Exercise Show that the Michael Line is not metrizable.
The Michael line is given by $\{U \cup A: U$ is open in $\mathbb{R}$ usual topology and $A \subset \mathbb{R} \setminus \mathbb{Q}\}$
Note that in a metric space, closed sets are $G_\delta$ sets, i.e. the countable intersection of open sets.
Proof
To see that the Michael line $\mathbb{M}$ is not metrizable, to derive a contradiction, suppose that $\mathbb{M}$ is metrizable. We may treat $\mathbb{M}$ as a metric space. Let $E$ be closed in $\mathbb{M}$. Since $E$ is closed, then $E$ is the countable intersection of $\mathbb{M}$-open sets, i.e.
$$E = \displaystyle\bigcap_{i \in \mathbb{N}} (U_i \cup A_i)$$
It follows then that the complelement $\mathbb{R} \setminus E$ is given by
$$\mathbb{R} \setminus E = \mathbb{R} \setminus \displaystyle\bigcap_{i \in \mathbb{N}} (U_i \cup A_i)$$
Since $E$ is $G_\delta$, then $\mathbb{R} \setminus E$ is $F_\sigma$, a countable union of closed sets.
$$\mathbb{R} \setminus \displaystyle\bigcap_{i \in \mathbb{N}} (U_i \cup A_i) = \displaystyle\bigcup_{i \in \mathbb{N}} \mathbb{R} \setminus (U_i \cup A_i)$$
I seem to have run into a wall at this point. Where can I derive a contradiction? Any advice would be of great value.
Best Answer
In the Michael line, $\mathbb{Q}$ is a closed set but not a $G_\delta$ set.
Proof:
$\mathbb{R}\setminus\mathbb{Q} $ is an open set, so $\mathbb{Q}$ is closed.
Suppose you have a pair of sequences $U_n$ and $B_n$,
such that $U_n$ is an open set in the usual Euclidean topology on $\mathbb{R}$ for all $n$
and $B_n\subset\mathbb{R}\setminus\mathbb{Q}$ for all $n$.
Suppose furthermore that $\mathbb{Q}\subset \bigcap_{n}(U_n\cup B_n)$.
Now, $\mathbb{Q}\subset \bigcap_{n}U_n$,
since each $B_n$ does not contain any rational number.
This means $\mathbb{Q}\subset U_n$ for all $n$,
which in turn means $U_n=\mathbb{R}$ for all $n$ because $\mathbb{Q}$ is dense in the usual Euclidean topology on $\mathbb{R}$.
The conclusion is $\bigcap_{n}(U_n\cup B_n)=\mathbb{R}$,
so that $\bigcap_{n}(U_n\cup B_n)$ is not equal to $\mathbb{Q}$.