# Is this set uncountable

1. Feb 13, 2013

### Bachelier

$S = \bigcup _{i=1}^{∞}\left\{{0,1}\right\}^i$

methinks yes because:

$S = \bigcup _{i=1}^{∞}\left\{{0,1}\right\}^i \equiv \left\{{0,1}\right\}^\mathbb{N}$

Last edited: Feb 13, 2013
2. Feb 13, 2013

### micromass

Staff Emeritus
This equality is false. Furthermore, the set on the right is uncountable. The set on the left is countable.

3. Feb 13, 2013

### Bachelier

So how do we look at $\left\{{0,1}\right\}^∞$?

4. Feb 13, 2013

### micromass

Staff Emeritus
What do you mean with $\infty$? The notation you are using now is not standard at all.

5. Feb 13, 2013

### jbunniii

As micromass said, this is false. The reason for this is that every element of the left hand side is an n-tuple for some n, i.e., a FINITE tuple such as (0, 1, 0, 1, 1, 0). On the other hand, every element of the right hand side is an infinite sequence, such as (0, 1, 0, 1, 1, 0, ...). Therefore the left hand side and right hand side actually contain no elements in common.

Last edited: Feb 13, 2013
6. Feb 13, 2013

### jfgobin

I think that $S = \bigcup _{i=1}^{∞}\left\{{0,1}\right\}^i$ is countable all right. The mapping with $\mathbb{N}$ is quite obvious.

7. Feb 13, 2013

### Bachelier

Thanks guys, yes it is kind of clear that $\bigcup_{i=1}^{∞}\left\{{0,1}\right\}^i$ is countable...I was just looking too much into it.

I believe my confusion was coming from misunderstanding the set: $\left\{{0,1}\right\}^\mathbb{N}$ which has the cardinality of the power set of $\mathbb{N}$.