**Question:**

*True/False?*

Let \(S\) be the set of all sequences \(\{a_1,a_2,…,a_n,…\} \) where each entry \(a_i \) is either \(0\) or \(1\). Then \(S\) is countable.

*Hint:*

What if instead of \(0\) and \(1\) the values were allowed to be any digit from \(0\) to 9? What would that correspond to?

**Discussion:**

Given a sequence \(\{a_1,a_2,…,a_n,…\} \) , we can associate it to the binary number \(0.a_1a_2…a_n… \). This association (or function if you like) is one-one, and onto the set of all real numbers having binary expansion in the form of \(0.something\), which is same as the set \((0,1)\), which is uncountable.

**Remark:**

One could use Cantor’s diagonalization argument as well to argue in this problem. If possible let \(x_1,x_2,…\) is an enumeration of the given set (of sequences) then consider the sequence \(\{b_1,b_2,…\} \) in \(S\) defined by \(b_i \ne x_i \). We get a contradiction because this is a sequence which is not in the enumeration but is a member of \(S\).

*Related*

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