Is $S_{\Omega}$ locally connected?

Question

I am now trying to show that $S_{\Omega}$ is locally connected (or not) This is how I tried to show:

Let $ S_{\Omega}$ denote the minimal uncountable well oredered set.

Let $a_0=minS_{\Omega}$

For given $a\in S_{\Omega}$, $a\not=a_0$, let $b$ denote the imediate successor of $a$. If $U$ is a neighborhood of $a$, then $U$ cannot be connected since $U\cap[a_0,a]=U\cap[a_0,b)$ is both open and closed so $U$ cannot be connceted unless $U\subset[a_0,a]$

The probelm is that I cannot show $U\not\subset[a_0,a]$

Is there a way to show this?

Or is there any better way to show that $S_{\Omega}$ is locally connected (or not)?

I'd really aprreciate your help. Thanks.

Answer 1

$S_\Omega$ (and indeed virtually all ordinal spaces) are quite far from being locally connected: they are in fact totally disconnected, meaning that the only connected subsets are the singletons. Here's a short proof.

• Proof. If $A \subseteq S_\Omega$ contains at least two points, consider $\alpha = \min A$. Note that $( - \infty , \alpha ] = ( - \infty , \alpha + 1 )$ is open in $S_\Omega$, and $( - \infty , \alpha] \cap A = \{ \alpha \}$, so $\{ \alpha \}$ is a relatively open subset of $A$. Since $S_\Omega$ is T₁ it follows that $\{ \alpha \}$ is also a relatively closed subset of $A$. So $A$ cannot be a connected subset of $S_\Omega$.

What this shows is that the only connected open subsets of $S_\Omega$ are the singletons of the isolated points. The isolated points of $S_\Omega$ are $0$ and all successor ordinals. So if $\omega \in S_\Omega$ denotes the minimal point with infinitely many predecessors, it follows that $\omega$ is not an isolated point, and so $\omega$ has no connected neighborhood in $S_\Omega$.