Is this set uncountable


by Bachelier
Tags: uncountable
Bachelier
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#1
Feb13-13, 04:44 AM
P: 376
##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}##
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micromass
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#2
Feb13-13, 05:07 AM
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Quote Quote by Bachelier View Post
##\bigcup _{i=1}^{∞}\left\{{0,1}\right\}^i \equiv \left\{{0,1}\right\}^\mathbb{N}##
This equality is false. Furthermore, the set on the right is uncountable. The set on the left is countable.
Bachelier
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#3
Feb13-13, 06:12 AM
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Quote Quote by micromass View Post
This equality is false. Furthermore, the set on the right is uncountable. The set on the left is countable.
So how do we look at ##\left\{{0,1}\right\}^∞##?

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Feb13-13, 07:33 AM
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Is this set uncountable


Quote Quote by Bachelier View Post
So how do we look at ##\left\{{0,1}\right\}^∞##?
What do you mean with [itex]\infty[/itex]? The notation you are using now is not standard at all.
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#5
Feb13-13, 12:38 PM
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Quote Quote by Bachelier View Post
##S = \bigcup _{i=1}^{∞}\left\{{0,1}\right\}^i \equiv \left\{{0,1}\right\}^\mathbb{N}##
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
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#6
Feb13-13, 02:37 PM
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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
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#7
Feb13-13, 10:46 PM
P: 376
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}##.


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