Prove $3\mathbb{Z}+1=\{6\mathbb{Z}+1\}\cup\{6\mathbb{Z}+4\}$ I was wondering if someone could confirm I have proven the following equality correctly.
Also, for part II should I have let $n\in \mathbb{Z}$ as opposed to $n\in 6\mathbb{Z}+1$ or was I correct?
Thank you for your help.

Recall for integers $a,b,a\mathbb{Z}+b=\{a\mathbb{Z}+b$| $z\in\mathbb{Z}\}$
Prove $3\mathbb{Z}+1=\{6\mathbb{Z}+1\}\cup\{6\mathbb{Z}+4\}$

NTS:
I) $3\mathbb{Z}+1 \subseteq \{6\mathbb{Z}+1\}\cup\{6\mathbb{Z}+4\}$
II)$\{6\mathbb{Z}+1\}\cup\{6\mathbb{Z}+4\}\subseteq 3\mathbb{Z}+1$
I) Suppose $n\in \mathbb{Z}$ then $\exists k\in \mathbb{Z}$ s.t. $n=2k$ or $n=2k+1$
For $n=2k$, $n= 3n+1$ =$3(2k)+1$=$6k+1$$\in 6\mathbb{Z}+1$, Thus $3\mathbb{Z}+1 \subseteq \{6\mathbb{Z}+1\}$
Now for For $n=2k+1$, $n=3n+1$ =$3(2k+1)+1$=$6k+3+1=6k+4$$\in 6\mathbb{Z}+4$, Thus $3\mathbb{Z}+1 \subseteq \{6\mathbb{Z}+1\}$
We have shown $3\mathbb{Z}+1 \subseteq \{6\mathbb{Z}+1\}\cup\{6\mathbb{Z}+4\}$
II) Now suppose $n\in 6\mathbb{Z}+1$ then $\exists l\in \mathbb{Z}$ s.t. $n=6l+1$
Then $n=6l+1=3(2l)+1\in 3\mathbb{Z}+1$ Thus $\{6\mathbb{Z}+1\}\subseteq 3\mathbb{Z}+1$
Suppose $n\in 6\mathbb{Z}+4$ then $\exists l\in \mathbb{Z}$ s.t. $n=6l+4$
Then $n=6l+4=3(2l+1)+1\in 3\mathbb{Z}+4$ Thus $\{6\mathbb{Z}+4\}\subseteq 3\mathbb{Z}+1$
We have shown $\{6\mathbb{Z}+1\}\cup\{6\mathbb{Z}+4\}\subseteq 3\mathbb{Z}+1$
Therefore by I and II we have proven $3\mathbb{Z}+1=\{6\mathbb{Z}+1\}\cup\{6\mathbb{Z}+4\}$.
 A: You may want to denote the second $n$ differently, since for e.g. $n=3n+1$ may create some confusion. Also a typo in I), you want to say that: “Thus $3\Bbb Z+1\subseteq(6\Bbb Z+1)\cup(6\Bbb Z+4)$” and not “Thus $3\Bbb Z+1\subseteq(6\Bbb Z+1)$.” As for the rest, everything seems fine. You could also opt for the notation $(6\Bbb Z+4)$ as opposed to $\{6\Bbb Z+4\}$, to avoid confusion. 
A: For every integer number $n$ of the form $3k+1$ we have that $k$ is either even or odd. In the first case, $k=2K$ and $n=6K+1$; in the second case, $k=2K+1$ and $n=6K+4$, so
$$ (3\mathbb{Z}+1) = (6\mathbb{Z}+1)\cup(6\mathbb{Z}+4).$$
A: That looks reasonable, but your working is cluttered.
Recall that for scalars $a,d$ and scalar sets $B,C$, then: $$\begin{align}a(B\cup C)+d =&  \{ax+d: x\in B\cup C\}
\\ =& \{ax+d: x\in B\}\cup\{ax+d: x\in C\}
\\ =& (aB+d)\cup (aC+d)\end{align}$$
Prove that in your own style, then use this general result and the fact that the set of integers is composed of odd and even integers: $$\Bbb Z= (2\Bbb Z)\cup (2\Bbb Z+1)$$
Hence by substitution you will obtain what was to be proven.

 $$\begin{align}3\Bbb Z+1  =&~ 3\big(2\Bbb Z \cup (2\Bbb Z+1)\big)+1\\[1ex] =&~ (6\Bbb Z+1)\cup (6\Bbb Z+3+1)\\[1ex] =&~ (6\Bbb Z+1)\cup(6\Bbb Z+4)\end{align}$$

$\blacksquare$
A: Hint $\ \ \Bbb Z = 2\Bbb Z \cup (2\Bbb Z\!+\!1)$
$\ \ \Rightarrow\, 3 \Bbb Z = 6\Bbb Z \cup (6\Bbb Z\!+\! 3)$
$\ \ \Rightarrow\, 3 \Bbb Z\! +\! 1 =  \ \ldots$
