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.
If we take your definition ($Y$ compact Hausdorff and $y_0 \in Y$ such that $i: X \rightarrow Y \setminus \{y_0\}$ is a homeomorphism), the answer is clear:
$i[X] = Y \setminus \{y_0\} \subset Y$ is compact and hence closed in $Y$ (as $Y$ is Hausdorff), and so $\{y_0\}$ is open (i.e. $y_0$ is an isolated point of $Y$). So $Y$ then consists of a (closed) topological copy of $X$ (namely $Y \setminus \{y_0\}$) and an isolated point $y_0$. Indeed $i[X]$ is not dense in $Y$. So indeed this is incompatible with the usual definition of a compactification:
Commonly, a (Hausdorff) compactification of a space $X$ is a pair $(Y, i)$ where $Y$ is compact Hausdorff and $i : X \rightarrow Y$ is an embedding (or equivalently, $i$ is a homeomorphism between $X$ and $i[X] \subset Y$) and $i[X]$ is dense in $Y$. In this definition, a one-point compactification $X$ is a Hausdorff compactification $(Y,i)$ such that moreover $Y \setminus i[X]$ is a singleton.
Two compactications $(Y,i)$ and $(Y',i')$ of $X$ are called equivalent when there is a homeomorphism $h: Y \rightarrow Y'$ such that $h \circ i = i'$ as maps on $X$. One then shows that $X$ has a one-point compactification iff $X$ is locally compact Hausdorff and non-compact and moreover all of them are equivalent in the above sense.
In this general definition, if $X$ is compact and $(Y,i)$ is a Hausdorff compactification, $i[X]$ is compact and so closed, so it can only be dense in $Y$, $i[X] = Y$ and $Y$ is just a homeomorph of $X$. So there is no one-point Hausdorff compactification, in the general sense, when $X$ is already compact.
So maybe your text does not use the more common definition that includes denseness, or it implicitly assumes all considered spaces are non-compact, in which case we have no problem.
Best Answer
Edit: Now that your post has been fixed, let me respond appropriately. First, let me clarify a few points.
In fact, every non-compact set admits a one-point compactification, in the following sense:
We also have the following:
The above are good exercises to prove.
Now, the rest depends on what you mean by $X\cup\{(1,1)\}$.
If you intend that $X\cup\{(1,1)\}$ be simply considered as a subspace of $\Bbb R^2$ (which it seems that you do), then we can construct the one-point compactification of $X\cup\{(1,1)\}$ by the Proposition. However, it will not be a Hausdorff space. Note that for any neighborhood $U$ of $(1,1)$ in $X\cup\{(1,1)\}$, and for any $K$ such that $U\subseteq K\subseteq X\cup\{(1,1)\},$ we have that $K$ has as part of its boundary (in $\Bbb R^2$) the open segment from $(1,1)$ to $(\alpha,1)$ for some $0<\alpha<1$, and this open segment is disjoint from $K$, so in particular, $K$ is not compact. (Why not?) Thus, $X\cup\{(1,1)\}$ is not locally compact (though it is Hausdorff), so by the Corollary, its one-point compactification is not Hausdorff (though the compactification does exist by the Proposition, and is $T_1$). Now, the one-point compactification of $X$ is homeomorphic to the closed triangular figure with vertices $(0,0),(1,0),(1,1)$. The one-point compactification of $X\cup\{(1,1)\}$ is the quotient space $$[0,1]^2/\bigl\{(x,1)\in[0,1]^2:x\ne 1\bigr\}.$$ I can't see any "nice" spaces homeomorphic to that. In particular (as kahen points out in his comment below), it isn't a pseudometrizable space. It does have some nice properties, though, such as uniqueness of sequence limits (as I lay out in the comments below), which not all $T_1$ spaces satisfy.
If, on the other hand, you intend that $X\cup\{(1,1)\}$ be topologized as a disjoint union--which in particular means that $(1,1)$ is an isolated point--then the one-point compactification of $X\cup\{(1,1)\}$ is simply the one-point compactification of $X$, together with another point that is isolated.