# The Topology of Open Intervals (-n, n) on the Set of Real Numbers

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 $\tau$.

We will now look at the topology of open intervals of the form $(-n, n)$ with $\emptyset$, $\mathbb{R}$ included on the set of real numbers.

Consider the collection, from $\mathbb{R}$, of open intervals of the form:

(1)Let's verify that $\tau$ is a topology on $\mathbb{R}$.

For the first condition we clearly see that $\emptyset, \mathbb{R} \in \tau$.

For the second condition, notice that:

(2)Therefore any arbitrary union $\displaystyle{\bigcup_{i \in I} U_i}$ for $U_i \in \tau$ for all $i \in I$ of these open intervals in $\tau$ will be the "largest" subset in the union in the nesting above and hence is contained in $\tau$.

For the third condition, from above, we see that any finite intersection $\displaystyle{\bigcap_{i=1}^{n} U_i}$ for $U_i \in \tau$ for all $i \in \{ 1, 2, ..., n \}$ of these open intervals in $\tau$ will be the "smallest" subset in the intersection in the nesting above and hence is contained in $\tau$.

Therefore $(X, \tau)$ is a topological space.