Reference: Hocking and Young -Topology p.54

I need to state some definitions and theorems to describe where I get stuck.


Let $p,q$ be points of a connected space $S$. We denote $E(p,q)$ the subset of $S$ consisting of the points $p$ and $q$ together with all cut points of $S$ that separates $p$ and $q$. The separation order in $E(p,q)$ is defined as follows. Let $x$ and $y$ be two points in $E(p,q)$. Then $x<y$ if either $x=p$ or $x$ separates $p$ and $y$ in $S$.


Theorem1. The separation order is a total ordering.

Theorem2. Let $S$ be a connected and $p,q$ be two points of $S$ such that $p\cup q$ is a proper subset of $E(p,q)$. Then the order topology is coarser than the subspace topology.

I understand these theorems but I have a trouble with the proof for the following theorem:

Theorem. Let $S$ be a compact connected Hausdorff space with just two non-cut points, $a$ and $b$. If $S=E(a,b)$, then the order topology defined by the order in $E(a,b)$ is the same as the topology in $S$.

Proof in the text:

Since open sets in the order tolology were shown to be open in $S$, we need only show that open sets in $S$ are unions of basis elements of the order topology. If this is not so, there is an open set $U$ in $S$ and a point $x$ in $U$, such that no order-basis element that contains $x$ lies in $U$. For verbal simplicity, suppose that $x$ is neither $a$ nor $b$, so that we need consider only basis elements of the form $(y,z)$. Using maximal principle, we obtain a collection of sets $(y_\alpha,z_\alpha)$, which is totally ordered by inclusion, and which has only $x$ as their intersection. The same is true of the sets $[y_\alpha,z_\alpha]$ and these are closed sets in $S$.

Why does the intesection contain only $x$? I have no idea why..


As shown in my answer to your previous question, there need not exist any such collection $(y_\alpha,z_\alpha)$ which is totally ordered by inclusion. However, you don't actually need them to be totally ordered; you only need them to be directed, so that the closed intervals $[y_\alpha,z_\alpha]$ (together with the set $S\setminus U$) have the finite intersection property. This is easy to arrange; in fact, you can just take the collection of all open intervals $(y,z)$ containing $x$.

By the way, there is a quicker way to state this argument. Let $T$ denote $S$ with the order topology; then Theorem 2 says that the identity map $S\to T$ is continuous. Since $S$ is compact and $T$ is Hausdorff, it follows automatically that it is a homeomorphism.

  • $\begingroup$ Oh right.. The identity map is definitely a homeomorphism. I don't understand why the author proved (and it is a false proof) it in such way. Thank you so much! $\endgroup$ – Rubertos Oct 25 '15 at 1:29
  • $\begingroup$ Well, (my corrected version of) their argument is just a (slightly convoluted) reformulation of the usual proof that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. $\endgroup$ – Eric Wofsey Oct 25 '15 at 1:30

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