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.

(iii) Your formula for the intersection is off:

$$(V\cup A) \cap (W \cup B) = (V \cap W) \cup (V \cap B) \cup (A \cap W) \cup (A \cap B)$$

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.

The union I totally agree with, of course.


$U_{\alpha}\cap U_{\beta}=(V\cup A)\cap (W\cup B)=((V\cup A)\cap W)\cup ((V\cup A)\cap B)= (V\cap W)\cup (A\cap W) \cup (V\cap B)\cup (A\cap B)\in \tau_M$


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