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?


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.

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  • $\begingroup$ I had never heard about this theorem, could you give me a reference where I can find it? $\endgroup$ – Math6952 Nov 8 '17 at 22:32
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    $\begingroup$ 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 $T_X\subset T_C\subset T_X.$ $\endgroup$ – DanielWainfleet Nov 9 '17 at 0:56

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