I could use some help on this problem:
Suppose $k_1$, $k_2$, $m$ are cardinals. Given that $ k_2 \geq k_1$ prove that $ k_2^m\geq k_1^m$ .
I know that I need to find a one to one function $f$ from $k_1^m$ to $k_2^m $ in order to prove that.
I also know that if $|A|=k_1$, $|B|=k_2$, $|C|=m$ and $k_2 \geq k_1$ there is a one to one function $g$ from $A$ to $B$.
Is someone can guide me or give me a direction. Thanks.
Let the sets $A,B,C$ have cardinalities $k_1, k_2, m$ respectively, and suppose there is an injection $f\colon A\stackrel{1-1}\longrightarrow B$. Then the following is the desired injection: $$ F\colon g\mapsto f\circ g\colon A^C \to B^C $$ Suppose $g,h\in A^C$ and $F(g)=F(h)$ — that is, $f\circ g = f\circ h$. Then for all $c\in C$, $f(g(c)) = f(h(c))$. Because $f$ is injective, it follows that for all $c\in C$, $g(c)=h(c)$, which is to say, $g=h$. So $F$ is an injection.