# Thread: Screenable and metacompactness of Moore Spaces

1. ## Screenable and metacompactness of Moore Spaces

Hello all mathematicians:

If $X$ is the upper half plane including the real axis $L$, we let each point above the real axis be open and take as a neighborhood basis of points $x$ a "V" with vertex at $x$, sides of slopes ±1 and height $1/n$. I showed that $X$ is a metacompact Moore space. But I need your help in order to show that $X$ is not screenable. I found in Heath's paper that $X$ is not screenable follow directly by a category arguement. But I don't catch the idea honestly. This topological space $X$ is called the tangent V topology.
Please guide me and every advise is highly appreciated.

2. Are you referring to Heath's paper "Screenability, pointwise paracompactness, .." in the CJM? He gives two arguments, the second referring to McAuley's paper. You don't like this one as well?

3. I have Heath's paper. In the page 766, Heath gave this example of non screenable Moore space. He said that we can show that tangent V topology is not screenable follow readily by a category arguement or by another method I didn't understand it. If you can halp me in order to show that tangent V topology is not screenable by a simple way, I will be very gratefull to you.

Thank you in advance

4. I don't off-hand have a proof. But according to Heath, it's not that complicated. First, do you know the meaning of a screenable space? Second, have you tried to use the fact that the irrationals are second category in the reals?

5. Hello Mr. ojones

Certainly I know what is it mean by screenable space, otherwise how could I ask about it ? In short, a topological space $X$ is called screenable if for each open covering $G$, there is a sequence $G_{n}$ of collections of pairwise disjoint open sets such that union of $G_{n}$ is a refinement of $G$.

6. OK, so you need to show that there's an open cover for which the condition fails. What about the set of V's whose vertex is irrational?

7. Thank you very much Mr. ojones. You gave me the arguement I shall use in order to prove that such a condition fails. That is by taking the set of V's whose vertex is irrationals and using the fact that irrationals are second category subset of the reals. Since this is the time of my final exam, I will complet the prove and write it properly at the end of my exams, and then I will write it here.

Thank you very very much Sir for your help and guidance.

8. I'm not saying this will work; it's just something to try.