Non-equivalent compactifications of $[0, \infty)$ I need to find non-equivalent compactifications for the interval $[0,\infty)$. I found the basic compactification which is analogy for inverse of stereographic projection only for one dimension and half of the circumference. But I cannot seem to understand how to add another points for it to be compactification. I understand there will be at least one more the Stone-Cech compactification, but I know there must be at least $4$ compactifications as I found it as an exercise.
 A: Here are some examples of compactifications of $[0,\infty)$.

*

*The closed topologist's sine-curve is a compactification whose remainder is a copy of $[0,1]$.


*The topologist's whirlpool is a compactification whose remainder is a copy of the circle $S^1$.


*Modifying the construction of the topologist's whirlpool you can see that each rose with non-zero integer values of $k$ is the remainder of a compactification: Let a ray spiral from outside around the rose such it comes closer and closer to the rose.


*The closed topologist's sine-curve can be obtained as the remainder of a compactification: Let a ray spiral around it such it comes closer and closer to it. In the $n$-th "circulation" of the ray we walk nearby the first $n$ oscillations of the sine-curve (seen from the right side) and then walk parallel to the $x$-axis above resp. below the infinitely many remaining oscillations.
These examples are easy to understand, but we can prove the very general
Theorem. For each connected compact metrizable $X$ there exists a compactification $c : [0,\infty) \to X'$ such that the remainder $X' \setminus c([0,\infty)) \approx X$.
To prove it, we can of course replace $[0,1)$ by the homeomorphic $[1,\infty)$.
Case 1. $X$ is a subset of a normed linear space $(E, \lVert - \rVert)$.
Let $B(e,r) = \{ e' \in E' \mid \lVert e' - e \rVert < r\}$ be the open ball with center $e \in E$ and radius $r$. These balls are open and convex (hence path-connected) subsets of $E$; they form a base for the topology on $E$. Note that therefore $E$ is locally path-connected.
Define
$$d_X : E \to [0,\infty), d_X(e) = \inf\{ \lVert e - x \rVert \mid x \in X \} .$$
This function is continuous and we have $d_X(e) = 0$ iff $e \in X$ because $X$ is compact. Define
$$U_n = d_X^{-1}([0,1/n)) .$$
These sets are open neigborhoods of $X$ such that $\overline U_{n+1} \subset U_n$ for all $n$ and $\bigcap_{n=1}^\infty U_n = X$. It is easy to verify that
$$U_n = \bigcup_{x \in X} B(x,1/n) .$$
Since all $B(x,1/n)$ are connected and have non-empty intersection with the connected set $X$, we see that $U_n$ is a connected open subset of $E$. Since $E$ is locally path-connected, we conclude that $U_n$ is path-connected.
Since $X$ is compact, it has a countable dense subset $D = \{x_1, x_2, x_3, \ldots \}$.
Choose a continuous path $\gamma_n : J_n = [n,n+1] \to U_n$ by travelling from $x_i$ to $x_{i+1+}$ for $i = 1,\ldots,n-1$ and then back from $x_n$ to $x_1$. The $\gamma_n$ define a continuous $\gamma : [1,\infty) \to E$ such that $\gamma \mid_{J_n} = \gamma_n$. Define
$$\Gamma : [1,\infty) \to E \times [0,1], \Gamma(t) = (\gamma(t),1/t) .$$
This map embeds $[1,\infty)$ into $E \times [0,1]$. Thus $M = \Gamma([1,\infty))$ is a copy of $[1,\infty)$ in  $E \times [0,1]$ which is disjoint from $X_0 = X \times \{0\}$. Note that $M$ is closed in $E \times (0,1]$ and that $\Gamma([n,\infty) \subset U_n \times (0,1/n]$ for all $n$.

*

*$\overline M  = M \cup X_0$:

(a) $D_0 = D \times \{0\} \subset \overline M$: By construction for each $x_i \in D$ there is a sequence $(t_n)$ in $[1,\infty)$ such that $t_n \to \infty$ and $\gamma(t_n) = x_i$. Then $\Gamma(t_n) = (x_i,1/t_n) \to  (x_i,0)$.
(b) $X_0 \subset \overline M$: Since $D_0 \subset \overline M$, we get $X_0 =  \overline D_0 \subset \overline M$.
(c) $\overline M \setminus  M = X_0$: Clearly $\overline M  \setminus  M \subset E_0 = E \times \{0\}$. So let $(e,0) \in \overline M$; we shall show that $e \in X$. Assume $e \notin X$. Then $e \notin \overline U_n$ for some $n$. The set $(E \setminus \overline U_n) \times [0,1]$ is an open neighborhood of $(e,0)$ in $E \times [0,1]$ not containing any $\Gamma(t)$ for $t \ge n$; thus $(e,0)$ cannot be a limit point of $M$.


