Prove the 1 point compactification of a Hausdorff space X is a compact space which contains X as a dense subspace

Cratylus

Cratylus

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I am using Alexandroff 1-point compactification. I am clueless how to prove it

Here is a summary of the definition:
Let X [imath]\cup[/imath]{p} with topology defined by 1) nbhds of points of X are the same as in the
topology on X and 2) U [imath]\subset[/imath] p[imath]\notin[/imath] X is a basic open nbhd of p iff p[imath]\in[/imath] U and p[imath]\notin[/imath] X -U.

Please help.
 
My text. The text uses a symbol for Alexandroff compactication that I cannot duplicate. nbhd is neighborhood. iff is if and only if
 
The exact symbol does not matter, but you still haven't addressed my question about about "p in U and p not in U", which is always false.

What do you find difficult to prove? That [imath]X\cup {p}[/imath] is compact? That [imath]X[/imath] is dense in [imath]X\cup {p}[/imath]? Both?

Which definition of compact space are you using?

Have you looked at the Wikipedia page? Is it helpful?
 
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def of compact.
A Subspace A of a topological space X is compact if every open cover of A has finite subcover

Here is the original def of one point compactification from my text.
A79D64E6-1BA9-4623-9E17-4BF3C8A3AF4C.jpeg

How to show X is compact and contains a dense subset.
I looked at Dugunji’s text so I can start it off to a point.
 
I tried Latex formatting, but it was messy,so I gave a pic

Let p be an object not in X and [imath]\tilde{ X}[/imath] be the set X [imath]\cup[/imath] {p} with a Hausdorff space..
 
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Here is the beginning of a proof that [imath]\tilde X[/imath] is compact:

Let [imath]\{U_s\}[/imath] be a collection of open sets covering [imath]\tilde X[/imath], where [imath]s\in S[/imath] and [imath]\bigcup_{s\in S} U_s = \tilde X[/imath]. We need to show that there is a finite subset [imath]T \subset S[/imath] such that [imath]\bigcup_{t\in T} U_t = \tilde X[/imath].

If we take an arbitrary [imath]U_0[/imath] from [imath]\{U_s\}[/imath] such that [imath]p\in U_0[/imath] then what can we say about [imath]\tilde X - U_0[/imath] ?
 
X-U0 is compact. If it is compact in T2 it is closed. Then by using closed Def of sub space F=A \cup X-U0
 
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Cratylus said:
X-U0 is compact. If it is compact in T2 it is closed. Then by using closed Def of sub space F=A \cup X-U0
Click to expand...
What do you mean? What is [imath]A[/imath]? Can you prove that [imath]\tilde X[/imath] is compact?
 
blamocur said:
What do you mean? What is [imath]A[/imath]? Can you prove that [imath]\tilde X[/im [/QUOTE][/imath]
Click to expand...

Since U0 is compact X is indeed an open subset of one-point compactification
 
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The answer there says a set is open if it is a subset of [imath]X[/imath] and is open, or (more relevantly to my question) if it is of the form [imath]{p}\cup U[/imath], where [imath]U\in x[/imath] has compact complement (I've replaced [imath]\infty[/imath] with [imath]p[/imath]).
 
I guess [imath]X-U_0[/imath] is not compact,or I don’t know the answer
 
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