elementary set theory - Does $a le b$ imply $a+cle b+c$ for cardinal numbers?

Let $a, b, c$ be cardinalities.

Prove or disprove:

If $a \le b$ then $a+c\le b+c$

I realize that $a \le b$ means that there's a bijection between A and B. But I don't really know what to do with the addition in the inequality.

Can I simply separate this into cases where one or two cardinalities are infinite while the others aren't and then just solve from there like numbers or known cardinalities like $\aleph_0$ ?

Thanks.

Answer

I will prove something stronger -

For cardinals $a,b,c$ and $d$, if $a\leq c$ and $b\leq d$, then $a+b\leq c+d$.

Proof: Let $A,B,C$ and $D$ be sets with the cardinals $a,b,c$ and $d$ respectively, and $A\cap B=C\cap D=\emptyset$.
If $a\leq c$ and $b\leq d$, then there exists injective functions $f:A\to C$ and $g:B\to D$. By defining new function $h:A\cup B\to C\cup D$ with

$$h(x)=\left\{\begin{matrix} f(x) & x\in A\\ g(x) & x\in B \end{matrix}\right.$$

That $h(x)$ is well defined(why?) and injective one(why?). From that, we get

$$|A\cup B|\leq|C\cup D|$$

i.e. $a+b\leq c+d$.