Prove or disprove that there are uncountably many disjoint open subset of reals......
According to me, this statement is true... via this a argument.... Divide reals into open intervals with consecutive irrationals as the end points.... Then since irrationals are uncountable so are the intervals...and these are disjoint and open too......
I am not sure whether this argument will work or not.... Please cross check it and comment on it....
Further I want to ask whether the same will be true if I replace the set of reals with any uncountable set , or not..... Thanks....
The phrase
is where the problem lies: there are never consecutive irrationals! Between any two irrationals we can always find a third. So your intervals can never be constructed in the first place.
As to solving the problem, here's a hint: can you think of a countable set $S$ of points which hits every open set? If so, what does this imply about any disjoint collection of open sets (given that each one must contain at least one point from the set $S$)?