Brushing Up on Set Theory Struggling with a simple proof: Let $X = \{3^n\mid n > 0\}$ and $Y = \{3n\mid n \geq 0\}$. Prove that $X$ is a subset of $Y$.
I tried Direct Approach and Contrapositive, but I'm not strong enough with my proof skills yet. Can someone show me some steps to doing this? I don't need a direct solution, just some steps to get me in the right direction so I have a good example.
 A: You need to show that if $x$ is an element of $X$, then $x$ is an element of $Y$ as well. Namely, if $x=3^n$, for some $n\in\Bbb N^+$, then there is some $k\in\Bbb N^+$ such that $x=3k$ (where $\Bbb N^+$ denotes the non-zero elements of $\Bbb N$, of course).
Can you think what $k$ should be?
A: You just need to know a few basic things.


*

*Firstly, to prove $X \subseteq Y$, you can instead consider fixed but arbitrary $x \in X$, and prove that $x \in Y$.

*Secondly, you need to know how to rewrite $\{f(x) \mid x \in X\}$ into a more useful form. Here's how you do it: assume $f : Y \leftarrow X$ is a function. Then:
$$\{f(x) \mid x \in X\} = \{y \in Y \mid \exists x \in X, y=f(x)\}.$$
The form on the right tends to be more useful than the one on the left.
Okay, so lets prove $X \subseteq Y$. We know that:


*

*$X = \{k \in \mathbb{N} \mid \exists a \in \mathbb{N} : k=3^a \;\&\; a>0 \}$

*$Y = \{k \in \mathbb{N} \mid \exists a \in \mathbb{N} : k=3a\}$
So consider $k \in X$, and fix some $a \in \mathbb{N}$ such that $k=3^a$ and $a>0.$ To show that $k \in Y$, we need to find $b \in \mathbb{N}$ such that $k=3b$. That is, we're trying to find $b \in \mathbb{N}$ such that $3^a=3b$. So toy with the equation $3^a = 3b$ until you've got $b$ in terms of $a$, and then prove rigorously that this choice of $b$ really does do the job.
A: $$X=\left \{ {3^n|n>0} \right \} $$
$$X=\left \{ {3 \cdot 3^{n-1}|n \gt 0} \right \} $$
Let: $m=3^{n-1}$, $m>0$, for $n>0$ and is a subset of $\left \{n \geq 0 \right \}$,
$$X=\left \{ 3 \cdot m|m\in \left \{ n|n\geq 0\right \} \right \} $$
