R has an uncountable number of distinct subsets.

It's a five part question leading up to the point that, if S is a subset of T, then if is dense then so is . 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 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.

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.

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 has an uncountable number of distinct subsets and 2^C distinct uncountable subsets

Quote:

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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 is dense then so is . 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 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.

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 has an uncountable number of distinct subsets and 2^C distinct uncountable subsets

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.

Sorry, that should read distinct dense subsets and I have no clue what he means by distinct. That was never defined.

I think the 2^C means of cardinality of the power set of though because I can prove that.

Informally, take a uncountable dense subset in and remove a countable number of points. The set will remain dense (I prove this.) Then the number of sets that can be created is equal to the power set of R because the set of removed points will form the power set.

Does this sound right?

That's not quite right. For instance, the set Q U (0,1) is an uncountable dense set, but if you remove the rationals (countably many points), it's no longer dense. The sets { Q U {x} | x in R-Q } are all dense, and there are uncountably many of them, but that's pretty trivial.

I wonder if the problem was meant to be about pairwise-disjoint dense subsets? I think that should be true. For a fixed x in R, let Q+x denote { q+x | q in Q }. Obviously every such set is dense, and the sets Q+x and Q+y are either equal or disjoint, depending on whether x-y is rational. No countable collection of these can cover R, since a countable union of countable sets is countable, and R is uncountable. So there must be uncountably many distinct ones, and for these sets, distinct is equivalent to disjoint.

hmm. OK Then what about a finite amount of points removed? That would work because the x removed would become a limit point of the new set.

I would agree that pairwise disjoint would make the most sense. Your method proves that uncountability but I need to show it's cardinality.

Yes, removing a finite number of points from a dense set will leave it dense (in R--other spaces need not have this property, but I think we're strictly working with R here, right?).

There's no way there are 2^C pairwise-disjoint subsets of R. Suppose there was such a collection, and use the Axiom of Choice to pick one point from each. Now you have 2^C distinct real numbers, which is absurd. Surely "distinct" is what is meant in that part of the question. Or we've guessed wrong about something else.

Well if you remove the "distinct" part of it I think I've proven it. So from there there are 2^C uncountable dense subsets of R. If we were to remove the ones that were not "distinct" we can remove up to at least C of them since 2^C-C=2^C. I say at least C because I don't know if there's a cardinality between C and 2^C. I'm not sure of a criteria that would remove aleph0 or C of them though.