Confusion about Alexandroff Extension I am reading about one point compactification, which is a topic that was omitted by my professor in my first topology course, but I need it now. For some reason I find it really confusing and I have several questions about it.
To begin, the book I use defines compactification in general as:

A compactification of space $X$ is a dense embedding $i: X \to
> Y$, where $Y$ is a compact space.

Then it defines one point compactification as:

A one point compactification of space $X$ is an open embedding $i: X
 \to Y$ (where $Y$ is compact), such that:

*

*The set $Y - i(X)$ contains exactly one point.

*If $K$ is a
compact closed subset of $X$, then $i(K)$ is closed in $Y$.


Now let's go to the questions I have.
In the definition of one point compactification, I find it reasonable that we want an embedding to a compact space and that we want to add exactly one point. I do not find it clear, however, why we demand that the embedding is open. What does it give us? What would happen if we didn't demand that? I am also not entirely sure why the second part is necessary.
Then, although it's not mentioned in the book, I assume it would be normal to verify that one point compactification is in fact compactification. I tried to show that in the following way, but with no success.

We want to see that embedding $i$ from one point compactification is
dense in X (since everything else is already true). That would mean
that $i(X)$ intersects every non-empty open subset of $Y$. Now, the
only subset that it doesn't intersect is $\{ \infty \}$ (the point
added in compactification). The question now is whether $\{ \infty \}$
is open. It's complement is $i(X)$. $X$ is always closed in $X$. So,
if it is also compact, then $i(X)$ is closed and then $\{ \infty \}$
is open. That wouldn't be good, since then $i(X)$ is not dense.
Therefore, what I get is that $X$ must not be compact in order for
compactification to be possible.

I didn't come to what I wanted, which is that one point compactification is in fact compactification. I would also appreciate if someone tells me how to actually prove that.
What I did realise, however, is that it is not possible to do one point compactification on a compact set. Is this conclusion I made correct?
Then, I am confused about another thing that my book states. It states:

A compact (Hausdorff) space is its own compactification, even more, it
is it's only compactification, since it cannot be dense in any larger
compact space.

This statement is given without justification, but I simply cannot see why it would be true. I'm not sure if I am missing something obvious. I am also not entirely sure if this statement is only true for Hausdorff compact spaces or all compact spaces.
I would be very grateful for any help or answer, thanks in advance!
 A: 
I do not find it clear, however, why we demand that the embedding is open. What does it give us? What would happen if we didn't demand that?

The motivation behind this definition is that, when the compactification is Hausdorff then actually the embedding is always open, and the image of compact is closed. So in Hausdorff case these properties are automatic.
However without those properties (i.e. in non-Hausdorff case) the one-point compactification (as in compact space with one more point and an embedding) does not have to be compactification at all (as in dense embedding). Moreover such construction would be possible for compact spaces, giving as a weird proper compactification of a compact space, which is a rather pathological situation, not what we want.

I assume it would be normal to verify that one point compactification is in fact compactification. I tried to show that in the following way, but with no success.

So you want to show that $i(X)$ is dense in $Y$. Assume it is not. Then $\overline{i(X)}=i(X)$ and thus the special $\infty$ point is open. On the other hand $i(X)$ is open and so $\infty$ is clopen. And thus the topology on $Y$ is the disjoint union topology of $i(X)$ and $\{\infty\}$ and thus $Y$ is compact if and only if $i(X)$ is, which is if and only if $X$ is. And so if we start with a non-compact $X$, then we get contradiction. Meaning $i(X)$ has to be dense in $Y$.
On the other hand if $X$ is compact, then $i(X)$ is closed (by property 2) in $Y$. Meaning $\overline{i(X)}=i(X)$ is not dense in $Y$, and thus the compactifcation is invalid.

What I did realise, however, is that it is not possible to do one point compactification on a compact set. Is this conclusion I made correct?

Yes, true.

A compact (Hausdorff) space is its own compactification, even more, it is it's only compactification, since it cannot be dense in any larger compact space.

