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Example 1:

Let $\tau_{disc}$ be the discrete topology

We know $(\mathbb{R},\tau_{disc})$ is not compact

If we add a point P so that we have $X=\mathbb{R} \cup \{P\}$ and define a new topology as $\tau_1=\tau_{disc} \cup \{X\setminus F\}$, where $F$ is $F\subseteq \mathbb{R}$ and finite, then the new space $(\mathbb{R},\tau_1)$ is compact.

The reason is that if we take any open covering, there would be an open set containing P which is almost everything but a finite quantity $t$ of points, then we can take t open sets, one for each point, and we have a covering made of $t+1$ open sets

Example 2

Let $\tau_{e}$ be the euclidean topology

We know $(\mathbb{R},\tau_e)$ is not compact

If we add a point $P$ so that we have $X=\mathbb{R} \cup \{P\}$ and define a new topology as $\tau_2=\tau_e \cup \{X\setminus K\}$, where $K$ is $K\subseteq \mathbb{R}$ and compact, then the new space $(\mathbb{R},\tau_2)$ is compact.

Why isn't enough to apply the same reasoning as in Example 1 and define K as finite, instead of compact?

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  • $\begingroup$ your topology is not Hausdorff (T2) nor T1 $\endgroup$
    – Exodd
    Jul 13 '20 at 15:20
  • $\begingroup$ @Exodd, what do you mean , can you elaborate? why do we require T2 or T1?, The euclidean topology is T2 $\endgroup$
    – mathlover
    Jul 13 '20 at 15:31
  • $\begingroup$ The one-point (or Alexandroff) compactification is one of the infinite ways we can compactify a space. But it has the nice property that when you apply it to a T2 space, it remains T2. The topology you described is compact, but not T2 $\endgroup$
    – Exodd
    Jul 13 '20 at 15:55
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It is certainly possible to give $\Bbb R\cup\{P\}$ the topology $$\tau_3=\tau_e\cup\{X\setminus F:F\text{ is a finite subset of }\Bbb R\}\;,$$ and the resulting space is indeed compact. It is not, however, Hausdorff: if $x\in\Bbb R$, there do not exist $U,V\in\tau_3$ such that $x\in U$, $P\in V$, and $U\cap V=\varnothing$. The space $\langle X,\tau_2\rangle$, on the other hand, is Hausdorff, and since it is compact and Hausdorff, it is even normal. (In fact it turns out to be homeomorphic to $S^1$, the unit circle, so it is even metrizable.) Thus, $\langle X,\tau_2\rangle$ is a much nicer space than $\langle X,\tau_3\rangle$. Thus, if we’re looking for a nice compact space that has $\Bbb R$ as a dense subspace, $\langle X,\tau_2\rangle\rangle$ is preferable to $\langle X,\tau_3\rangle$. (Some people even make Hausdorffness part of the definition of compactness, so for them the space $\langle X,\tau_3\rangle$ isn’t compact.)

In any case, these examples are almost certainly setting you up for the definition of the one-point (or Alexandroff) compactification. A compactification of a space $X$ is an embedding of $X$ into a compact Hausdorff space $Y$ as a dense subspace. Your examples embed $\langle\Bbb R,\tau_{\text{disc}}\rangle$ and $\langle\Bbb R,\tau_e\rangle$ as dense subsets of compact Hausdorff spaces, so they are example of compactifications. My example above is an embedding of $\langle\Bbb R,\tau_e\rangle$ into $\langle X,\tau_3\rangle$ as a dense subset, and $\langle X,\tau_3\rangle$ is compact, but it’s not Hausdorff, so this is not an example of a compactification of $\langle\Bbb R,\tau_e\rangle$.

Your examples are not just compactifications: they are compactifications in which only one point has been added to the original space, hence the name one-point compactification. It turns out that such a compactification of a space $\langle X,\tau\rangle$ exists if and only if $X$ is a locally compact Hausdorff space, and in that case the one-point compactification is defined exactly as in Example 2: the open nbhds of the new point $P$ are the sets of the form $\{P\}\cup(X\setminus K)$, where $K$ runs over all compact subsets of $X$.

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  • $\begingroup$ So, in example 2 how can I always extract a finite subcover of any open covering?. My guess is that I can take$ K$ as a singleton ($K=\{x_0\}$), so, it is compact because it is closed and bounded and so $\{P\} \cup X\setminus K$ already contains all points of set except for $\{x_0\}$, so it is enough to take another open set to cover that point, and therefore I manage to cover everything with two sets. But this argument would hold If I didn't have neighborhoods of $P$ sufficiently big, for instance if $K$ were an infinite compact, would it? $\endgroup$
    – mathlover
    Jul 13 '20 at 18:02
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    $\begingroup$ @mathlover: If $\mathscr{U}$ is an open cover of $X$, there is some $U_P\in\mathscr{U}$ such that $P\in U$. Let $K=X\setminus U_P$; $K$ is a compact subset of $\Bbb R$. Let $\mathscr{U}_K=\{U\in\mathscr{U}:U\cap K\ne\varnothing\}$; then $\{U\cap\Bbb R:U\in\mathscr{U}_K\}$ is a cover of $K$ by sets that are open in $\Bbb R$, so some finite subcollection $\{U_1\cap\Bbb R,\ldots,U_n\cap\Bbb R\}$ covers $K$. Then $\{U_P\}\cup\{U_1,\ldots,U_n\}$ is a finite subcollection of $\mathscr{U}$ that covers $X$. $\endgroup$ Jul 13 '20 at 18:08
  • $\begingroup$ ..."there is some $U_P\in \mathscr{U}$ such that $P\in U$"... did you mean such that $P\in U_P$?, otherwise I don't know what $U$ is here and how do you know $K$ has the form $K=X\setminus U_P$? $\endgroup$
    – mathlover
    Jul 13 '20 at 18:34
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    $\begingroup$ @mathlover: Yes, that should have been $P\in U_P$. As for the other question, remember what nbhds of $P$ look like: they are of the form $\{P\}\cup(\Bbb R\setminus K)$ for compact sets $K$. Thus, $U_P=\{P\}\cup(\Bbb R\setminus K)$ for some compact $K\subseteq\Bbb R$, and clearly $X\setminus U_P=K$. $\endgroup$ Jul 13 '20 at 18:39
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    $\begingroup$ @mathlover: If you think about it for just a moment, it should be clear that $\{P\}\cup(X\setminus K)=\{P\}\cup(\Bbb R\setminus K)$ for any compact $K\subseteq\Bbb R$. And of course it’s a definition: it’s the definition of the members of $\tau_2$ that contain $P$. $\endgroup$ Jul 13 '20 at 18:53
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The 2nd version is a general construction for any (Haussdorf) topological space $X$, and is called the 'one point compactification'.
If we have an open cover $U_i$ of $X\cup\{P\}$, then $P$ is also covered by some base set $U_i=(X\cup\{P\})\setminus K$, but then the rest must cover the compact $K$ and thus a finite subcover can be selected.

Observe that the 1st version is a special case of the above, as in a discrete topological space the singleton sets are all open, thus they form an open cover, so exactly the finite subsets are the compact ones.

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  • $\begingroup$ I still don't see why K must be compact, if I want something to cover most of the points along with P, why don't we just take K finite as in the first version? $\endgroup$
    – mathlover
    Jul 13 '20 at 16:25
  • $\begingroup$ Also, Should $F$ and $K$ be subsets of the initial set $\mathbb{R}$ or subsets of $X=\mathbb{R}\cup \{ P\}$? $\endgroup$
    – mathlover
    Jul 13 '20 at 16:29

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