The following is a well known result from point-set topology:
Theorem: Suppose $\mathbb R^n$ is the Euclidean-$n$-space and $\infty$ be a symbol not contained in $\mathbb R^n$(By Russel's Paradox).Consider the set $X=\mathbb R^n\cup \{\infty\}$ by prescribing the open sets as follows:
Type-I For $U\subset \mathbb R^n$ open in $\mathbb R^n$,the set $U\subset X$ is declared as open in $X$.
Type-II If $\infty \in U$,then $U$ is declared open in $X$ if and only if $U=\mathbb R^n\setminus K$ for some compact subset $K\subset \mathbb R^n$.
Then $X$ is a Compact Hausdorff(T_2) topological space and this topology is called the one point compactification topology on $X=\mathbb R^n\cup\{\infty\}$ or the Alexandroff extension of $\mathbb R^n$.
I have shown that $X$ is Hausdorff.I want to show that $X$ is compact.I can take resort to the stereographic projection $\phi:S^n\setminus\{N\}\to \mathbb R^n$ extending continuously to $\hat\phi:S^n\to \mathbb R^n\cup \{\infty\}$ sending $N\mapsto \infty$ under $\hat\phi$.Then we can show that this map is a continuous surjection onto $X$ from the sphere $S^n$.Now since $S^n$ is a compact topological space,so $X$ being its continuous image,is again compact.
But I want to avoid using the stereographic projection because it is from an exercise given in Fosters' book and they have not introduced stereographic projection before this problem.In fact,the very next problem is about stereographic projection.So this sequence of suggests that we should be able to do the first one without taking help of the latter.Also,it is to be noted that Alexandroff extension is more general for any $Y$ Hausdorff non-compact topological space.So,it makes sense to think that there is a general way to show $X=Y\cup\{\infty\}$ should be compact with this prescribed topology on $X$.
Does anybody have any idea to approach directly(without stereographic projection)this question that he/she might suggest to me?
Thanks in advance!
