# General Topology: Topologies

## Topologies on R

Now that we've gotten a few abstract definitions under our belt, let's get some actual topologies to work with. The set of real numbers provides a fine starting point.

The **standard topology** on $\mathbb{R}$ is defined by the basis whose elements are all the bounded open intervals of $\mathbb{R}$. Formally, the standard topology is defined by the basis

$$\mathcal{B}_0 = \{ (a, b) : a, b \in \mathbb{R} \}.$$

The standard topology "standard" because it is the topology that is assumed when discussing "the" topology on $\mathbb{R}$ in the absence of any additional context. Other names such as the **typical** or **usual** topology refer to the standard topology.

Two additional topologies are the lower limit and upper limit topologies. The **lower limit** **topology** on $\mathbb{R}$, denoted $\mathbb{R}_l$ is defined by the basis $\mathcal{B}_1$ of all left-closed intervals:

$$\mathcal{B}_1 = \{ [a, b) : a, b \in \mathbb{R} \}.$$

Likewise, the **upper limit topology** is defined by the basis $\mathcal{B}$ of all right-closed intervals:

$$\mathcal{B}_2 = \{ (a, b] : a, b \in \mathbb{R} \}.$$

All three of these topologies are made in terms of intervals. A more exotic topology is the K-topology on $\mathbb{R}$. Let $K = \{ \frac{1}{n} : n \in \mathbb{N} \}$. Then the **K-topology** is defined by the basis consisting of all open intervals in $\mathbb{R}$ along with all differences between open intervals an $K$. Formally, the basis is defined as

$$\mathcal{B}_3 = \mathcal{B}_0 \cup \{ (a, b) - K : a, b \in \mathbb{R} \}.$$

The K-topology on $\mathbb{R}$ is denoted $\mathbb{R}_K$.

## Problems

Confirm that the basis $\mathcal{B}_0$ for the standard topology on $\mathbb{R}$ is a basis.

For the first condition, let $x \in \mathbb{R}$. Then $(x-1, x + 1) \in \mathcal{B}_0$ and $x \in (x-1, x+1)$. For the second condition, let $(a, b)$ and $(c, d)$ be basis elements in $\mathcal{B}_1$. Without loss of generality, assume $a < c$, and let $x \in (a, b) \cap (c, d)$. Then there is a basis element $(c, b)$ such that $x \in (c, b) \subseteq \left((a, c) \cap (b, d)\right)$.

Confirm that the basis $\mathcal{B}_1$ for the lower limit topology on $\mathbb{R}$ is a basis.

For the first condition, let $x \in \mathbb{R}$. Then $[x, x + 1) \in \mathcal{B}_1$ and $x \in [x, x+1)$. For the second condition, let $[a, b)$ and $[c, d)$ be basis elements in $\mathcal{B}_1$. Without loss of generality, assume $a < c$, and let $x \in [a, b) \cap [c, d)$. Then there is a basis element $[c, b)$ such that $x \in [c, b) \subseteq \left([a, c) \cap [b, d)\right)$.

Confirm that the basis $\mathcal{B}_2$ for the $K$-topology is a basis.

The proof is identical to the proof that $B_0$ is a basis.

Consider the standard, lower limit, and $K$ topologies on $\mathbb{R}$. Which ones are comparable to one another? Of the comparable ones, which are finer?

Consider a basis element $[a, b)$ in the lower limit topology. We can construct the basis element $(a, b)$ in the standard topology by taking a countably infinite union of closed sets:

$$(a, b) = \bigcup\limits_{i \in \mathbb{N}} \left[ a + \left(\dfrac{b - a}{2}\right)\left(\dfrac{1}{n}\right), b\right).$$

Therefore each $(a, b)$ in the standard topology is also in the lower limit topology, thus the lower limit topology is finer than the standard topology. However, is is strictly finer, or are the two in fact the same? We note that the reverse process of constructing a half-closed interval from open intervals does not work - there is no way to construct $[a, b)$ out of open intervals. Thus the lower limit topology is strictly finer than the standard topology.

The basis for the $K$ topology contains all of the basis elements for the standard topology, so the standard topology is a subset of, and thus coarser than, the $K$ topology. But is it strictly coarser? Yes. Consider the basis element $B = (-1, 1) - K$. Then there is no interval (a, b)$ that contains $0$ and lies within $B$.

It remains to be seen whether the lower limit and $K$ topologies are comparable, and if so, whether they are the same or one is strictly finer than the other. Consider again the basis element $B = (-1, 1) - K$ in the $K$-topology. There is no interval that contains $0$ and lies within $B$, so there are open sets in the $K$ topology that cannot be constructed from open sets in the lower limit topology. Likewise, consider the open set $C=[2, 3)$ in the lower limit topology. Because there is no way to construct a half-open interval out of open intervals or open intervals that are missing points outside $C$, $C$ is not in the $K$-topology. Therefore the lower limit and $K$-topologies are not comparable.