Topology - How do they know this is open?

[quasar987]

Homework Statement

The lemma sets out to show that if A in R^n is compact and x_0 is in R^m, then A x {x_0} is compact in R^n x R^m.

They say, "Let $\mathcal{U}$ be an open cover of A x {x_0} and

$$\mathcal{V}=\{V\subset \mathbb{R}^n:V=\{y:(y,x_0)\in U\}, \ \mbox{for some} \ U\in \mathcal{U}\}$$

Then $\mathcal{V}$ is an open cover of A."

How do they know that given some U in $\mathcal{U}$, the associated V is open?

Edit: In fact, consider the following counter example: Let n=m=1, A=[-½,½], x_0=0. Then A x {x_0} is just the segment [-½,½] considered in the R² plane. Let $\mathcal{U}=\{B_n(0,0)\}_{n\in\mathbb{N}}$ (the collection of open balls centered on the origin of radius n). Then, let V_n be the set V associated with B_n(0,0) as described above, i.e. $V_n=\{y\in\mathbb{R}:(y,0)\in B_n(0,0)\}$=[-½,½], a set that is not open in R. ah!

 

[morphism]

V_n = (-n, n)

V_n $\cap$ A = [-1/2, 1/2] (for n>0)

 

[quasar987]

Hm, yes. "" I wish I were smarter than the book.

Meanwhile I think I found the answer. Given an U and it's associated V, for any point y in V, there is an open set O_y in R^n and another one O'_y in R^m such that (y,x_0) is in O_y x O'_y and O_y x O'_y $\subset$ U. And actually, we can write

$$V=\bigcup_{y\in V}O_y$$

which is open as an arbitrary union of opens.

 

[matt grime]

Why are you introducing O_y's? If curly(U) is some cover of A x {x}, then projecting into the first coordinate gives an open cover of A, take a finite subcover, and pull back.

 

[quasar987]

I'm introducing the O_y because otherwise, I was not convinced that the projection was open.

After all, as far as I understand, if U is open in a topological space X x Y, it does not mean that it can be written as the product of an open of X with an open of Y.