Point-Set Topology Question: Convergence on Open and Closed Intervals

[Poopsilon]

I debated whether to put this in this sub-forum or in the Topology & Geometry sub-forum, but I decided I'd give you guys the first crack at it:

Take the union of all open intervals on the real numbers which do not include the number 1, call this union A. Then take the union of all closed intervals on the real numbers which do not include the number 1, call this union B. I am fairly confident that every point in B is a point in A, but I cannot decide if the reverse is true.

I mean I kind of want to say it is because I could take the infinite union of the intervals [0 k/k+1] and this would become the interval [0,1), not closed I know but as Rudin clearly states infinite unions of closed sets need not be closed.

To give some context, this basically came up as an ancillary consideration in one of the problems on uniform convergence in chapter 7 of Baby Rudin. If a sequence of functions appears to converge uniformly on any bounded interval, and is continuous on any bounded interval not including 1, where there is a vertical asymptote, than this means any closed interval should suffice.
With just point-wise convergence I can just set an interval to be (a,1), but I have to use that infinite union trick with closed intervals, does this make sense, am I just telling tales out of school or what?

Sorry if this should go in the Homework Forum, it was a borderline case.

 

[Fredrik]

Homework, topology, analysis...all three options make sense. If x is a member of A, it's also a member of a closed interval [a,b] that contains x but not 0, so x is a member of B. If x is a member of B, it's also a member of an open interval (a,b) that contains x but not 0, so x is a member of A.

 

[Poopsilon]

Ok, that makes sense ( I assume you mean 1 not 0). In regards to the function I discussed, I still must describe the sets upon which f converges uniformly in terms of closed or bounded intervals, correct? I still would not be able to say that f converges uniformly on any open interval, even though every point of A is a point of B and visa versa.

If this is in fact the case I assume it comes from the nonsensicalness of discussing uniform convergence on single points; instead it must be discussed with respect to sets. Rudin makes this remark explicit with regard to point-wise continuity vs uniform continuity, but does not expound upon the analogous issue of point-wise vs uniform convergence.