Let S be an infinite set and A a finite subset. Prove that $|S| = |S -A|$. I don't know if my solution is correct. This is what I have so far:
Let $S = \{ s_1, s_2,.....\}$ and $S-A = \{a_1, a_2,.....\}$ where all the elements are arranged in an fixed order.
let $f(x): S \to A , f(s_i) = a_i$
if I prove $f$ is bijective will my solution be correct?

Let $\mathfrak{a}=|A|$ and $\mathfrak{s}=|S|$, and let $\alpha$ denote an ordinal. We then have that $|S\sim A|=|\{\alpha:\mathfrak{a}<\alpha<\mathfrak{s}\}|$, and from here you should be able to construct a bijective function as you suggest.  
As a hint for constructing such a bijection, observe that $$\mathbb{F}=\langle\alpha+1:\alpha\in\omega\rangle\cup(\omega,0)$$ is a injective function mapping $\omega+1$ onto $\omega$ -- pretty much all 'finite collapse' bijections between infinities and natural numbers can be constructed in this fashion, and all transfinite cardinalities contain $\omega$ as a subset to be safely used for this purpose.
