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

That is what I was inferring too
U is a subset of X+ is a basic open nbhd
of p iff p is in U and X+ -U is a compact subset of X
 
So we agree now that [imath]\tilde X - U_0[/imath] is compact.

My next hint: look again at my post #8 and consider the sub-collection of [imath]\{U_s\}[/imath] such that [imath]p\notin U_s[/imath].
 
We have a finite sub cover of. X/T which with the
addition of T is finite sub cover of X+
 
Sorry, I don't understand your notation. X is a topological space and S is a set of indices for the collection of open subsets -- how can you use the difference operation on them.
 
We have to show there is finite set T[imath]\subset[/imath] S from post #8
It looks like a finite subcover to me
 
Cratylus said:
We have to show there is finite set T[imath]\subset[/imath] S from post #8
It looks like a finite subcover to me
Click to expand...
Can you show that there is a finite "subcover" ?

I find it difficult to figure out from your terse and cryptic statements what you understand and what you don't. Consequently, it does not seem that I am actually helping you much. Can you write a complete proof that [imath]\tilde[/imath] is compact?
 
blamocur said:
Can you show that there is a finite "subcover" ?

I find it difficult to figure out from your terse and cryptic statements what you understand and what you don't. Consequently, it does not seem that I am actually helping you much. Can you write a complete proof that [imath]\tilde[/imath] is compact?
Click to expand...
I gave a link to one such proof. Here: http://www.math.buffalo.edu/~badzioch/MTH427/_static/mth427_notes_18.pdf

I was hoping someone would do the proof cause I am not experienced in this topic

Since X-U0 is compact ,it is closed
and in Hausdorff,compact sets are closed
then there exists
 
I don't need links to proofs. I thought you needed help in proving yourself the title statement of your post. The purpose of this forum is not to supply solutions and proofs but to help with understanding and solving math problems. The fact that you are not experienced in this topic is not in itself a problem: if you want to get experience we are happy to help, but you really have to try, make some mistakes, ask some silly questions and in the end develop the experience and the understanding needed to do these proofs. Are you interested in going this route, or is your only goal to find a proof which you can copy-paste to your homework?
 
I seem to be asking silly questions. I am trying to prove the proposition in the title.

I am not a student. I am 59. I graduated years
ago. I am doing this for fun and am autodidact

yah I am interested
 
I am somewhat rusty on topology myself, but I don't see why it is necessary for [imath]X[/imath] to be a Hausdorff space. I'll try to write down a proof and see if I missed something.
 
I am rustier than I thought: [imath]X[/imath] cannot be an arbitrary space. At the very least it must be true that every compact subset is closed, and Hausdorff topologies have this property.

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 $T \subset S$ such that [imath]\bigcup_{t\in T} U_t = \tilde X[/imath].

Claim 1: If we take an arbitrary [imath]U_0[/imath] from [imath]\{U_s\}[/imath] such that [imath]p\in U_0[/imath] then [imath]X-U_0[/imath] is compact.
Indeed, [imath]X-U_0 = \tilde X-U_0[/imath], but the latter is compact because [imath]U_0[/imath] is open in [imath]\tilde X[/imath].

Claim 2: If [imath]U[/imath] is open in [imath]\tilde X[/imath] then [imath]X\cap U[/imath] is open in [imath]X[/imath].
To prove we notice that [imath]X-U[/imath] is compact and, since [imath]X[/imath] is Hausdorff, it is also closed. Which means that [imath]X\cap U = X-(X-U)[/imath] is open.

Claim 3: [imath]\tilde X[/imath] is compact.
To prove this we need to find that finite subset [imath]T\subset S[/imath] such that [imath]\{X_t\}[/imath] covers all of [imath]\tilde X[/imath]. First we notice that [imath]\{X\cup U_s\}[/imath] is a covering of [imath]X[/imath] because each [imath]X\cup U_s[/imath] is open in [imath]X[/imath] according to Claim 2. Now pick an arbitrary open subset [imath]U_0[/imath] from [imath]\{U_s\}[/imath] which contains [imath]p[/imath]. Since [imath]X-U_0[/imath] is compact there is a finite index set [imath]T[/imath] such that [imath]X-U_0 \subset \cup_{t\in T} U_t[/imath], i.e. [imath]X-U_T[/imath] is a covering for [imath]X-U_0[/imath]. But [imath]\{U_t\}[/imath] together with [imath]U_0[/imath] is a covering for [imath]\tilde X[/imath] -- q.e.d.

Claim 4: If [imath]X[/imath] is not compact then [imath]X[/imath] is dense in [imath]\tilde X[/imath].
If [imath]X[/imath] is not compact the single point subset [imath]\{p\}[/imath] of [imath]\tilde X[/imath] is not open. But then any neighborhood of [imath]p[/imath] intersects with [imath]X[/imath], q.e.d.
 
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