Originally Posted by

**magus** It's a five part question leading up to the point that, if S is a subset of T, then if $\displaystyle S$ is dense then so is $\displaystyle T$. That much was easy but part four took me some time and I have no idea what part five is asking. I wrote up a proof for four but I wanted to make sure it was correct before I moved on.

I also have the problem of getting the next element in countable set not necessarily $\displaystyle \mathbb{Z}$ so I've assumed a method for the object of a member of such a set named previous which returns the value of the next element. This makes sense since every member of an infinite set has a next element.

With the latex partially down this will be difficult to describe but here we go.

$\displaystyle \mathbb{R}$ has an uncountable number of distinct subsets.

Let i be an element of I (an infinite countable set) and j be an element of J (an uncoubtable set) and S_i be a countable dense subset. There exists an element x_j_i in R such that x_j_i is not in S_i

Let S_i.next() = S_i U {x_j_i}

Because the S_i is a subset of S_i.next() there are a infinite countable number of dense subsets. However note that any combination of x_j_i can be used which means that, since it is of an uncountable index, there are an uncountable number of possible S_i.

What do you mean by "distinct subsets"? Do you mean "disjoint subsets"? Then it's easy: the set { {x} / x in R } is an uncountable

set of pairwise disjoint subsets of R. Q.E.D.

Tonio

Unfortunately I can't render this in latex either but the symbol in part five is a 2 to the power of a gothic looking C

The question goes:

Again using (iii) prove that $\displaystyle \mathbb{R}$ has an uncountable number of distinct subsets and 2^C distinct uncountable subsets