# Set exercise

1. Sep 12, 2013

### aaaa202

1. The problem statement, all variables and given/known data
The problem is attached as a picture.

2. Relevant equations
...

3. The attempt at a solution
I have been trying a lot to prove this without any really fruitful approach. At first I thought that the statement was false, or that you could at least construct a sequence of rationals such that V=(0,1) the following way:
Let a1 be a real in (0,1). Since the rationals are dense there exists a rational number b1 such that d(a1,b1)<ε. Let this be the first rational number in the sequence. Now let a2 be another real. Because the rationals are dense there exists a rational b2 such that d(a2,b2)<ε/2 etc. etc. and by successive use of this method I could generate the whole (0,1) with my approach. But this fails because (0,1) is not countable. So I'm open for any other approach to this problem.

#### Attached Files:

• ###### exercise.png
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2. Sep 12, 2013

### Zondrina

Can you always construct a sequence from the enumeration that converges to an arbitrary $x \in (0,1)$?

Also, showing it's a proper subset isn't too bad. Can you think of an element in $(0,1)$ that isn't in $U$?

3. Sep 12, 2013

### gopher_p

The statement that you are trying to prove is false (or, rather, not necessarily true) even after fixing it so that epsilon is fixed and the last line refers to V rather than U.

Edit: The problem, as stated, is very poorly posed. Could you maybe copy it verbatim from the source?