Proving Properties of Open Sets in Real Numbers

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Homework Statement



A subset U \subseteq R is called open if, for every x \in U, there is an open interval (a, b) where x \in (a, b) \subseteq U.

(a) Show that, in the above de definition, the numbers a, b may be taken
as rational; that is, if x \in U, there is an open interval (c, d) where
x \in (c, d) \subseteq U and where c, d \in Q.

(b) Show that any open set U is a union of (possibly in finitely many)
intervals (a, b) where a, b \in Q.

(c) How many open subsets of R exist?

Homework Equations





The Attempt at a Solution



i don't have much idea, the idiot prof hasnt even covered most of the stuff in class.
 
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For (a), let m := (b - a) / 2 be the midpoint of the interval. Do there exist rational numbers c and d such that a < c < m < d < b?

For (b), here's a hint:
U = \bigcup_{x \in U} x.
 
Hi, thank you for the hints but I am still stuck on part c. any ideas?
 
I take it that means that you did a and b.

I haven't given c much though myself. You could start by counting how many open intervals there are, for which it suffices to count intervals of the form (c, d) with c and d rational. Then how many unions can you take?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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