How to prove that the set of real numbers can be written as the union of uncountably many pairwise disjoint subsets, each of which is uncountable?

The problem is taken from http://www.math.ucla.edu/grad/handbook/hbquals.shtml.

## Qual exam problem

### Re: Qual exam problem

Here's a nice solution for this problem that a friend of mine came up with a while back. First, it's enough to do this in the unit interval $$(0,1)$$. Begin by doing it in the unit square, $$(0,1)\times(0,1)\subset \mathbb{R}^2$$; just consider all horizontal lines for instance. Next, choose a bijection $$f: (0,1)\times(0,1) \to (0,1)$$. For example, you could map $$(x_0.x_1x_2\dots,y_0.y_1y_2\dots)$$ to $$x_0.y_0x_1y_1\dots$$. (Of course, you have to define the decimal expansion in a unique way, but this isn't too tough.) Now just consider the images of all the pairwise disjoint and uncountable sets you started with; this will be another collection of pairwise disjoint uncountable sets, and their union will be $$(0,1)$$. So you're done.

By the way, for future reference, a good site for this type of question is math.stackexchange.

By the way, for future reference, a good site for this type of question is math.stackexchange.

### Re: Qual exam problem

ns2675, thank you so much!

### Re: Qual exam problem

I think I might have another solution.

Let S be the power-set of the rationals after removing those sets which are unbounded from above.

Note that each set in S has a supremum in the Reals.

Let S_x be elements of S whose supremum is the real number x.

(1) S_x is uncountable.

Now choose two different real numbers x and y.

(2) You can see that S_x and S_y are disjoint.

(3) Note that Union(S_r | r in Reals) = S, since S_r's partition S.

Also that S has the same cardinality as the Reals.

Then we can make the bijection f(s) from S to the Reals.

Now the images f(S_r) are uncountable by (1) and disjoint by (2). Furthermore, there are uncountably many of them (one for each Real number r) and together, they cover the Reals (3).

Let S be the power-set of the rationals after removing those sets which are unbounded from above.

Note that each set in S has a supremum in the Reals.

Let S_x be elements of S whose supremum is the real number x.

(1) S_x is uncountable.

Now choose two different real numbers x and y.

(2) You can see that S_x and S_y are disjoint.

(3) Note that Union(S_r | r in Reals) = S, since S_r's partition S.

Also that S has the same cardinality as the Reals.

Then we can make the bijection f(s) from S to the Reals.

Now the images f(S_r) are uncountable by (1) and disjoint by (2). Furthermore, there are uncountably many of them (one for each Real number r) and together, they cover the Reals (3).