One-point compactification of $S_{\Omega}$ is homeomorphic with $\bar S_{\Omega}$.

Question

One-point compactification of $S_{\Omega}$ is homeomorphic with $\bar S_{\Omega}$.

Let $X$ be a topological space. Then the One-point compactification of $X$ is a certain compact space $X^*$ together with an open embedding $c : X \to X^*$ such that the complement of $X$ in $X^*$ consists of a single point, typically denoted $\infty$.

Let $X$ be a well-ordered set. Given $\alpha \in X$, let $S_{\alpha}$ denote the set $S_{\alpha} = \{x \mid x \in X \text{ and }x < \alpha \}$. It is called the section of $X$ by $\alpha$.

I am finding difficulty to do the problem!

Answer 1

Consider $\bar{S_\Omega}$ with its order topology. Since $S_\Omega$ is a locally compact and $T_2$ space and $\bar{S_\Omega}$ is a $T_2$ and compact space it is enough to prove that $\bar{S_\Omega} - \{\Omega\} \cong S_\Omega$ to prove that it is homeomorphic to its one-point compactification. (By your notation I guess you're using Munkres book, this is Theorem 29.1. This is the same result Henno is mentioning in his comment.)

Define the natural embedding $c:S_\Omega\to\bar{S_\Omega}$ considering both spaces with the order topology. It's easy to show that $c$ is a homeomorphism to its image ($\bar{S_\Omega} - \{\Omega$}).

Answer 2

I will give it a try...

Let $X=(\Omega,<)$ and notice this is your $S_\Omega$,

Hence $X=\{set\space of \space ordinals \space smaller \space than \space \Omega \}$, with the mentioned order '<'.

Let $Y=X\cup\{\Omega\}\cong X\cup\{\infty\}$ and notice $X=[0,\Omega)$ is an open set of Y (considering the order topology).

Now consider $\overline{X}=\cap C_\alpha$ where $\{C\alpha\}$ are all sets closed in Y, which contains $X$.

It is easy to see that $\overline{X}=Y$ because the only closed set which contains $X$ is exactly $Y=[0,\Omega]$.

So we have: $X=[0,\Omega)=\Omega$, $\overline{X}=[0,\Omega)\cup\{\Omega\}=[0,\Omega]=Y\cong X\cup\{\infty\}$

And there you have it...