Being Picky 01

Hausdorff Spaces are Sober

Our main reference is

📚 Jorge Picado & Aleš Pultr Frames and Locales: topology without points. Springer Basel, Series Frontiers in Mathematics, Vol. 28, xx + 400 pp., 2012 (ISBN: 978-3-0348-0153-9).

On page 2, there is a demonstration for the fact that

A Hausdorff space $X$ is always sober.

Being a little picky, I have found the demonstration on the reference book, although very easy… unnecessarily complicated. This is a very small post where I am going to demonstrate this fact in a much simpler way. It is basically the same, but simpler.

Let $U \subsetneq X$ be a meet-irreducible open set. Since $X$ is Hausdorff, given two distinct $x,y \in X$, there are two disjoint open neighbourhoods $V$ and $W$ of $x$ and $y$. Now, $V \cap W = \emptyset \subset U$. And since $U$ is meet-irreducible, at least one must be contained in $U$:

This implies that, $U \subsetneq X$ must contain every point except for one. It is, therefore, of the shape $X \setminus \{x\}$ for some $x \in X$.

🥳 🍻 😎

Cheers!!!

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