The statement is false. For any topological space $X$ define $X^*=X\sqcup\{\infty\}$ with topology given by: $U\subseteq X^*$ is open if and only if $U\subseteq X$ is open or $U=X^*$. Then $X^*$ is always a non-Hausdorff compactification of $X$, always "larger", even when $X$ already was compact Hausdorff. In fact $X$ embeds into $X^*$ as an open subset, however the property 2. is violated.
The statement is only true when we consider Hausdorff compactifications. Because if $i:X\to Y$ is an embedding with $X$ compact and $Y$ Hausdorff, then $i(X)$ has to be closed in $Y$, and thus $i(X)=Y$ by density.
A: We begin with a few preparations.
Let $X$ be a space with topology $\mathcal T_X$,  $X^+ = X \cup \{\infty\}$ with a point $\infty \notin X$ and $i :X \to X^+$ be the inclusion function. Define
$$\mathcal U_\infty = \{ Y \setminus C \mid C \subset X \text{ is compact and closed in } X\} ,$$
$$\mathcal T^+ = \mathcal T_X \cup \mathcal U_\infty .$$
It is easy to verify that $\mathcal T^+$ is a topology on $X^+$ such that $X^+$ becomes a compact space and $i$ becomes an open embedding (note that for $Y \setminus C \in \mathcal U_\infty$ we have $(Y \setminus C) \cap X  = X  \setminus C \in \mathcal T_X$). This space is called the Alexandroff extension of $X$. It is defined for all spaces $X$.
Clearly the Alexandroff extension is a one-point compactification in the sense of your definition.
Unfortunately it is not always a compactification in the sense of your general definition. In fact, it is a compactification if and only if $X$ is non-compact: If $X$ is compact, then $Y$ is the topological sum of $X$ and a one-point space and $X$ is not dense in $Y$; if $X$ is non-compact, then each $V \in \mathcal U_\infty$ intersects $X$ which shows that $X$ is dense in $Y$.
Let us call an embedding $i : X \to Y$ a compact one-point hull of $X$ if $Y$ is compact and $Y \setminus i(X)$ is a one-point set. Clearly each one-point compactification of $X$ is a compact one-point hull of $X$.
Let us consider an arbitrary compact one-point hull. Of course we may assume w.l.o.g. that $Y = X \cup \{\infty\}$ with a point $\infty \notin X$ and that $i :X \to Y$ is the inclusion function.
Let $\mathcal T_X, \mathcal T_Y$ denote the topologies of $X, Y$. Since $X$ is a subspace of $Y$, for each $U \in \mathcal T_X$ at least one of the sets $U, U^+ = U \cup \{\infty\}$ must be contained in $\mathcal T_Y$. If $U^+ \in \mathcal T_Y$, then $Y \setminus U^+$ is closed in $Y$, thus compact. Clearly $Y \setminus U^+ \subset X$, i.e. $Y \setminus U^+ = (Y \setminus U^+) \cap X$ is closed in $X$. This shows that $U^+ = Y  \setminus (Y \setminus U^+) \in \mathcal U_\infty$. We conclude that
$$\mathcal T_Y \subset \mathcal T^+ .$$

*

*If $X$ is non-compact, then $X$ is dense in $Y$: Each open neigborhood of $\infty$ must have the form $Y \setminus C$ with a compact closed $C \subset X$. Since $C \ne X$, we see that $Y \setminus C$ intersects $X$. In other words, if $X$ is non-compact, then $Y$ is a compactification of $X$ in the sense of your general definition.


*If $X$ is compact, then $X$ may be dense or non-dense in $Y$. For example, $X$ is open but not dense in its Alexandroff extension. If we take $\mathcal T_Y = \mathcal T_X \cup \{Y\}$, then $X$ is dense (and open) in $Y$.


*$X$ is open in $Y$ if and only if $\mathcal T_X \subset T_Y$. This is obviuos.


*Here is an example where $X$ is not open in $Y$:  Take $X = \mathbb R$ with its standard topology and $Y = \mathbb R \cup \{\infty\}$ with $\mathcal T_Y = \{ U \in \mathcal T_X \mid \xi \notin U \} \cup \mathcal U_\infty$, where $\xi \in \mathbb R$ is fixed.
Let us now show that your definition of a one-point compactification of a space $X$ (which is a particular case of a compact one-point hull of $X$) specifies precisely the Alexandroff extension. We may again assume that $Y = X \cup \{\infty\}$ with a point $\infty \notin X$ and that $i : X \to Y$ is the inclusion function. From the above considerations we know that

*

*$\mathcal T_X \subset \mathcal T_Y \subset \mathcal T^+$.

*Since all closed compact subsets $C$ of $X$ are also closed in $Y$, all $Y \setminus C$ are open neigborhoods of $\infty$.  Thus $\mathcal U_\infty \subset \mathcal T_Y$.

*We conclude that $\mathcal T_Y = \mathcal T^+$.

Note that a space $X$ may have many distinct compactifications with a one-point remainder (see 4. above).
Compact spaces $X$ have a trivial compactification with empty remainder, but also have compactifications with a one-point remainder (see 2. above).
This is also true for compact Hausdorff spaces $X$, but if we require that compactifications are Hausdorff, then $X$ only has the  trivial compactification (because no compact space can be dense in bigger Hausdorff space).
