A compactification theorem issue

[radou]

So, in Munkres' chapter about the Stone-Čech compactification, there is a Lemma at the beginning of the chapter the proof of which seems a bit unclear to me. The Lemma states:

Let X be a space and let h : X --> Z be an imbedding of X in the compact Hausdorff space Z. Then there exists a corresponding compactification Y of X with the property that there is an imbedding H : Y --> Z that equals h on X. This compactification is uniquely determined up to equivalence.

Now, the first part of the proof is perfectly clear; given the map h, let Y0 denote the closure of h(X) in Z. Since Y0 is a subspace of a Hausdorff space, it is Hausdorff, and since it is closed, it is compact. Hence Y0 is a compactification of h(X).

But now it says: "We now construct a space Y containing X such that the pair (X, Y) is homeomorphic to (X0, Y0). Let us choose a set A disjoint from X that is in bijective correspondence with the set Y0\X0 under some map k : A --> Y0\X0."

This seems an easy step, but how do we know that such a set A exists? What guarantees us the legicimacy of this step? Is this equivalent to saying that, given some set S, we can choose a set S' such that these two can be put in bijective correspondence?

 

[micromass]

Hmm, it is indeed a very easy step, but it seems to be quite tricky to prove it rigourously...
I know a rigourous proof of this, but it requires some set theory (some ordinals and cardinals). I don't think it is that easy to come op with a rigourous proof of the statement, but I'll think of it...

 

[micromass]

Let me present you the proof of this step, in the case where $$Y_0\setminus X_0$$ and X are both finite.

Since X is finite, the intersection $$X\cap \mathbb{N}$$ is finite. So the complement $$\mathbb{N}\setminus X$$ is infinite. Now, assume that $$|Y_0\setminus X_0$$|=n[/tex]. Then choose n elements from $$\mathbb{N}\setminus X$$. Call the set of these n elements A. Then A is disjoint from X, by construction. And A contains n elements, thus can be put in bijective correspondace with $$Y_0\setminus X_0$$.

Now, this proof can be extended to cover all cases of X and $$Y_0\setminus X_0$$. But the problem is that we need a replacement for $$\mathbb{N}$$. This is exactly where cardinals and cardinality comes in... But there is no easy way to explain this...

It's a bit the same with theorem 29.1 on page 183. In step 2, Munkres asks you to pick an element $$\infty$$ that is not an element of X. But how can we find such an element, how do we know for certain that such an element exists. Munkres doesn't really explain these things, because he can't. We need some more set theory in order to prove such things... But it's quite unsatisfactory that we can't even explain such easy steps...

 

[radou]

Hmm, ok...

The problem with such steps is either my lack of experience or simply I'm trying to think too rigorously...

The next step of this proof is a bit unclear to me, too. It says (after constructing the bijection H):

"Then topologize Y by declaring U to be open in Y iff H(U) is open in Y0."

Y = X U A, by definition. How do we know what these open sets look like? Or should I think of all the sets open in Y0 first, and in means of them defining the open sets in Y, which must in turn again be open in Y0?

Again, I'm confused.

 

[radou]

Of course, it's clear what open sets in X are. But when we adjoin A to X...

 

[micromass]

Hmm, ok...

The problem with such steps is either my lack of experience or simply I'm trying to think too rigorously...

Well, you can't think too rigorously in math, so that's not the problem. I actually think that this is a great question. I remember that I asked a very related question when I studied topology for the first time. The professors couldn't answer my questions too, but instead they told me to have some patience and wait for the set theory course.

I suspect that Munkres knows that the step is tricky, but he hopes that nobody will notice it. All I can tell you is to wait till you learn some set theory...

[/QUOTE]
The next step of this proof is a bit unclear to me, too. It says (after constructing the bijection H):

"Then topologize Y by declaring U to be open in Y iff H(U) is open in Y0."

Y = X U A, by definition. How do we know what these open sets look like? Or should I think of all the sets open in Y0 first, and in means of them defining the open sets in Y, which must in turn again be open in Y0?

Again, I'm confused.[/QUOTE]

Well, be definition, the topology on Y is $$\mathcal{T}=\{U\subseteq Y~\vert~H(U)~\text{is open in}~Y_0\}$$. So what do they open sets in Y look like? They are just subsets of Y such that their image is open. There's not really much more one can say about it.
Let's check that $$\mathcal{T}$$ is a topology. Let's check the intersection axiom. Take U and V in $$\mathcal{T}$$, then H(U) and H(V) are open in Y0. This implies that $$H(U)\cap H(V)$$ is open in Y0. Since H is bijective, this implies that $$H(U\cap V)=H(U)\cap H(V)$$ is open in Y0. Thus, $$U\cap V\in \mathcal{T}$$.
So, how would one construct the open sets in Y. Since H is bijective, this is easy. First take all the open sets in Y0. Then take all the pre-images of such a sets. These are all the open sets in Y.

 

[radou]

OK, but why do we need the "iff" formulation then?

If the open sets in Y are basically just preimages of open sets in Y0. I assume this is because we require a homeomorphism, right? Since another way to define a homeomorphism is exactly this way "U is open iff f(U) is open"

 

[micromass]

OK, but why do we need the "iff" formulation then?

Hmm, good point. Since Munkres simply defines the open sets, I suppose that the "iff" could be changed in an "if". I suppose we could also say "We call U open if H(U) is open". This is, in fact, how I would write it...

If the open sets in Y are basically just preimages of open sets in Y0. I assume this is because we require a homeomorphism, right? Since another way to define a homeomorphism is exactly this way "U is open iff f(U) is open"

Correct. In order for H to be a homeomorphism, we need (among other things) that the preimages of open sets are open. So, what we do is just define a set open if it has this property. This forces that H will be continuous.
This kind of construction is fairly important in topology. It even has a name: the "initial topology" or the "weak topology".

 

[radou]

OK micromass, thanks a lot.