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Can I have feedback on my proof, please?

Prove the Michael line topology, $T_\mathbb{M}=\{U \cup F: U$ is open in $\mathbb{R}$ and $F\subset \mathbb{R}\setminus \mathbb{Q}\}$ is $T_{2}$ (Hausdorff).

Let $a,b \in \mathbb{R}$. WLOG, let $a<b$. Let $x \in (a,b)$. Then $a<x<b$. Clearly, $(-\infty , x), (x, \infty)$ are open disjoint subsets of $T_\mathbb{M}$. Furthermore, $a \in (-\infty , x)$ and $b \in (x, \infty)$, thus, $T_\mathbb{M}$ is $T_{2}$ or Hausdorff.

PerpetualStudent
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    Or simply note that the topology is finer (has more opens) than the standard topology – Hagen von Eitzen Nov 04 '18 at 18:19
  • Ah-ha! That might work, but I have not been introduced to the term finer. Does my proof hold water the way it stands? – PerpetualStudent Nov 04 '18 at 18:21
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    Your proof is correct, and it really just uses Hagen von Eitzen's point. The standard topology on $\mathbb{R}$ is Hausdorff, and so if you can find such open neighborhoods for $a$ and $b$ in the standard topology, those same open neighborhoods will also exist in the Michael line topology, since it has all of the open sets of the standard topology and more. – Kevin Long Nov 04 '18 at 18:22
  • Your proof does precisely that - that's why I wrote "Or simply ..." or I could have written "In a more fancy language ..." – Hagen von Eitzen Nov 05 '18 at 21:12

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Your proof is correct: you separate the points with sets that are standard-open (in the reals) and because the Michael line topology has as its topology a superset of the usual topology (because we can always take $F = \emptyset$), these standard-open sets are still open in the new topology and are as required.

Henno Brandsma
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