I am trying to prove this theorem which states : Let $(Y,$$\sigma$$)$ be a compact $T_2$ space and $y$$\in$$Y$.Let $X=Y-${$y$}. Then$ Y$ is the one point compactification of $X$.
My Try________________
$Y$ is $T_2$. {$y$} is closed in $T_2$ as $T_2$ space implies $T_1$ also and hence $Y-${$y$} $\in$ $\sigma$.Let $U$$\in$$\sigma$ such that $y$ $\notin$$U$ $\implies$$U$$\subset$$X$
Now we define the subspae topology $(X,$$\tau$$)$ as $X$$\cap$$U$ $\in$ $\tau$,for all $U$ $\subset$ $X$
Thus clearly $y$ $\notin$$U$$\implies$$U=X$$\cap$$U$$\in$$\tau$
If $y$$\in$ $U$ $\in$ $\sigma$
Then $Y-U$ is closed in $(Y,$$\tau$$)$
$\implies$ $Y-U$$\cup$$\{$ $p$ $\}$ is closed in $Y$, union of two closed sets $\implies$ $Y-$[$U$ $\cap${$p$}$^c$ $]$ is closed in $Y$
$\implies$ $U$ $\cap${$p$}$^c$ is open in $Y$
$\implies$ $U$$\cap${$p$}$^c$ is open in $(X$,$\tau$) $\implies$ $X$$\cap$[$U$$\cap$ {$p$}$^c$] $^c$ is closed in $X$
$\implies$X$\cap$$U$$^c$(=$Y$$\cap$$U$$^c$) is closed in $X$
Now I have to show $Y$$\cap$$U$$^c$ is compact in $X$.
As $Y$ is given compact $Y$$\cap$$U$$^c$ is ccompact in $Y$.
Let $Y$$\cap$$U$$^c$ has a collection of open cover $(X$$\cap$$U$$_A$) due to subspace topology,
Now as $X$ is oopen in$Y$ topology all those open set of above collection {$A$} are open in $Y$.Now we have a open cover of $Y$$\cap$$U$$^c$ in $Y$.Since every closed subset of compact space is also compact $Y$$\cap$$U$$^c$ is compact in Y and hence has a finite subcollection of {$A$}to cover it.
This finite subcolection are all in ($X$,$\tau$) also. Thus $Y$$\cap$$U$$^c$ is compact in $X$ and closed in $X$.
and hence $Y$ is compactification of $X$ but i used the property of $T_1$ space to assume {$y$} as closed,infact my procedure tells that the above theorem is valid even in $T_1$ space But my teacher has given me example where the above one compactness does not occur in $T_1$ space.So surely there is something which I have missed in above procedure,Please help me to find out my misstake.I will be happy if someone give me a hint to prove it. Thanks for reading.