# Cardinality of a subset of [0,1]

1. Aug 5, 2013

### ribbon

1. The problem statement, all variables and given/known data
What is the cardinality of the set of all numbers in the interval [0, 1] which
have decimal expansions with a finite number of non-zero digits?

2. Relevant equations

3. The attempt at a solution
I say its still c? Am I correct, there is no way I can pair this set with the natural numbers.

Define
f(1) = 0.1
f(2) = 0.02
f(3) = 0.003
f(4) = 0.0004
...
But then we must pair elements of N to 0.2, 0.3, 0.4... and 0.03, 0.04, 0.05, and 0.004, 0.005,...

This seems weak though no?

Last edited: Aug 5, 2013
2. Aug 5, 2013

### micromass

Your set is a subset of $\mathbb{Q}$, isn't it?

3. Aug 5, 2013

### verty

@Ribbon: Your argument doesn't make any sense to me. You seem to be arguing that there are a countable number of sets equinumerous with the naturals. Is this what you are doing and what are you trying to achieve by it?

4. Aug 5, 2013

### tiny-tim

hi ribbon!
hint: can you pair the set of all numbers with two non-zero digits with the natural numbers?

5. Aug 5, 2013

### vela

Staff Emeritus
Yes, your argument doesn't work. All you've shown is that you failed at your first attempt at finding a bijection between N and the set. Perhaps if you were more clever, you could find such a bijection.

6. Aug 5, 2013

### ribbon

Could I go,
f(1) = 0.11
f(2) = 0.12
f(3) = 0.13
...

and skip numbers like 0.2, 0.3? But then what happens to all the numbers in between I'm missing?

7. Aug 5, 2013

### ribbon

It seems so, but the unit interval has cardinality c and at first glance it appeared to me that we were excluding a finite set from an infinite one, and if I am not mistaken with cardinal arithmetic, c would remain the cardinality. But it seems to be otherwise now.

8. Aug 5, 2013

### LCKurtz

Hey Tiny, what happened to your fish avatar? Did he try to swim through a screen?

9. Aug 5, 2013

### tiny-tim

but you're not even trying to count any numbers with more than 2 decimal places!

try again, but with binary numbers (instead of decimals), it's easier
it wasn't a proper underwater camera!

10. Aug 5, 2013

### ribbon

Hmm, ok well could the sequence be 0.1, 0.101, 0.010101... till I hit that n (n being the finite number of non zero digits?

11. Aug 6, 2013

### LCKurtz

So, have you just walked away from the thread where we were discussing that topic?

12. Aug 6, 2013

### pasmith

Your set is infinite: if $x$ is in that set, then so is $10^{-n}x$ for all $n \in \mathbb{N}$. Adding any number of zeroes to the beginning of the decimal expansion does not change the number of non-zero digits in the expansion.

13. Aug 6, 2013

### HallsofIvy

You do not recognize this as the subset of all rational numbers whose denominators have only powers of 2 and powers of 5 as factors? If $$x= 0.a_1a_2\cdot\cdot\cdot a_n$$ has only "n" non-zero digits, then $$10^nx= a_1a_2\cdot\cdot\cdot a_n$$ and, finally, $$x= \frac{a_1a_2\cdot\cdot\cdot a_n}{10^n}$$. You may be able to do a lot of "cancelling" to reduce the denominator but 10= 2(5) so there will never be any factors other than "5" or "2" in the denominator.