Question

In: Advanced Math

We denote by N^∞= {(a_1, a_2, a_3, . . .) : a_j ∈ N for all...

We denote by N^∞= {(a_1, a_2, a_3, . . .) : a_j ∈ N for all j ∈ Z^+}. Prove that N^∞is uncountable.

Please answer questions in clear hand-writing and show me the full process, thank you (Sometimes I get the answer which was difficult to read).

Expert Solution

To prove that is uncountable we will provide an one-to-one function from to .

We know that every real number in the interval have unique binary representation generated by the numbers 0 and 1 that is for each , can be written as of the form where each is either 0 and 1 .

is defined by ,

Now we will prove that is one-to-one .

Let ,

for all

So the map is one-to-one .

As the set is uncountable and there exist a one-to-one map from to . Hence the set is uncountable .

Colby Messinger answered 2 years ago

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Question

We denote by N^∞= {(a_1, a_2, a_3, . . .) : a_j ∈ N for all...

Solutions

Expert Solution

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