Can we define a topology on $\mathbb{R}^2$ using lines instead of circles or squares?

Question

Does it make sense to define a topology on the coordinate grid $\mathbb{R}^2$ using the horizontal and vertical lines as "building-blocks"?

• $B_{1, y} = \{ (x,y): y \in \mathbb{R} \}$ vertical lines are closed
• $B_{2, x} = \{ (x,y): x \in \mathbb{R} \}$ horizontal lines are closed

The definition of basis of Topology is a collection of open sets:

• the base elements cover $X = \mathbb{R}^2$
• let $B_1$ and $B_2$ be the basis elements and $I = B_1 \cap B_2$ the there exists a $B_3$ such that $z \in B_3 \subseteq I = B_1 \cap B_2$.

Basically my question is that if we use a basis other than squares or circles can we change the topology of $\mathbb{R}^2$. Here we coudl write one possible basis for the Euclidean topology:

• $B_{(x_0, y_0), \epsilon} = \{ (x^2 - x_0)^2 - (y - y_0)^2 < \epsilon \} $

if we have two bases $(\mathbb{R}^2, \mathcal{B}_1)$ $ (\mathbb{R}^2, \mathcal{B}_2)$ can these two copies of the plane be homeomorphic.

This matches the definition on p. 78 of Munkres. We can define the topology generated by a basis $\mathcal{B}$. A subset $U \subseteq X$ is open if for any $x \in U$ there is a basis element $B \in \mathcal{B}$ such that $x \in B \subset U$. On the next page, he shows that from the basis $\mathcal{B}$ the collections of sets $U$ is a topology.

Example: check that $U_1 \cap U_2 \in \mathcal{T}$ (the intersection of open sets is open). Given $x \in U_1 \cap U_2$ we can find $x \in B_1 \subset U_1$ and $x \in B_2 \subset U_2$ and then $x \in B_3 \subset B_1 \cap B_2 \subset U_1 \cap U_2$. Our open sets might be irregular and yet the basis elements are quite standard.

Possible I have to change my question to define a basis of open sets rather than closed. We do have the result that if $U$ is open then $X \backslash U$ is closed.

---

Related

Why $B_3 \subset B_1 \cap B_2$ in the definition of a basis for a Topology in Munkres' Book?

Answer 1

So we consider the coarsest topology on $ℝ^2$ such that horizontal and vertical lines are closed. We will characterize closed sets and the closure, and we will mention some properties of this topology.

Preliminaries.

• $A ⊆ ℝ^2$ is thin if every two distinct points in $A$ are distinct in both coordinates.
• $A_x := \{y ∈ ℝ: (x, y) ∈ A\}$ is the vertical slice through $A$,
• $A^y := \{x ∈ ℝ: (x, y) ∈ A\}$ is the horizontal slice through $A$,
• $C(x, y) := (\{x\} × ℝ) ∪ (ℝ × \{y\})$ is the cross at $(x, y)$.

Proposition 1. Let $A ⊊ ℝ^2$. The following conditions are equivalent.

1. $A$ is closed.
2. $A$ is a finite union of horizontal lines, vertical lines, and points.
3. Every horizontal and vertical slice through $A$ is full or finite, and $A$ contains no infinite thin subset.

Proof. 2) ⇒ 1) clear. 1) ⇒ 3) since every horizontal or vertical line satisfies 3) and since sets satisfying 3) are stable under finite unions and arbitrary intersections. 3) ⇒ 2) follows from the following lemma.

Lemma. Let $A ⊆ ℝ^2$ be a set containing no infinite thin subset. Let $V := \{x ∈ ℝ: A_x \text{ infinite}\}$, $H := \{y ∈ ℝ: A^y \text{ infinite}\}$, $F := \{p ∈ A: C(p) ∩ A \text{ finite}\}$. We claim that $V, H, F$ are finite sets, that $A ⊆ (ℝ × H) ∪ (V × ℝ) ∪ F =: B$, and that 3) ⇒ ($A = B$) ⇒ 2).

Proof. If $V$ or $H$ is infinite, then we can easity find an infinite thin subset. For every $p ∈ F$ the intersection $C(p) ∩ F$ is finite, so if $F$ is infinite, we can inductively construct an infinite thin subset by picking $p_n ∈ F \setminus ⋃_{i < n} C(p_i)$. Clearly $A ⊆ B$. 3) ⇒ $(A = B)$ since if $x ∈ V$, then $A_x = ℝ$, and similarly for $H$. $(A = B)$ ⇒ 2) since $B$ satisfies 2).

Proposition 2. Let $A ⊆ ℝ^2$. If $A$ contains an infinite thin subset, then $\overline{A} = ℝ^2$. Otherwise, $\overline{A} = B$ (the notation from the Lemma).

Proof. The density follows from Proposition 1. $B$ is a closed superset of $A$ by 2) and the Lemma. $B ⊆ \overline{A}$ by 3).

Note that this topology is $T_1$, but not $T_2$. In fact it is anti-$T_2$ or hyperconnected: every two non-empty open sets intersect. It is easy to see that the topology is (quasi)compact and separable. However it has weight continuum.

Also note that the whole construction can be generalized to a product $X × Y$ of any sets. If $X$ is finite, then $X × Y$ will be the sum of $X$-many copies of $Y$ with the cofinite topology. Similarly if $Y$ is finite. In particular, if both $X$ and $Y$ are finite, then $X × Y$ is discrete. If both $X$ and $Y$ are infinite, then everything works as above (and the weight is $\max(|X|, |Y|)$).