1
$\begingroup$

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 $\mathbb{R}$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.