One-point compactification problem

[radou]

Homework Statement

Show that the one-point compactification of N (the naturals) is homeomorphic with the subspace {0} U {1/n : n is in N} of R.

The Attempt at a Solution

If we show that N is homeomorphic with {1/n : n is in N}, then this homeomorphism extends to the one-point compactifications of these spaces.

First of all, I assume we have the discrete topology on N? And on {1/n : n is in N}, too, right? Since if {1/n : n is in N} is a subspace of R, then the open sets in the subspace topology are all the intersections of the open sets of R with {1/n : n is in N}, which form its power set.

Let f(n) = 1/n. Clearly this is a bijection from N to {1/n : n is in N}. Clearly it's a homeomorphism, since if U is open in N, f(U) is open in {1/n : n is in N}, and backwards.

Since {0} U {1/n : n is in N} is the one-point compactification of {0} U {1/n : n is in N}, it must be homeomorphic to the one-point compactification of N.

 

[radou]

By the way, this seems like a simple example, but I want to make sure if my reasoning is correct.

The set {1/n : n is in N} is locally compact and Hausdorff, so it has a one-point compactification.

Hausdorff: for any 1/n1 and 1/n2, the neighborhoods {1/n1} and {1/n2} are open and disjoint.

Locally compact: for any 1/n, the subspace {1/n} is compact and contains the neighborhood {1/n} of 1/n.

 

[micromass]

Yes, it seems all fine.

But why is $$\{0\}\cup\{1/n~\vert~n\in \mathbb{N}\}$$ the one-point compactification of $$\{1/n~\vert~n\in \mathbb{N}\}$$. It's not to hard to see, but you still have to show it...

 

[radou]

Yes, good point.

Well, first of all, {1/n : n is in N} is not compact - take, for example, the open cover by sets of type

U1 = R\{1/n : n is a positive integer}
U2 = R\{1/(n+1) : n is a positive integer}
U3 = R\{1/(n+2) : n is a positive integer}, etc.

The family {Un} covers {1/n : n is in N}, but has no finite subcover.

A space has a one-point compactification iff it is locally compact Hausdorff, and not compact.

 

[micromass]

Yes, so you've proven that $$X=\{1/n~\vert~n\in \mathbb{N}\}$$ is noncompact, locally compact and Hausdorff. Thus the space has a one-point compactification. This is the space $$\{0\}\cup \{1/n~\vert~n\in \mathbb{N}\}$$ with the following topology

$$\mathcal{T}=\{A~\vert~X\cap A~\text{open and}~(0\in A~\Rightarrow~X\setminus A~\text{compact})\}$$

This is, by definition, the topology of $$\{0\}\cup \{1/n~\vert~n\in \mathbb{N}\}$$. But this space already carries a topology (the Euclidian topology). So you still need to show that these two topologies coincide...

 

[radou]

Hm, just a second.

Aren't the open sets in a one-point compactification Y of the space X the sets which are open in X, and the sets of the form Y\C, where C is a compact subspace of X?

The compact subspaces of {0} U {1/n : n is in N} are its finite subsets, right? Isn't it obvious that this topology coincides with the discrete topology?

 

[micromass]

The space {0} U {1/n : n is in N} is not really discrete... the singleton {0} is not open...

But anyway, yes I know it's trivial that {0} U {1/n : n is in N} is the one-point compactification of {1/n : n is in N}. But since you didn't mention it, I thought you forgot it...

 

[radou]

Hmm, I'd like to make sure I fully understand this.

The "main theorem" about compactification states the following:

Let X be a space. Then X is locally compact Hausdorff iff there exists a space Y such that i) X is a subspace of Y, ii) Y\X consists of a single point, iii) Y is compact Hausdorff.

(Further on, if Y and Y' are two such spaces, then they are uniquely determined up to a homeomorphism)

Now, in the proof of this theorem, the space Y is constructed from the space X by adjoining some object ∞ that is not in X and forming the set Y = X U {∞} (by the way, is this "∞" an allusion to the compactification of the reals? why the infinity symbol? and why the term "object"? obviously this can be very general)

Now, the topology on Y is defined as I mentioned in post #6, and it is proved that this is in fact a topology on Y. Then it is showed that X is a subspace of Y, that Y is compact, and that Y is Hausdorff.

The thing I don't understand about this:

What is the object that we add to X in order to compactify it? Can we add any object to X? Since from that point on, in the proof of the theorem, everything seems to follow easily, no matter what this object is. Or what set this element belongs to.

This may seem like a stupid question, but could we compactify {1/n : n is in N} with some other point? So that this compactification is homeomorphic to {0} U {1/n : n is in N}?

Sorry if I wrote a bunch here, but I really want to straighten this out. :)

 

[micromass]

Ow I see. Well if you just use that theorem, then you have proven everything already in your OP. Ignore my other posts.

About that $$\infty$$ you adjoin to X. Well, you can just take $$\infty$$ to be everything, as long as you don't take anything in X. It doesn't matter what you take, it ends up being homeomorphic anyway.

So for $$X=\{1/n~\vert~n\in \mathbb{N}\}$$, you could adjoin 0 to X. But you could as well adjoin -1 to X to obtain the one-point compactificiation. But note: if you adjoin -1, then the space $$X\cup \{-1\}$$ will carry two different topologies: the one-point compactification and the Euclidean topology. Only if you adjoin 0, these two topologies will coincide...

 

[radou]

So for $$X=\{1/n~\vert~n\in \mathbb{N}\}$$, you could adjoin 0 to X. But you could as well adjoin -1 to X to obtain the one-point compactificiation.

OK.

But note: if you adjoin -1, then the space $$X\cup \{-1\}$$ will carry two different topologies: the one-point compactification and the Euclidean topology. Only if you adjoin 0, these two topologies will coincide...

Hm, just a second, by the Euclidean topology you mean the topology induced by the Euclidean metric, i.e. in this case d(x, y) = |x - y|? I don't see how, aren't the open sets in our subspace sets from the power set family?

 

[micromass]

Yes, the Euclidian topology on $$\{-1\}\cup\{1/n~\vert~n>0\}$$ is discrete, thus every set is open.
But this set with this topology is not compact, thus this topology can not be the one-point compactification.
You can however put a one-point compactification topology on $$\{-1\}\cup\{1/n~\vert~n>0\}$$, but this will be different from the subspace topology.

 

[radou]

OK.

So, basically, if we consider the one-point compactification topology, then the set is compact, right?

 

[micromass]

Yes, the one-point compactification must always be compact!

 

[radou]

OK, micromass thanks a lot for your patience, as always! :)