Will $\kappa_1, \kappa_2, m$ cardinals. Given $\kappa_1 \leq \kappa_2$. prove: $\kappa_1 \cdot m \leq \kappa_2 \cdot m$

Question

Will $\kappa_1, \kappa_2, m$ cardinals. Given $\kappa_1 \leq \kappa_2$. prove: $\kappa_1 \cdot m \leq \kappa_2 \cdot m$.

Hi, I would be happy if someone could help me with this. What I did until now:I replaced the cardinals with sets: $|K_1|=\kappa_1$, $|K_2|=\kappa_2$, $|M|=m$. From what is given stems there is a injection $f:K_1→K_2$. Now I need to prove there is a injection $g:K_1⋅M \to K_2⋅M$, from multiplication of cardinals→ $g:K_1\times M → K_2\times M$. Now how do I show that? I just started to learn this subject so would be happy to get a complete answer. Thanks!

Answer 1

If you have $f \colon A \to B$, then a very natural way to define $g\colon A\times C \to B\times C$ is $$g(a,c)=(f(a),c).$$

This is Cartesian product of functions $g= f\times id_C$.

Can you show that if $f$ is injective then $g$ is injective? (Or, more generally, if $f_1 \colon X_1\to Y_1$ and $f_2 \colon X_2 \to Y_2$ are injective, then $f_1\times f_2 \colon X_1\times X_2 \to Y_1\times Y_2$ is injective.)

Let $g:=f_1\times f_2$. If $g(x_1,x_2)=g(x'_1,x'_2)$ then we have $f_1(x_1)=f_1(x'_1)$ and $f_2(x_2)=f_2(x'_2)$. (This follows from the definition of cartesian product of functions.) Now it only remains to use the fact that the functions $f_{1,2}$ are injective.

---

It is relatively easy to show that cardinal multiplication is commutative. Using this fact and the inequality from the question we can get:

If $\kappa_1\le\kappa_2$ and $\lambda_1\le\lambda_2$ then $\kappa_1\cdot\lambda_1 \le \kappa_2\cdot\lambda_2$.

This fact is also shown here: Proof of cardinality inequality: $m_1\le m_2$, $k_1\le k_2$ implies $k_1m_1\le k_2m_2$.