Prove if $B$ is infinite and $F$ is finite, then $B-F$ is infinite.
I understand conceptually what is going on. $B$ is infinite so no matter how big of a finite set $F$ you take from it, its still going to be infinite.
I am struggling constructing a proof. I started with contradiction assuming $B-F$ is finite, but cannot seem to draw the conclusion I need.
Now I am doing a proof be induction increasing the number of elements in $F$ but I am unsure if that is a valid technique.
On
I'll use the fact that a set $X$ is infinite iff there is an injection from the natural numbers $\mathbb{N}$ into $X$.
Since $B$ is infinite, we have such an injection, say $\sigma : \mathbb{N} \hookrightarrow B$. The set $S := \sigma^{-1}(F)$ is a finite subset of $\mathbb{N}$, let $\mathbb{m} := \max(S)$.
Set $\tau : n \mapsto n + \mathbb{m} + 1$.
$\tau : \mathbb{N} \mapsto \mathbb{N} \setminus S$ is an injection (this verifies our intuition that $\mathbb{N} \setminus S$ is finite).
The map $\sigma \circ \tau$ is an injection from $\mathbb{N}$ into $B \setminus F$. Hence, $B \setminus F$ is infinite.
If $B-F$ is finite, then $B$ is the union of two finite sets $B-F$ and $F$.