Is this set uncountable

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}##

micromass

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

Bachelier

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

micromass

What do you mean with [itex]\infty[/itex]? The notation you are using now is not standard at all.

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.

jfgobin

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

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}##.