Michael Line is not Lindelöf

Question

As we know, the Michael line is not Lindelöf. The links

• https://dantopology.wordpress.com/2012/10/28/michael-line-basics/#:~:text=Result%202-,To%20see%20that,-is%20not%20Lindelof

and

• https://math.stackexchange.com/questions/1471133/the-michael-line-is-not-lindel%c3%b6f#:~:text=HINT%3A%20Find%20an%20uncountable%20closed%20discrete%20subset%20of%20the%20Michael%20line

tell us that if we can find an uncountable, closed, and discrete subset of the Michael line, then the Michael line is not Lindelöf. But why is this the case? What is it about finding a subset like this that makes the Michael line not Lindelöf?

Answer 1

Given a topological space $(X,\tau)$ and an uncountable closed discrete set $C\subseteq X$, then we can consider the following open cover of $X$: $$\{\{x\} \mid x \in C\} \cup \{X\setminus C\}.$$ If $X$ were Lindelöf, then there would be a countable subcover, but such a subcover, being countable, cannot cover $C$, hence the above cover doesn't have any countable subcover and $X$ is not Lindelöf.

Answer 2

Every closed subspace of a Lindelöf space is also a Lindelöf space. But an uncountable discrete space is not Lindlöf.