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Let $S$ be a set inductively defined as follows:

$2$ belongs to $S$, and if $x$ belongs to $S$ and $y$ belongs to $S$ then $xy$ belongs to $S$.

A/ $S = \{2n \,|\, n \text{ belongs to } \mathbb{N}\}$

B/ $S = \{2(n+1) \,|\, n \text{ belongs to } \mathbb{N}\}$

C/ $S = \{2^n \,|\, n \text{ belongs to } \mathbb{N}\}$

D/ $S = \{ 2^{n+1} \,|\, n \text{ belongs to } \mathbb{N}\}$

I am not sure about the answer. Anyone helps please?

Thanks

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    set of all $2^n$ such that $n\in \mathbb{N}$ – 2017-02-28

2 Answers 2

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If $\mathbb{N}=\{1,2,3,\cdots\}$ is your convention, then it's C.

First note that $S$ contains (positive) powers of $2$ only (you could prove this by induction).

Then note that $S$ actually contains all those powers of $2$ since you can multiply $2$ by $2$ the number of times you want.

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First, try to write out some of the elements in $S$ by starting with $2$ and then repeatedly applying the rule to generate some elements in $S$.

In order to prove the answer let $S_0 = \{2\}$ and define $S_1$ to be the elements obtained by a single application of the rule starting with elements in $S_0$. In general, define $S_n$ to be the elements obtained by $n$ applications of the of the rule starting with elements in $S_0$. From there, it should be straightforward to use strong induction to prove the result.