3
$\begingroup$

Let $ \Omega$ be the set of ordinales with $\omega_1$ being the first uncountable ordinal. Denote $\Omega ( \omega) = \mathbb{N} \cup \lbrace \omega \rbrace$, where $\omega$ is the first infinite ordinal. When $\Omega(\omega)$ is given its order topology, the points of $\mathbb{N}$ are isolated and the point $\omega$ has for basic nhoods the sets $\lbrace n,n+1,...\rbrace \cup \lbrace \omega \rbrace $.

The product space $\Omega \times \Omega(\omega)$ will be denoted T$^{*}$. Considere the subspace $T=T^{*}-\lbrace (\omega_1,\omega) \rbrace$ of T$^{*}$.

Show that the one-point compactification of $T$ is $T^{*}$.

I know that I must prove that the topology on $T^{*}$ coincides with the topology obtained by the compactification.

I think it is enough to study the basic neighborhoods of the elements of $\Omega \times \Omega(\omega)$.

However I've some problems understanding how I must do that.

Can anyone help me?

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Theorem: if $X$ is compact Hausdorff and $p \in X$ is a non-isolated point of $X$, then $X$ is the one-point compactification of $X\setminus\{p\}$ (or $X - \{p\}$ if you prefer that). This is the essential uniqueness of the one-point compactification.

$\endgroup$
2
  • $\begingroup$ I had never heard about this theorem, could you give me a reference where I can find it? $\endgroup$ Commented Nov 8, 2017 at 22:32
  • 1
    $\begingroup$ To the proposer: It's easy to prove. It likely appears as an exercise in many texts.Use the fact that if $U$ is open in $X$ in the (given) compact Hausdorff topology on $X$ then$X\setminus U$ is compact. .... It could get confusing as you are looking at 3 spaces : (i). $(X,T_X)$ where $T_X$ is the given topology, (ii) $(Y, T_Y)$ as a subspace of $(X,T_X)$, where $Y=X\setminus \{p\}$ and (iii) $(X,T_C)$, the 1-point compactification of $(Y,T_Y).$.... Prove that $T_X\subset T_C\subset T_X.$ $\endgroup$ Commented Nov 9, 2017 at 0:56

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.