Proving $A$ is a subset of $B$ I'm trying to understand the proof behind showing a set is a subset of another set, but I'm struggle to do so.
Can some one help using this example to show:
$A \subseteq B$?
Here
$A = \{x | x = 4n + 3,  n \in \mathbb{N}\}$,
$B = \{x | x = 2n+1,  n \in \mathbb{N}\}$.
 A: It doesn't hold that $B \subseteq A$, but $A \subseteq B$.
In order to show this, we pick a $x \in A$. Then $x$ is of the form $x=4n+3$ for some $n \in \mathbb{N}$.
$x=4n+3=2(2n+1)+1$.
$n \in \mathbb{N}$, so $2n+1$ also belongs to $\mathbb{N}$.
So $x$ is of the form $2m+1, m \in \mathbb{N}$ and thus $x \in B$.
A: Consider the following sets $A$={$1,2,3,4,5$}, $B$= {$1,2,3$}. Then we can see that every element (an element is a member of the set) of $B$ is also in $A$ Then we say that $B$ is a subset of $A$. More generally speaking, if every element of set $B$ is in a set $A$ then we say $B$ is a subset of $A$ and denote it by $B \subset A$  
With regards to your question we have to show that every element in A is in B (You have your relation the wrong way around). So pick an element in $A$ which is of the form $4n+3$ where $n$ is any natural number, then we can show that $4n+3 = 2(2n+1)+1$, where $2n+1$ is our natural number and you see that every element in A takes the form of an element in $B$ and hence it also belongs to $B$ and thus we conclude it is a subset of $B$
