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November 23rd, 2011, 12:08 PM   #1
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Prove that f: N -> {0,1} is uncountable

I need to prove that f: N -> {0,1} is uncountable.

Can anyone help me with this proof?

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November 23rd, 2011, 12:46 PM   #2
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Re: Prove that f: N -> {0,1} is uncountable

Look here for an answer to your question. ... 637AAgxmSm

b) the set of all functions from N to {0,1} is uncountable

Any such function f:N->{0,1} can be seen as a sequence {a(n)}, n=0,1,2,..., where a(n) is 0 or 1.
From such a sequence, let's build the number 0.a(0)a(1)a(2)a(3)a(4)... which is not a product, but the 0s and 1s written one after the other.
So, to a function we associate a sequence, which in turn can be associated to a real number in [0,1] written in binary form. This association is almost unique (with the same problem we have for repeated 1s, like 0.01111111.... which is the same as 0.1, but this isn't a problem).
In this way our set of functions is (almost) uniquely mapped to the reals in [0,1], which is an uncountable set.
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