigcup \mathcal{S}\) is open.

Proof of (III):

If \(\mathcal{S}\) is empty (\(|\mathcal{S}| = 0\)), its intersection \(\bigcap \mathcal{S}\) is \(X\) itself, which is an open set.

If \(\mathcal{S}\) is non-empty, then let \(\mathcal{S} = \{U_1, U_2, \dots, U_n\}\) and let \(x \in \bigcap \mathcal{S}\). By the definition of an intersection, we know that \(x \in U_i\) for every \(i \in \{1, 2, \dots, n\}\).

Since \(\mathcal{S}\) is a collection of open sets, we know that each \(U_i\) is open. Since \(x \in U_i\) for every \(i \in \{1, 2, \dots, n\}\), it follows that \(U_1, \dotsc, U_n\) are neighborhoods of \(x\). Therefore, one of the defining properties of a neighborhood system tells us that \(\bigcap \mathcal{S} = U_1 \cap U_2 \cap \dots \cap U_n\) is a neighborhood of \(x\).

Since this is true for all \(x \in \bigcap \mathcal{S}\), it follows that \(\bigcap \mathcal{S}\) is open.

We need to prove two things:

• (I) If \(N\) is a neighborhood of \(x\), then there exists some open set \(U\) with \(x \in U\) and \(U \subseteq N\).
• (II) If there exists some open set \(U\) with \(x \in U\) and \(U \subseteq N\), then \(N\) is a neighborhood of \(x\).

Proof of (I):

Let \(U\) be the subset of \(X\) containing all points which have \(N\) as a neighborhood:

Since \(N\) is a neighborhood of \(x\) by assumption, it immediately follows that \(x \in U\). Furthermore, one of the defining properties of a neighborhood system tells us that if \(p \in X\), then \(p\) is contained in all of its neighborhoods, i.e. if \(N \in \mathcal{N}(p)\), then \(p \in N\). Therefore, \(U \subseteq N\).

We have thus shown that \(x \in U\) and \(U \subseteq N\). Finally, we must show that \(U\) is open.

Let \(p \in U\). By the definition of \(U\), we know that \(N \in \mathcal{N}(p)\). The defining property of a neighborhood system tells us that there exists some \(M \in \mathcal{N}(p)\) with \(M \subseteq N\) such that \(N \in \mathcal{N}(p')\) for all \(p' \in M\). The condition \(N \in \mathcal{N}(p')\) is the defining property of \(U\) and so \(p' \in U\) for all \(p' \in M\). Therefore, \(M \subseteq U\). Since \(M\) is a neighborhood of \(p\) and since \(M \subseteq U\), one of the defining properties of a neighborhood system tells us that \(U\) is also a neighborhood of \(p\). Since this holds for all \(p \in U\), it follows that \(U\) is open.

Proof of (II):

The definition of an open set tells us that \(U\) is a neighborhood of \(x\). Therefore, one of the defining properties of a neighborhood system tells us that \(N\) must also be a neighborhood of \(x\), since \(U \subseteq N\).

Let \(\mathcal{N}: X \to \mathcal{P}(\mathcal{P}(X))\) be defined according to the theorem which identifies neighborhoods and open sets:

We need to prove three things:

• (I) \(\mathcal{N}\) is a neighborhood system on \(X\).
• (II) The open sets w.r.t. \(\mathcal{N}\) are precisely the elements of \(\mathcal{U}\).
• (III) \(\mathcal{N}\) is unique.

Proof of (I):

We need to prove the four defining properties of a neighborhood system:

• If \(N \in \mathcal{N}(x)\), then \(x \in N\): Let \(x \in X\) and let \(N \in \mathcal{N}(x)\). By the definition of \(\mathcal{N}\), there exists some \(U \in \mathcal{U}\) with \(x \in U \subseteq N\). Therefore, \(x \in N\).
• If \(N \in \mathcal{N}(x)\) and \(N \subseteq M \subseteq X\), then \(M \in \mathcal{N}(x)\): Let \(x \in X\) and let \(N \in \mathcal{N}(x)\). By the definition of \(\mathcal{N}\), there exists some \(U \in \mathcal{U}\) such that \(x \in U \subseteq N\). Since \(N \subseteq M\), we have \(U \subseteq M\) and so \(x \in U \subseteq M\). Therefore, \(M \in \mathcal{N}(x)\).
• If \(N_1, \dotsc, N_p \in \mathcal{N}(x)\), then \(N_1 \cap \cdots \cap N_p \in \mathcal{N}(x)\). Let \(x \in X\) and let \(N_1, \dotsc, N_p \in \mathcal{N}(x)\). By the definition of \(\mathcal{N}\), there exist some \(U_1, \dotsc, U_p \in \mathcal{U}\) such that \(x \in U_1 \subseteq N_1, \dotsc, x \in U_p \subseteq N_p\). Consequently, \(x \in U_1 \cap \cdots \cap U_p \subseteq N_1 \cap \cdots \cap N_p\). From the third assumption for \(\mathcal{U}\), it follows that \(U_1 \cap \cdots \cap U_p \in \mathcal{U}\). Therefore, \(N_1 \cap \cdots \cap N_p \in \mathcal{N}(x)\).
• If \(N \in \mathcal{N}(x)\), then there exists some \(M \in \mathcal{N}(x)\) with \(M \subseteq N\) and \(N \in \mathcal{N}(y)\) for each \(y \in M\): Let \(x \in X\) and let \(N \in \mathcal{N}(x)\). By the definition of \(\mathcal{N}\), there exists some \(U \in \mathcal{U}\) such that \(x \in U \subseteq N\). Let \(M = U\). We have \(x \in U \subseteq M\) and so \(M \in \mathcal{N}(x)\). For each \(y \in M\) (i.e. \(y \in U\)), we have \(y \in U \subseteq N\) and so \(N \in \mathcal{N}(y)\).

