Define ~ on x by (a1,a2, ...) ~ (b1, b2, ...) if ai = bi for 1 <= i <= k. Find a bijection between X/~ and N x N x .... x N Here is the full question written out: 
My professor defined a bijection f: N x N x ... N -> X/~ as f((a1, ..., ak)) = (a1, ..., ak, 0, 0, ...). I do not understand how she got this bijection. 
Here is her full answer if that helps:

 A: I will present an alternative solution to this problem. In the first place, recall that $X / \sim$ is the set of all equivalence classes induced by $\sim$, that is, $X / \sim$ is the collection of all the sets of the form
$$[(a_1,a_2,\dots)] = \{ (b_1,b_2,\dots) \in X :\, a_i = b_i \textrm{ for all } i=1,\dots,k \}$$
where $(a_1,a_2,\dots) \in X$. So, define $f : (X/\sim) \to \mathbb{N}^k$ by the rule
$$f \big( [(a_1,a_2,\dots)] \big) = (a_1,a_2,\dots,a_k).$$
We claim that $f$ is, in fact, a function. To see this, it suffices to prove that
$$\textrm{if $[(a_1,a_2,\dots)] = [(b_1,b_2,\dots)]$ then } f \big( [(a_1,a_2,\dots)] \big) = f \big( [(b_1,b_2,\dots)] \big), \tag{1}$$
but this is easy, since $[(a_1,a_2,\dots)] = [(b_1,b_2,\dots)]$ tell us that $(a_1,a_2,\dots) \sim (b_1,b_2,\dots)$ and then $a_i=b_i$ for $i=1,\dots,k$, that is, $(a_1,a_2,\dots,a_k) = (b_1,b_2,\dots,b_k)$. Hence, $f$ is well-defined.
Now, to prove that $f$ is a bijection, we need to prove that $f$ is injective and that is surjective. To see the injectivity, show that the converse of $(1)$ is also true. The surjectivity is also very easy, given $(c_1,c_2,\dots,c_k) \in \mathbb{N}^k$, the sequence $(c_1,c_2,\dots,c_k,0,0,\dots) \in X$ satisfies that
$$f \big( [(c_1,c_2,\dots,c_k,0,0,\dots)] \big) = (c_1,c_2,\dots,c_k)$$
showing that the image of $f$ is the whole $\mathbb{N}^k$, you agree?
Let me know if you need help showing the injectivity.
