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Prove that the Michael line is not Lindelöf.

The Michael line is the space with underlying set $\mathbb{R}$ having the topology generated by the base $$\mathcal{B} = \{ U \subseteq \mathbb{R} : U\text{ is open in the usual metric topology on }\mathbb{R} \} \cup \{ \{ x \} : x \in \mathbb{R} \setminus \mathbb{Q} \}.$$

We know the Michael line is Hausdorff, but why isn't it Lindelöf?

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HINT: Find an uncountable closed discrete subset of the Michael line. The result proved in this answer may be useful.

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  • $\begingroup$ will uncountable Euclidean closed set consisting entirely of irrational numbers work $\endgroup$ – Ramu Oct 10 '15 at 15:52
  • $\begingroup$ @Ramu: Yes, it will, and you can use the linked answer to see why there is one. $\endgroup$ – Brian M. Scott Oct 10 '15 at 16:40
  • $\begingroup$ thanks. can u give me hint on proving that int of metric space = int of topology. i m confused as int is just a point in metric where as it is a union of points in topology $\endgroup$ – Ramu Oct 11 '15 at 19:59
  • $\begingroup$ @Ramu: Do you mean interior? The interior of a set $A$ in a metric space $\langle X,d\rangle$ is a set: it’s $$\left\{x\in U:\exists r>0\big(B_d(x,r)\subseteq A\big)\right\}\;,$$ where $B_d(x,r)$ is the open $d$-ball of radius $r$ and centre $x$. $\endgroup$ – Brian M. Scott Oct 13 '15 at 21:02

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