Proof of (II):

Let \(\mathcal{U}_{\mathcal{N}}\) be the collection of open sets w.r.t. the neighborhood system \(\mathcal{N}\). By definition, \(O \in \mathcal{U}_{\mathcal{N}}\) if and only if \(O \in \mathcal{N}(x)\) for each \(x \in O\). We need to show that \(\mathcal{U}_{\mathcal{N}} = \mathcal{U}\).

Let \(U \in \mathcal{U}\) and let \(x \in U\). We have \(x \in U \subseteq U\) and so the definition of \(\mathcal{N}\) tells us that \(U \in \mathcal{N}(x)\). This holds for every \(x \in U\) and so \(U \in \mathcal{U}_{\mathcal{N}}\), i.e. \(\mathcal{U} \subseteq \mathcal{U}_{\mathcal{N}}\).

Let \(O \in \mathcal{U}_{\mathcal{N}}\). By the definition of \(\mathcal{U}_{\mathcal{N}}\), for each \(x \in O\), we have \(O \in \mathcal{N}(x)\). Therefore, using the definition of \(\mathcal{N}\), we get that for each \(x \in O\), there exists some \(U_x \in \mathcal{U}\) such that \(x \in U_x \subseteq O\). We can express \(O\) as the union \(O = \bigcup_{x \in O} \{x\}\). Since \(\{x\} \subseteq U_x \subseteq O\) for each \(x \in O\), we have \(O = \bigcup_{x \in O} U_x\). The second assumption for \(\mathcal{U}\) implies that \(\bigcup_{x \in O} U_x \in \mathcal{U}\) and so \(O \in \mathcal{U}\), i.e. \(\mathcal{U}_{\mathcal{N}} \subseteq \mathcal{U}\).

Since \(\mathcal{U} \subseteq \mathcal{U}_{\mathcal{N}}\) and \(\mathcal{U}_{\mathcal{N}} \subseteq \mathcal{U}\), we have \(\mathcal{U} = \mathcal{U}_{\mathcal{N}}\).

Proof of (III):

Suppose \(\mathcal{N}'\) is another neighborhood system on \(X\) which generates the exact same open sets \(\mathcal{U}\) as \(\mathcal{N}\). We need to show that \(\mathcal{N}(x) = \mathcal{N}'(x)\) for all \(x \in X\).

Let \(N \in \mathcal{N}(x)\). By the definition of \(\mathcal{N}\), there exists some \(U \in \mathcal{U}\) such that \(x \in U \subseteq N\). By assumption, \(U\) must be open w.r.t. \(\mathcal{N}'\). Therefore, \(U\) is considered a neighborhood of \(x\) under \(\mathcal{N}'\), i.e. \(U \in \mathcal{N}'(x)\). Since \(U \subseteq N\) and \(U \in \mathcal{N}'(x)\), one of the defining properties of neighborhood systems tells us that \(N \in \mathcal{N}'(x)\). Thus, \(\mathcal{N}(x) \subseteq \mathcal{N}'(x)\).

Let \(N \in \mathcal{N}'(x)\). Let \(V = \{y \in X \mid N \in \mathcal{N}'(y)\}\). Let \(z \in V\). By the definition of \(V\), we know that \(N \in \mathcal{N}'(z)\). One of the defining properties of neighborhood system tells us that there exists some \(M \in \mathcal{N}'(z)\) with \(M \subseteq N\) such that \(N \in \mathcal{N}'(w)\) for all \(w \in M\). However, this is the defining condition for being an element of \(V\). Therefore, \(w \in V\) for all \(w \in M\) and so \(M \subseteq V\). Since \(M\) is a neighborhood of \(z\) and since \(M \subseteq V\), another of the defining properties of neighborhood systems tells us that \(V\) must be a neighborhood of \(z\) w.r.t. \(\mathcal{N}'\). This holds for all \(z \in V\) and so \(V\) is open w.r.t. \(\mathcal{N}'\), i.e. \(V \in \mathcal{U}\). We see that \(x \in V\), i.e. \(x \in V \subseteq N\), and since \(V \in \mathcal{U}\), it follows that \(N \in \mathcal{N}(x)\) by the definition of \(\mathcal{N}\). Therefore, \(\mathcal{N}'(x) \subseteq \mathcal{N}(x)\).

Since \(\mathcal{N}(x) \subseteq \mathcal{N}'(x)\) and \(\mathcal{N}'(x) \subseteq \mathcal{N}(x)\), we get that \(\mathcal{N}'(x) = \mathcal{N}(x)\).

The above holds for all \(x \in X\) and so \(\mathcal{N} = \mathcal{N}'\).

TODO

TODO