*$X' = \overline M$ is compact:

Let $\{V_\alpha\}$ be a cover of $X'$ by open  $V_\alpha \subset E \times [0,1]$. For each $z \in X'$ choose open $U_z \subset E$ and open $J_z \subset  [0,1]$ such that $P_z = U_z \times J_z \subset V_{\alpha(z)}$ for some $\alpha(z)$. We shall show that the open cover $\{P_z \}$ has a finite subcover $\{P_{z_i}\}$; then $\{V_{\alpha(z_i)} \}$ is a finite subcover of $\{V_\alpha\}$. Since $X_0$ is compact, we find finitely many $z_1,\ldots,z_m \in X'$ such that $X_0 \subset \bigcup_{i=1}^m P_{z_i}$. Then $U = \bigcup U_{z_i}$ is an open neighborhood of $X$ in $E$ and $J = \bigcap J_{z_i}$ is an open neigborhood of $0$ in $[0,1]$ such that $X_0 \subset U \times J \subset \bigcup_{i=1}^m P_{z_i}$. We have $U_n \subset U$ for $n \ge n_0$. Choose $n \ge n_0$ such that $[0,1/n] \subset J$. Then $\Gamma([n,\infty)) \subset U_n \times (0,1/n] \subset U \times J$. The set $\Gamma([0,n])$ is compact, thus it is covered by finitely many $P_{z_{m+1}}, \ldots, P_{z_n}$. This shows that $X' \subset \bigcup_{i=1}^n P_{z_i}$.
We conclude that $c : [1,\infty) \to X', c(t) = \Gamma(t)$, is the desired compactification of $[1,\infty)$.
Case 2. Let $(X,d)$ be an arbitary metric space. It is well-know $(X,d)$ isometrically embeds into a normed linaer space; thus case 1 proves the theorem. See Show that any compact metric space $X$ can be isometrically embedded into $C([0,1])$, the space of continuous functions over $[0,1]$ and its answer.
Remark.
If $X$ is a path-connected compact metrizable space, the above proof can be substantially simplified. In fact, we do not need a bigger space $E \supset X$ to find a descending sequence of path-connected neighborhoods $U_n$; instead of paths in $U_n$ we can take paths in $X$. Compactness of $X'$ is trivial because $X'$ is closed in $X \times [0,1]$ which is compact.
Update.
As David Hartley noticed in a comment, the approach in the above remark can easily be generalized to the following
Theorem. For each compact space $X$ containing a separable dense path-connected subspace there exists a compactification $c : [0,\infty) \to X' \subset X \times [0,1]$ such that the remainder $X' \setminus c([0,\infty)) \approx X$.
This theorem applies to each compactification $c : [0,\infty) \to X'$: The space $X'$ is compact and contains $c([0,\infty))$ as a separable dense path-connected subspace. Thus there exists a compactification $c' : [0,\infty) \to X''$ such that $X'' \setminus c'([0,\infty)) \approx X'$.
This gives non-geometric proofs of examples 1 and 4.
Moreover we get the following design procedures for concrete examples:

*

*Let $f : [0,1] \to X$ be a continuous surjection such that $f(0) = f(1)$. This map $f$ has a periodic extension $F : [0,\infty) \to  X, F(t) = f(t-n)$ for $t \in [n,n+1]$. Define
$$f^* : [0,\infty) \to X \times [0,1], f^*(t) = (f(t),1/(t+1)) .$$
Then
$$c_f : [0,\infty) \to X' = X \times \{0\} \cup f^*([0,\infty)), c_f(t) = f^*(t)$$
is a compactification with remainder $X \times \{0\} \approx X$.
This is how example 2 (topologist's whirlpool) and example 3 are obtained.


*Let $f : [0,1] \to X$ be a continuous surjection. Define $\tilde f : [0,1] \to X, \tilde f(t) = \begin{cases} f(2t) & t \le 1/2 \\ f(2(1-t)) & t \ge 1/2 \end{cases}$. The map $\tilde f$ satisfies the assumption of 1. and gives rise to a compactification.
This is how example 1 (closed topologist's sine-curve) is obtained.
