Let S be an uncountable subset of [0,1]. Prove that there is a real number c in [0,1] such that both S∩[0,c] and S∩[c,1] are uncountable.
It seems obvious, but what is the proof?
thanks.
well, if $\displaystyle S=[a,b] \text{ or } (a,b) \subseteq [0,1]$, you can take the midpoint of $\displaystyle S$.
so, $\displaystyle c= \frac{b-a}{2}$
since $\displaystyle c$ divides $\displaystyle S$ into two, countability of $\displaystyle [a,c]$ and $\displaystyle [c,b]$ (or the open set) are the same..
if both $\displaystyle [a,c]$ and $\displaystyle [c,b]$ are countable, then $\displaystyle S=[a,c]\cup [c,b]$ is also countable which is a contradiction.
thus, both must be uncountable. and $\displaystyle S\cap [0,c] = [a,c]$ and $\displaystyle S\cap [c,1] = [c,b]$
As "both $\displaystyle [a,c]$ and $\displaystyle [c,b]$ are countable" fails, $\displaystyle [a,c]$ is uncountable OR $\displaystyle [c,b]$ is uncountable. you say "both must be uncountable", why?
The set $\displaystyle S$ can be the irrationals in $\displaystyle [0,1]$ i.e. $\displaystyle S$ may contain no subintervals.
the fact that you cut S in the middle part makes all the works..
to illustrate more, if you stay in the middle and YOU are in the middle, and since you know you are in the middle, there will be equal number from your left and from you right.. that is the time when you could say that they have the same countability (cardinality).. get the point?
as per S can be irrationals, you can always write [a,b]\Q.