Does $a \le b$ imply $a+c\le b+c$ for cardinal numbers?

Question

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 1

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

Answer 2

You realize wrong. The definition of $a\leq b$ is that there is an injection from a set $A$ of cardinality $a$ into a set $B$ of cardinality $b$. We also define $a+b$ to be the cardinality of $A\cup B$, where $A$ and $B$ have cardinalities $a$ and $b$ respectively, and $A\cap B=\varnothing$.

To show that the statement is correct, you have to start by taking three disjoint sets, $A,B,C$ and assume that there is an injection $f\colon A\to B$. Then you have to show that there exists an injection $g\colon A\cup C\to B\cup C$. If you cannot prove that, then you should try and construct a counterexample from the point of impasse.