The problem arises from the exercise 29.8 of the book "Topology" by Munkres:

Show that the one-point compactification of $\mathbb{Z}_{+}$ is homeomorphic with the subspace $\{ 0 \} \cup \{ 1/n \mid n \in \mathbb{Z}_{+} \}$ of $\mathbb{R}$.

According to the definition of the "one-point compactification",

Definition: If $Y$ is a compact Hausdorff space and $X$ is a proper subspace of $Y$ whose closure equals $Y$, then $Y$ is said to be a compactification of $X$. If $Y-X$ equals a single point, then $Y$ is called the one-point compactification of $X$.

we have that $X$ has the one-point compactification if and only if $X$ is a locally compact Hausdorff space that is not itself compact. In this case, the single point $Y-X$ is a limit point of $X$. Therefore, the task is to find such limit point.

However, how can we find a limit point of $X$ when we have no idea of what the bigger space $Y$ is? And what is the one-point compactification of $\mathbb{Z}_{+}$?

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    $\begingroup$ Well, you know {$0$} is the closure of {$1/n $} , and it seems like the map $n \rightarrow 1/n$ is a homeomorphism between $\mathbb Z_{+}$ and {$1/n$} $\endgroup$
    – user99680
    Feb 27, 2014 at 14:07
  • $\begingroup$ @user99680 So, the single point is $\infty$? But, how could it be the limit point of $\mathbb{Z}_{+}$? In addition, is the topology on $\mathbb{Z}_{+}$ the usual order topology? $\endgroup$
    – hengxin
    Feb 27, 2014 at 14:10
  • $\begingroup$ since the map is a homeomorphism, both topologies are equivalent, so that it seems $\mathbb Z_{+}$ has the discrete topology. And yes, then the single point is $\infty$ $\endgroup$
    – user99680
    Feb 27, 2014 at 14:23
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    $\begingroup$ @user99680 Following your comment and the answer of MJD, I realize that $\infty$ is the single point we add in the construction of the one-point compactification of $X$ such that $\infty$ is the limit point of $X$. Thanks. $\endgroup$
    – hengxin
    Feb 27, 2014 at 14:32
  • $\begingroup$ no problem; $\infty$ is a limit point in the compactification topology; not the original one. $\endgroup$
    – user99680
    Feb 27, 2014 at 14:34

1 Answer 1


I think you are reading the wrong part of your book. That definition (top of p. 185 in my edition) is immediately preceded by a Theorem 29.1, which is a long explanation of how one can take a noncompact locally-compact Hausdorff space $X$ and embed it into a compact space $\def\Y{X^\ast}\Y$ that has exactly one point more than $X$. This larger space $\Y$ is the one-point compactification of a space $X$, sometimes called the Alexandroff compactification of $X$.

The one-point compactification $\Y$ consists of $X\cup \{\infty\}$, where $\infty$ is some new point that is not an element of $X$. The topology is as follows:

  • If $G$ is an open subset of $X$, then $G$ is also an open set of $\Y$
  • If $C$ is a compact subset of $X$, then $\{\infty\}\cup(X\setminus C)$ is an open set of $\Y$

Theorem 29.1 shows that if $X$ is a Hausdorff space that is locally compact but not compact, then $\Y$, with the topology described above, is a compact Hausdorff space of which $X$ is a subspace. Theorem 29.1 also shows that any one-point compactification of $X$ must be homeomorphic to the $\Y$ described above, so that the one-point compactification of $X$ is essentially unique. This is the construction Munkres wants you to consider.

This construction is the topological formalization of the idea of taking an infinite space $X$ and "adding a point at infinity". We do this, for example, with the complex numbers, to obtain the Riemann sphere. (Ignore this example if you don't know about the Riemann sphere.) A simpler example is that the one-point compactification of $\Bbb R$ is (homeomorphic to) $S^1$, the circle: the two ends at infinity are brought together and joined at the new point $\infty$.

I hope the question makes more sense in this light.

A word of advice: In some subjects, and on standardized tests, you can read the question first, then go back and skim the material looking for something pertinent, and then answer the question without reading all the material. In advanced mathematics, this strategy will not work. You have to adopt a different strategy. First read over the entire chapter, very slowly, taking time to understand and digest each sentence before you move on to the next one. This may take several days, or more. Then do the exercises.

  • $\begingroup$ Therefore, the single point $X^{\ast} - X$ when $X$ is a locally compact Hausdorff is not necessarily a number (in this example). We just add a point $\infty$ and construct the topology $X^{\ast}$ and in the construction, $\infty$ is the limit point of $X$. Is this understanding right? Special thanks for your advice. $\endgroup$
    – hengxin
    Feb 27, 2014 at 14:25
  • $\begingroup$ That is exactly correct. The one-point compactification $\Bbb Z_+^*$ of $\Bbb Z_+$ contains all the integers, and also a special point $\infty$. $\endgroup$
    – MJD
    Feb 27, 2014 at 14:27
  • $\begingroup$ @hengxin But I should add that since Munkres wants you to show that $\Bbb Z_+^\ast$ is homeomorphic to a certain subspace of $\Bbb R$, the homeomorphism you construct will naturally map $\infty$ to some element of $\Bbb R$. $\endgroup$
    – MJD
    Feb 27, 2014 at 14:54
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    $\begingroup$ Yes, the function $f(n) = 1/n$ mapping the special point $\infty$ to $0$ works. Thanks. $\endgroup$
    – hengxin
    Feb 28, 2014 at 1:48
  • $\begingroup$ Sorry but let me ask you the question. When we need to prove that 1-point compactification of $\mathbb{N}$ is homeomorphic to $\{0\}\cup \{1/n: n\geq 1\}$. What topologies do we consider on $\mathbb{N}$ and $\{0\}\cup \{1/n: n\geq 1\}$? $\endgroup$
    – ZFR
    Dec 1, 2018 at 23:19

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