Why is the discrete topology called discrete?

Recall from the Topological Spaces page that a set $X$ and 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 two rather trivial topologies known as the discrete topologies and the indiscrete topologies.

Here, the notation "$\mathcal P[X] = \{ Y : Y \subseteq X \}$" represents the power set of $X$ or rather, the set of all subsets of $X$.

Let's verify that $[X, \tau] = [X, \mathcal P[X]]$ is indeed a topological space. Since $\emptyset \subseteq X$ and $X \subseteq X$, we clearly have that $\emptyset, X \subseteq \mathcal P[X]$, so the first condition holds.

Now consider any arbitrary collection of subsets $\{ U_i \}_{i \in I}$ from $\mathcal P[X]$ for some index set $I$. Suppose that $\displaystyle{\bigcup_{i \in I} U_i \not \in \mathcal P[X]}$. Then $\displaystyle{\bigcup_{i \in I} U_i \not \subseteq X}$ and so there exists an element $\displaystyle{x \in \bigcup_{i \in I} U_i}$ such that $x \not \in X$. Say that $x \in U_j$ for some $j \in I$. Then $U_j \not \subseteq X$, which contradicts the fact that $\{ U_i \}_{i \in I}$ is an arbitrary collection of subsets from $\mathcal P[X] = \{ U : U \subseteq X \}$. Therefore $\displaystyle{\bigcup_{i \in I} U_i \in \mathcal P[X]}$.

Lastly, consider any finite collection of subsets $U_1, U_2, ..., U_n$ of $\mathcal P[X]$. Suppose that $\displaystyle{\bigcap_{i=1}^{n} U_i \not \in \mathcal P[X]}$. Then $\displaystyle{\bigcap_{i=1}^{n} U_i \not \subseteq X}$, so there exists an $\displaystyle{x \in \bigcap_{i=1}^{n} U_i}$ such that $x \not \in X$. So $x \in U_j$ for all $j \in \{1, 2, ..., n \}$. But then $U_j \not \subseteq X$ for all $j \in \{ 1, 2, ..., n \}$ which contradicts the fact that $U_1, U_2, ..., U_n$ are a collection of subsets of $\mathcal P[X]$. Therefore $\displaystyle{\bigcap_{i=1}^{n} U_i \in \mathcal P[X]}$.

Since all three conditions for $\tau = \mathcal P[X]$ hold, we have that $[X, \mathcal P[X]]$ is a topological space.

One again, let's verify that $[X, \tau] = [X, \{ \emptyset, X \}]$ is indeed a topological space.

For the first condition, we clearly see that $\emptyset \in \{ \emptyset, X \}$ and $X \in \{ \emptyset, X \}$.

For the second condition, the only possible unions are $\emptyset \cup \emptyset = \emptyset \in \{ \emptyset, X \}$, $\emptyset \cup X = X \in \{ \emptyset, X \}$, and $X \cup X = X \in \{ \emptyset, X \}$.

For the third condition, the only possible intersections are $\emptyset \cap \emptyset = \emptyset \in \{ \emptyset, X \}$, $\emptyset \cap X = \emptyset \in \{ \emptyset, X \}$, and $X \cap X = X \in \{ \emptyset, X \}$.

Since all three conditions for $\tau = \{ \emptyset, X \}$ hold, we have that $[X, \{ \emptyset, X \}]$ is a topological space.