One-point compactification problem I am a little bit confused about that question. 
$A=[0,1)\cup [2,3).$
Prove that the one-point compactification $A^{+}$ of $A$ is homeomorphic to a closed interval.
I think it is possible to choose $[0,1]$ and show that $f:[0,1]\rightarrow A^{+}$ is homeomorphic, but I am not sure.
Is it a right way to do? And if it is then I have shown that the map is bijective and don't really know how to show that the map $f$ is continuous and that the inverse is continous too. Maybe you have another way to solve the problem. 
Thank You
 A: Thank You all for your answers. I now write my answer and hope someone would give me some feedback. 
A=[0,1)U[2,3). Then I chose f:[0,2]->A+. Want to show that the map is bijectiv and continous. 
Bijectiv because:
f:[0,2]->A+
[0,1)->[0,1)
(1,2]-> [2,3)
1-> x_0 
where x_0 is the point at infinity in the one-point compactification A+ of A. 
Is it right?
Continous:
Take U open in [0,2] and want to show that the image is open in A+. 
But from here I am confused?
A: Theorem: 

If $X$ is a locally compact Hausdorff space and $Y$ is compact Hausdorff such that for some point $p \in Y$, $X$ is homeomorphic to $Y\setminus \{p\}$, then $Y$ is homeomorphic to the one-point compactification of $X$.

Proof: Let $h: X \to Y\setminus \{p\}$ be the promised homeomorphism.
Define $h': \alpha(X) = X \cup \{\infty\} \to Y$ by $h'(x) = h(x)$ for $x \in X$ and $h'(\infty) = p$. Then $h'$ is clearly a bijection. To see it is continuous, let $O \subseteq Y$ be open. If $p \notin O$, then $O \subseteq Y\setminus \{p\}$ and this set is open in that subspace so that $h'^{-1}[O] = h^{-1}[O]$ is open in $X$ and so open in $\alpha(X)$. If $p \in O$ then $Y \setminus O \subseteq Y\setminus \{p\}$ is compact and so is $C:= h^{-1}[Y \setminus O]$ as $h$ is a homeomorphism, and $h'^{-1}[O] = \{\infty\} \cup (X\setminus C)$ is also open in $\alpha(X)$ (by the definition of the topology on the one-point compactification $\alpha(X)$). So $h'$ is a continuous bijection from a compact space to a Hausdorff space hence a homeomorphism.
Then note that $(1,2] \simeq [2,3)$ (via $f:[2,3) \to (1,2]; f(x) = -x + 4$) and so 
$$[0,1) \cup [2,3) \simeq [0,1) \cup (1,2] = [0,2] \setminus \{1\}$$
so $[0,2]$ is homeomorphic to the one-point compactification of the first space.
A: Let $\infty$ be the point at infinity in the one-pint compactification B of A. Define $f:[0,2] \to B$ by $f(2)=\infty$, $f(x)=x$ if $0 \leq x <1$ and $f(x)=x+1$  if  $1 \leq x <2$. This f is a bijection and it it easy to show from the definition of the topology on B that it is continuous. Its inverse is automatically continuous since the domain and range are compact and Hausdorff.
A: Let $f(x)=x$ for $0\leq x<1$ and $f(x)=4-x$ for $2\leq x<3.$ Then $f:A\to B=[0,1)\cup (1,2]$  is a homeomorphism, and $id_B:B\to [0,2]$ is a one-point compactification of $B.$
