# General Topology: Topologies

## The Order Topology

Consider a totally ordered set $X$ that has at least two elements. Let $a, b \in X$, and let $\mathcal{B}$ be the collection of subsets of $X$ that contains

• All open intervals $(a, b)$.
• All right open intervals $[a_0, b)$ where $a_0$ is the smallest element of $X$, if it exists.
• All left closed intervals $(a, b_0]$ where $b_0$ is the largest element of $X$, if it exists.

Then $\mathcal{B}$ is a basis for a topology on $X$, which is called the order topology on $X$. If $a_0$ does not exist, then there are no sets of the second kind, and if $b_0$ does not exist, then there are no sets of the third kind. Note that the standard topology on $\mathbb{R}$ is the same as the order topology on $\mathbb{R}$.

A subset $Y$ of $X$ is convex if for each pair of points $a, b \in Y$ where $a < b$, the interval $(a, b) \subseteq Y$.

## Problems

1. Prove that the set generated by $\mathcal{B}$ is in fact a topology.

To show that the set generated by $\mathcal{B}$ is a topology, we must show that $\mathcal{B}$ is in fact a basis.

1. Let $x \in X$. If $x$ is the smallest element in $X$, then it is included in basis elements in $\mathcal{B}$ by the second property. If it is the largest element in $X$, then it is included in basis elements in $\mathcal{B}$ by the third property. If it is neither the smallest nor the largest element in $X$, then there exist some $y, z \in X$ such that $y < x < z$. Then $x \in (y, z) \in \mathcal{B}$.

2. Let $B_1, B_2 \in \mathcal{B}$ such that $x \in B_1 \bigcap B_2$. If $x$ is the smallest element of $X$, then $B_1$ is of the form $[x, a)$, and $B_2$ is of the form $[x, b)$, where $a, b \in X$ and neither equals $x$. Without loss of generality, assume that $a \leq b$. Then $B_1 \bigcap B_2 = B_1$. Thus $x \in B_1 \subseteq B_1 \bigcap B_2$. A similar argument holds if $x$ is the largest element of $X$. If $x$ is neither the largest nor the smallest element, then without loss of generality, let $B_1 = (a, b)$ and $B_2 = (c, d)$ where $a \leq c$ and $b \leq d$. Then $c < x < b$. Let $B_3 = (c, b)$. Then $B_3 \in \mathcal{B}$ and $x \in B_3 \subseteq B_1 \bigcap B_2$.

2. Let $X = [0, 1]$ be a subspace of $\mathbb{R}$. Determine whether each of the following sets is open in $X$, $\mathbb{R}$, both, or neither:

1. $(0.5, 1)$

2. $[0.5, 1)$

3. $(0.5, 1]$

4. $[0.5, 1]$

5. $(1, 1)$

6. $[1, 1]$

7. $(1, 1]$

1. $(0.5, 1)$ is open in both $X$ and $\mathbb{R}$.

2. $[0.5, 1)$ is open in neither $X$ nor $\mathbb{R}$.

3. $(0.5, 1]$ is open in $X$ but not in $\mathbb{R}$.

4. $[0.5, 1]$ is open in neither $X$ nor $\mathbb{R}$.

5. $(0, 1)$ is open in both $X$ and $\mathbb{R}$.

6. $[0, 1]$ is open in $X$ but not in $\mathbb{R}$.

7. $(0, 1]$ is open in $X$ but not in $\mathbb{R}$.

3. Let $X = [0, 1]$ be a subspace of $\mathbb{R}$. Determine whether $P_0 = \{ x : 0 < x < 1 \text{ and } \frac{1}{x} \notin \mathbb{N} \}$ is open in $X$, $\mathbb{R}$, both, or neither.

Let $K_n = (\frac{1}{n+2}, \frac{1}{n+1})$ for $n > 0$. Clearly $K_n$ is open in $\mathbb{R}$ for all $n \in \mathbb{N}$. Let $x \in P_0$. Then $x \in K_p$, where $p = \lfloor\frac{1}{x}\rfloor - 1$. Therefore $P_0 = \bigcup\limits_{n \in \mathbb{N}} K_n$ is a union of open sets and thus open in $\mathbb{R}$. Because $K_n \bigcap X = K_n$ for all $n \in \mathbb{N}$, $P_0$ is open in $X$, too.

4. Let $X$ be a set and let $Y \subset X$ be convex. Is $Y$ necessarily an interval or a ray?

No. Here are two counterexamples:

1. Let $X = \mathbb{Q}$ and let $Y = \{ x : x^2 < 2 \}$. Then $Y$ is convex in $X$ but there are no $a, b \in X$ such that $Y = (a, b)$.

2. Let $X = \mathbb{R} \times \mathbb{R}$ with the dictionary order, and let $Y = \{ (3, y) : y \in \mathbb{R} \}$. Then $Y$ is convex, as $((3, a), (3,b))$ is an interval in $X$ for any $a < b \in X$, but there are no $c, d \in X$ such that $Y = (c,d)$.