Discrete Math - Question on Sets

Question

Been trying for a while to figure out how to solve this question. Let A = {n ∈ N | n ≥ 1 and n = 4j − 3 for some j ∈ N} and B = {n ∈ N | n ≥ 0 and n = 2k + 1 for some k ∈ N}. Prove that A ⊆ B. Here, N = {0, 1, 2, 3, . . . }.

Answer 1

Take an arbitrary element $x \in A$. Then, by the definition of set $A$, we can

$$x = 4j - 3$$

for some $j \in \mathbb{N}$. Moreover, since $4j - 3 \geq 1$ and $x \equiv 1 \pmod{2}$, we know that $x$ is an odd positive integer. This is precisely the requirement an element needs to satisfy in order to be in set $B$. Thus, $x$ is also in set $B$.

Since we have shown that an arbitrary element of $A$ also lies in $B$, we conclude that $A \subseteq B$. So we're done.