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I am having trouble doing this theorem. I have a weak grasp Alexandrov compactification

Here is the version compactification I am using

Let X be a space not compact and let p be an object that is not a member of X The Alexandroff one—point compactification of X, will be denoted here by X+ and is defined to be X $\cup $ {p} win topology defined by:

  1. nbhds of points of X are the same as in the topology on X
  2. U={p} $\cup$ X\K ,K$\cup$ X is compact

Things I know

  1. sets in T2 are closed

  2. Any Subspace of T2 is T2

My Try

To see X+ is compact,let $U_{{n_0}} $ $\in $ U be a set such that p $\in U_{{n_0}} $ By def. of topology of X+ we get X+\ ${U_n}$ =K where K$\subset $X is a compact set. Compactness of K gives us K $\subset$ $\bigcup$ $U_{{n_i}}$for some set $U_1$,…$U_{{n_i}}$ $\subset$U. So { $U_{{n_i}}$ } is a finite a subcover for our set.

Since we have a compactification then{p} $\cup$ (X-K),K$\subset$X is compact set {p} is open .

Ref:(https://math.stackexchange.com/a/971138/1070930)

If X\K is closed,it’s compact, it is T2

Since we have T2 space,it’s Subspace is T2,so closed

From here https://math.stackexchange.com/a/971839/1070930

I can imply denseness.

I hope my attempt doesn’t look like to much gibberish

Dazed and Confused
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  • Your proof that $X^+$ is compact, is fine. – Berci Jun 25 '22 at 22:44
  • Thanks @Berci . – Dazed and Confused Jun 25 '22 at 22:46
  • I will disagree slightly with Berci. The open cover $U$ of $X+$ is composed of open sets in $X+$, not in $X$. So while you are fine with picking one set in $U$ containing $p$, whose complement is a set $K$ compact in $X$. The remaining sets in $U$ need not be open in $X$ - some or even all of them may also contain $p$ - so you cannot claim they are an open cover of $K$ in $X$. This only requires a small adjustment to your proof to solve, but you do need that adjustment. – Paul Sinclair Jun 26 '22 at 21:53
  • Um,can you help out with that part,please @PaulSinclair – Dazed and Confused Jun 26 '22 at 22:11
  • Let $V$ be an open set in $X+$ containing $p$. What can you say about $V \setminus {p}$ as a subset of $X$? – Paul Sinclair Jun 26 '22 at 22:13
  • V is. an open subset of X+ contained in X . So V$\subset$ X is open? – Dazed and Confused Jun 27 '22 at 00:00

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