$(0,2)$ is a locally compact Hausdorff space. Then what will be that one point which will make $(0,2)$ a compact Hausdorff space? If $X$ is locally compact Hausdorff , there exists a compact Hausdorff space $Y$ such that $Y-X$ consists of a single point. This theorem is taken from Mukresh. But I have a doubt. 
$(0,2)$ is a locally compact Hausdorff space. Then what will be that one point which will make $(0,2)$ a compact Haudorff space?
 A: Let $T_X$ be a locally compact, non-compact  Hausdorff topology on $X.$ Let $p$ be some (any) point not in $X$ and let  $Y=\{p\}\cup X.$ Let $C$ be the set of all compact subsets of $X,$ with respect to the topology $T_X.$  Define a topology on $Y$ as $T_Y=T_X\cup \{Y\setminus c:C\in C\}.$
It is straightforward to show that $T_Y$ is a compact Hausdorff topology on $Y,$ and that the topology on $X,$ as a subspace of $Y,$ is $T_X.$ (And also that $X$  is dense in $Y$ and open in $Y.$)
It may be convenient to take a space $(X', T_{X'})$  that is homeomorphic to $(X,T_X),$  where $(X', T_{X'})$  is a dense open subspace of  a compact Hausdorff space $(Y', T_{Y'}),$ such that $Y'\setminus X'$ contains a single point $p$.
For example  with $X=(0,2),$ the function  $f(x)=e^{\pi i (x-1)}$  maps $X$ homeomorpically to $X'=Y'\setminus \{-1\},$ where $ Y'= \{z\in \Bbb C:|z|=1\}.... $
....  so let $p=-1$  and let $Y=X\cup \{-1\}.$ And extend the function $f:X\to X'$ to $f:Y\to Y'$ by letting $f(p)=p.$ 
Now let $U\subset Y$ be open in $Y$ iff $\{f(u):u\in U\}$ is open in $Y'.$
As to what this topology on $Y=(0,2)\cup \{-1\}$ "looks like":  If $U \subset Y$ then $U$ is open in $Y$ iff 
(i). $U\cap (0,2)$ is open in $X=(0,2),$ and 
(ii). if $(-1)\in U$ then $U\supset (0,r_0)$ and $U\supset (r_2,2)$ for some $r_0,r_2\in (0,2).$  
