# Open sets in the product topology

1. Jul 5, 2013

### dumbQuestion

In reading about the Tube Lemma, an example is given where the Tube Lemma fails to apply: namely, the euclidean plane constructed as R X R. The Tube Lemma does not apply here because R is not compact. The example given is as follows:

Consider R × R in the product topology, that is the Euclidean plane, and the open set N = { (x, y) : |x·y| < 1 }. The open set N contains {0} × R, but contains no tube, so in this case the tube lemma fails.

I understand why N contains no tube (because as we go out along the x in the + and - direction and also as we go up and down along the y axis, the region N "hugs" these axis tighter and tighter and tighter, with no bounds on how tight it gets around the axis). My problem is I don't understand how I should know just by looking at it that N is an open set in R X R. I know that open sets in R X R in this product topology are generated by unions of basis elements, and the basis elements are generated by taking cartesian products of open sets in R. (so for example, the cartesian product of (1,2) X (2,3) would be a basis element in the product topology R X R). So I guess I could look at some bounded open region in R X R and say, yeah, well this is clearly an open set in the product topology. Or I could look at the union of a bunch of open bounded regions and it would be obvious these too were open sets since they are unions of basis elements. But when I start to look at an unbounded region, I am kind of "afraid" to trust my intuition here. Does this work because the topology generated by a basis is made up of any arbitrary union of basis elements? So I can keep "unioning" basis elements that are smaller and smaller down the axis to get that open set N?

EDIT: Also, why wouldn't that example for the tube lemma not applying not apply to the case where we have R X [-1,1] with the subspace topology inherited from R X R? Because here [-1,1] is compact, but if we still take N = N = { (x, y) : |x·y| < 1 } then it keeps "hugging" the x - axis as it gets further and further out and I don't see how we could get a tube in there?

Last edited: Jul 5, 2013
2. Jul 5, 2013

### micromass

Staff Emeritus
The easiest way is to take $f:\mathbb{R}^2\rightarrow \mathbb{R}:(x,y)\rightarrow |x-y|$. Then $N = f^{-1}( (-\infty, 1) )$ and is open as the inverse image of an open set under a continuous map.