# Proof that the Michael line is a topology.

May I have feedback on my proof?

Prove that the Michael Line, $$\tau_M = \{U \cup F: U$$ is open in the usual topology on $$\mathbb{R}$$ and $$F \subset \mathbb{R} \setminus \mathbb{Q} \}$$, is a topology on $$\mathbb{R}$$.

We need to show that $$\tau_M$$ satisfies the following properties:

i. $$\emptyset \in \tau_M$$

ii $$X \in \tau_M$$

iii. If $$U \in \tau$$ and $$V \in \tau$$, then $$U \cap V \in \tau_M$$.

iv. If $$U_i \in \tau, \ \forall i \in I$$, then $$\bigcup_{i \in I} U_i \in \tau_M$$.

Proof. Let $$X = \mathbb{R}$$. Let $$\tau$$ be the usual topology on $$\mathbb{R}$$. Note that $$\emptyset$$ and $$\mathbb{R} \in \tau$$ and $$\emptyset \subset \mathbb{R} \setminus \mathbb{Q}$$.

i. $$\emptyset = \emptyset \cup \emptyset \ \in \tau_M.$$

ii. $$\mathbb{R} = \emptyset \cup \mathbb{R} \ \in \tau_M.$$

iii. Let $$U_\alpha, U_\beta \in \tau_M.$$ Then there exists $$V, W$$ open sets and $$A, B \in \mathbb{R} \setminus \mathbb{Q}$$ such that $$U_\alpha = V \cup A$$ and $$U_\beta = W \cup B$$.

Then, $$V \cap W \in \tau$$ and $$A \cap B \in \mathbb{R} \setminus \mathbb{Q}$$.

Therefore, we have $$U_\alpha \cap U_\beta = (V \cup A) \cap (W \cup B) = (V \cap W) \cup(A \cap B) \in \tau_M$$.

iv. Let $$U_i \in \tau_M$$ for all $$i \in I$$, such that $$U_i = V_i \cup A_i$$ with $$V_i \in \tau$$ and $$A_i \in \mathbb{R} \setminus \mathbb{Q}.$$

Note that $$\bigcup_{i \in I} V_i \in \tau$$ and $$\bigcup_{i \in I} A_i \subset \mathbb{R} \setminus \mathbb{Q}.$$

Then, $$\bigcup_{i \in I} U_i = \bigcup (V_i \cup A_i) = \left( \bigcup_{i \in I} V_i \right) \cup \left(\bigcup_{i \in I} A_i \right) \in \tau_M.$$

$$\therefore \tau_M$$ is a topological space on $$\mathbb{R}$$.

For (ii) I would write $$\mathbb{R} =\mathbb{R} \cup \emptyset$$ where $$\mathbb{R}$$ is open in the usual (in fact any) topology on $$\mathbb{R}$$ and $$\emptyset \subseteq \mathbb{P}$$ (where the latter is the set of irrationals) instead of the other way round. Nitpick, really.
$$(V\cup A) \cap (W \cup B) = (V \cap W) \cup (V \cap B) \cup (A \cap W) \cup (A \cap B)$$
The first set is open in the usual topology , the last three are either subsets of $$A \subseteq \mathbb{P}$$ or $$B \subseteq \mathbb{P}$$, so together a subset of $$\mathbb{P}$$ as well, so the intersection is of the right form again.
$$U_{\alpha}\cap U_{\beta}=(V\cup A)\cap (W\cup B)=((V\cup A)\cap W)\cup ((V\cup A)\cap B)= (V\cap W)\cup (A\cap W) \cup (V\cap B)\cup (A\cap B)\in \tau_M$$