Is C Closed, Open, or Neither? Understanding the Topology of R^2 and the Set C

[pivoxa15]

Homework Statement

Consider R^2 and the set C={(x,y)|x in Q, y in R}
Is C closed, open or neither?

The Attempt at a Solution

C consists an infinite union of closed sets (infact straight lines in the plane). Is an infinite union of closed sets not closed? If so why. C is clearly not open.

 

[StatusX]

An infinite union of closed sets is not necessarily closed for the same reason an infinite intersection of open sets is not necessarily open.

If the set is closed, sequences in the set with a limit have a limit in the set. Is this true?

 

[pivoxa15]

Good point.

Any infinite continued fraction equals an irrational number

Consider an infinite sequence of finite continued fractions that converge in the limit to an irrational number. The fact that each element in the sequence is a finite fraction means it is a rational number hence on the x-axis but the limit of this sequence is an irractional number which is not in C so C is not closed.

C is not open because one can draw a ball around any point in C and not every point in the ball is in C.

 

[StatusX]

Yea, although it's slightly easier to just note that any real number is the limit of a sequence of rationals (say, the truncated decimal expansions). Also, you need to show that there is no ball around the point contained in C, not just that some ball isn't.

 

[pivoxa15]

So I should have said for any ball with radius r>0, there will be points in the ball not in C.

 

[matt grime]

If a set is closed it is its own closure. WHat is the closure (in the metric topology) of the rationals in the reals? By *definition* it is all of the reals, so the rationals are not closed. (we'll assume you can assume that there are non-rational real numbers).