The Initial and Final Segment Topologies

The Initial and Final Segment Topologies

Recall from the Topological Spaces page that a set $X$ an a collection $\tau$ of subsets of $X$ together denoted $(X, \tau)$ is called a topological space if:

• $\emptyset \in \tau$ and $X \in \tau$, i.e., the empty set and the whole set are contained in $\tau$.
• If $U_i \in \tau$ for all $i \in I$ where $I$ is some index set then $\displaystyle{\bigcup_{i \in I} U_i \in \tau}$, i.e., for any arbitrary collection of subsets from $\tau$, their union is contained in $\tau$.
• If $U_1, U_2, ..., U_n \in \tau$ then $\displaystyle{\bigcap_{i=1}^{n} U_i \in \tau}$, i.e., for any finite collection of subsets from $\tau$, their intersection is contained in $\tau$.

On The Topology of Open Intervals on the Set of Real Numbers page we saw that if $\tau = \emptyset \cup \mathbb{R} \cup \{ (-n, n) : n \in \mathbb{Z}, n \geq 1 \}$ then $(X, \tau)$ is a topological space.

We will now look at two topologies known as the initial segment and final segment topologies.

The Initial Segment Topology

 Definition: Consider the set of natural numbers $\mathbb{N}$. The collection of subsets $\tau = \{ \emptyset, \mathbb{N} \} \cup \{ \{1, 2, ..., n \} : n \in \mathbb{N} \}$ from $\mathbb{N}$ is called the Initial Segment Topology.

Let's verify that $(\mathbb{N}, \tau)$ is indeed a topology.

For the first condition, clearly $\emptyset, \mathbb{N} \in \tau$ by the definition of $\tau$.

For the second condition, notice that:

(1)
\begin{align} \quad \emptyset \subset \{ 1 \} \subset \{1, 2 \} \subset ... \subset \{1, 2, ..., n \} \subset ... \subset \mathbb{N} \end{align}

Therefore any arbitrary union $\displaystyle{\bigcup_{i \in I} U_i}$ where $U_i \in \tau$ for all $i \in I$ will be the "largest" subset in the collection (or $\mathbb{N}$ itself) with respect to the nesting above, and hence is contained in $\tau$.

For the third condition, any finite intersection $\displaystyle{\bigcap_{i=1}^{n} U_i}$ where $U_i \in \tau$ for all $i \in \{1, 2, ..., n \}$ will be the "smallest" subset in the collection with respect to the nesting above, and hence is contained in $\tau$.

Therefore $(\mathbb{N}, \tau)$ is a topological space.

The Final Segment Topology

 Definition: Consider the set of natural numbers $\mathbb{N}$. The collection of subsets $\tau = \{ \emptyset, \mathbb{N} \} \cup \{ \{n, n+1, ...\} : n \in \mathbb{N} \}$ from $\mathbb{N}$ is called the Final Segment Topology.

Let's verify that $\tau$ is indeed a topology.

For the first condition, we have that once again $\emptyset, \mathbb{N} \in \tau$ by the definition of $\tau$.

For the second condition, notice that:

(2)
\begin{align} \quad \emptyset \subset ... \subset \{ n, n+1, n+2, ... \} \subset \{n - 1, n, n+1, ... \} \subset ... \subset \{2, 3, 4, ... \} \subset \{ 1, 2, 3, ... \} = \mathbb{N} \end{align}

Therefore, any arbitrary union $\displaystyle{\bigcup_{i \in I} U_i}$ for $U_i \in \tau$ for all $i \in I$ is the "largest" subset in the union (or $\mathbb{N}$) with respect to the nesting above, so it is contained in $\tau$.

For the third condition, any finite intersection $\displaystyle{\bigcap_{i=1}^{n} U_i}$ for $U_i \in \tau$ for all $i \in \{1, 2, ... n \}$ is the "smallest" subset in the intersection with respect to the nesting above, so it is contained in $\tau$.