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!

Do you have any idea how to solve it? What have you tried so far? — Ove Ahlman, Aug 10 '15 at 12:18
Well, what does $\cdot$ mean when dealing with cardinals? What does $\le$ mean when dealing with cardinals? Please add your thoughts and attempts to the post. This helps us gauge your experience level (so we don't waste your time telling you things you already know) and identify possible areas of misunderstanding (so we can better tailor our answers to your needs). — Cameron Buie, Aug 10 '15 at 12:21
Replace the cardinals with sets: you have $A,B,C$, and an injection $f : A \to B$. You want to cook up an injection $g : A \times C \to B \times C$. How might you do it? — Ian, Aug 10 '15 at 12:34
For some basic information about writing math at this site see e.g. here, here, here and here. — Martin Sleziak, Aug 11 '15 at 13:42

score 1 · Answer 1 · edited Apr 13 '17 at 12:21

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 $k_1m_1\le k_2m_2$.

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

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