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Recall that the Michael line is the real line with the topology $\{U\cup F:U$ is open in usual topology on $\mathbb{R} $ and $ F\subseteq\mathbb{R} - \mathbb{Q}\}$.

To show that the Michael line is not Lindelöf, I want to construct an open cover of the Michael line and verify that such open cover does not have any countable open cover. I’ve already looked up the following information: a solution and Michael Line Basics (Result 2).

After reading the information above, the first idea that comes to mind is to apply the property that the set $\mathbb{R} - \mathbb{Q}$ is uncountable. I try the cover $\{\{x\}:x\in\mathbb{R} - \mathbb{Q} \}\cup\{\mathbb{Q}\}$ at first, but it seems that $\mathbb{Q}$ is not open in the Michael line. I would like to attempt modifying the collection in the last part (i.e., the collection $\{\mathbb{Q}\}$) to meet my requirements. But I don't know how to modify it. I would like to ask how to modify it.

In addition, if the open cover which is of the form $\{\{x\}:x\in\mathbb{R} - \mathbb{Q} \}\cup\{...\}$ is impracticable, please give some other kinds of open covers.

Moishe Kohan
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ChungJunLin0217
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    Hint: try covering $\mathbb{Q}$ with open sets which rapidly decrease in size – Giorgio Genovesi May 28 '25 at 22:27
  • I like this, but there's some measure theory necessary here to argue that their union must miss an uncountable set of irrationals. Might there be a more elementary approach? – Steven Clontz May 29 '25 at 02:31
  • @GiorgioGenovesi A single rational point is not open in Michael line. If I have misunderstood your perspective, I would appreciate it if you could provide more specific guidance. – ChungJunLin0217 May 29 '25 at 05:14
  • That was my first thought, but I haven’t come up with a better example so far. If you have any better examples or suggestions, I’d be glad to discuss them with you. @StevenClontz – ChungJunLin0217 May 29 '25 at 05:19
  • Hint: in the Euclidean topology, $\mathbb R \setminus \mathbb Q$ is homeomorphic to $\mathbb N^{\mathbb N}$. Hence, $\mathbb R \setminus \mathbb Q$ contains a subset $A$, which is homeomorphic to $2^{\mathbb N}$. – Ulli May 29 '25 at 05:59

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I will write out the proof in detail. Fix an enumeration of the rational numbers $(q_i)_{i\in \mathbb{N}}$ and for each $i\in \mathbb{N}$ let $U_i=(q_i-2^{-i},q_i+2^{-i})$ which will have measure $\mu(U_i)=2^{-i+1}$. So $\mathbb{Q}\subseteq \bigcup_{i\in \mathbb{N}}U_i=U$ and: $$\mu(U)=\mu(\bigcup_{i\in \mathbb{N}}U_i)\leq \sum_{i\in \mathbb{N}} 2^{-i+1}=4$$ So $\mu(\mathbb{R}\setminus U)>0$, which implies $\mathbb{R}\setminus U$ is uncountable and contains only irrational points. Let your covering be: $$\{U\}\cup\{\{r\}: r\in \mathbb{R}\setminus U\}$$ the elements are pairwise disjoint so they do not admit any proper subcovering.

Alternatively, if you want to avoid using measure theory, you can construct a Cantor set containing only irrational numbers. The only such constructions that comes to mind is quite tedious.

Giorgio Genovesi
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