Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Recently, I'm reading Engelking's book. In this book, it always uses as the example the space $A(m)$, see for instance the example 1.1.8, page 15.


For this space, I'm not very clear, for the description is complex to me. Sometimes I view it as a sequence with a limit point when $m=\omega$. But I am not sure I'm completely right. So, could anybody knowing this example help me to describe it more clearly? Thanks ahead.

share|cite|improve this question
It would help if you could scan or type the description, in case anyone doesn't own the book. – Kevin Carlson Aug 13 '12 at 3:07
@Paul It denotes the one-point compactification of the discrete space of cardinality $m$. – azarel Aug 13 '12 at 3:09
@KevinCarlson I try to find the link for the book, however, I cannot find it now. – Paul Aug 13 '12 at 3:15
@azarel Why does it denote the one-point compactification? Could you explain more? It will be welcome if you write it as an answer:) – Paul Aug 13 '12 at 3:17
Compare this topology with the construction given in the Alexandroff Compactification Theorem (3.5.11, pp.169-170) on an infinite discrete space. (If $X_0$ denotes the subspace $X \setminus \{ x_0 \}$, note that $X_0$ is discrete, and then look at what the neighbourhoods of $x_0$ are.) – arjafi Aug 13 '12 at 3:28
up vote 1 down vote accepted

Let $D$ be the discrete space of cardinality $\kappa$, and let $x_0$ be a point not in $D$. Let $A=D\cup\{x_0\}$ be the one-point compactification of $D$, with $x_0$ as the added point. The compact subsets of $D$ are precisely the finite sets, so open nbhds of $x_0$ in $A$ have the form $\{x_0\}\cup(D\setminus F)$ for finite $F\subseteq D$. Let $\mathcal{O}$ be the topology on $A$; what, exactly, is $\mathcal{O}$?

Let $S\subseteq A$. If $x_0\notin S$, then $S\subseteq D$, so $S\in\mathcal O$ iff $S$ is open in $D$. But $D$ is discrete, so $S$ is open in $D$ $-$ all subsets of $D$ are open in $D$ $-$ and therefore $S\in\mathcal{O}$. If $x_0\in S$, then $S$ must contain an open nbhd of $x_0$, so there must be a finite $F\subseteq D$ such that $\{x_0\}\cup(D\setminus F)\subseteq S$. But then $$A\setminus S\subseteq A\setminus\Big(\{x_0\}\cup(D\setminus F)\Big)=D\setminus(D\setminus F)=F\;,$$ so $A\setminus S$ is finite. In other words, $S\in\mathcal O$ iff either $x_0\notin S$, or $x_0\in S$ and $A\setminus S$ is finite. Finally, this is the same as saying that $S\in\mathcal O$ iff $x_0\notin S$, or $A\setminus S$ is finite, which is exactly the definition of the topology $\mathcal O$ given in Example 1.1.8. Thus, $A(\kappa)$ is, as azarel said, the one-point compactification of the discrete space of cardinality $\kappa$.

$A(\omega)$ is indeed homeomorphic to $\omega+1$, a simple sequence with a limit point. If $\kappa>\omega$, it’s similar, but it’s not a sequence: it’s a set of $\kappa$ isolated points and one limit point, $x_0$, with the property that every open nbhd of $x_0$ contains all but finitely many of the isolated points. This means, among other things, that if $S$ is an infinite set of isolated points, then $x_0\in\operatorname{cl}S$: every open nbhd of $x_0$ must contain all but finitely many points of $S$. It also means that $A(\kappa)$ is compact: once you’ve covered the point $x_0$, there are at most finitely many points left to be covered. (Of course this is also clear from the fact that $A(\kappa)$ is a one-point compactification!)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.