Let $X$ be a non-compact connected Hausdorff space in which every point has a compact neighborhood. Show $X'=X\cup\{\infty\}$ is compact and connected, $X'$ takes on the one point compactification, where $X'$ denotes the Alexandroff one-point compactification of $X$.
Another question I'm solving to prepare for an exam. To show $X'$ is compact, what do I take as the open cover of $X'$? And I am looking for an outline of why $X'$ is connected
How do you topologize $X'=X\cup\{\infty\}$?
If you take the topology given by defining a set to be open if, and only if it is either an open subset of $X$ or if it contains $\infty$ and its complement is compact, then $X'$ is called the Alexandroff compactification, and under the conditions you impose it is compact.
For connectedness, take $A,B\subseteq X'$ an open partition. Then notice that $A\cap X,B\cap X$ form an open partition of $X$, so one of them (wlog $A\cap X$) must be empty. This implies $A=\{\infty\}$. But this singleton cannot be open (as this would happen if, and only if $X$ were compact to start with).
