cardinal product $mn=m$ if $n\leq m$?

Question

If $m,n \geq \aleph_0$ and $m\geq n$, then which is right? $mn=m$ or $mn\leq m$? Why if the first not correct? For examply, if the weight of a topological space $\omega(X)=n$, then the product $\omega(X^m)\leq m$ or $\omega(X)=m$?

Answer 1

The product of two infinite cardinals is the maximum, so if $m \ge n \ge \aleph_0$, $mn = m$.

If $w(X) = n$ (weight uses w = $w$ not omega= $\omega$ as its symbol) then $w(X^m)$ is the weight of an infinite power, so this weight is indeed $mn = m$ (the standard base built from basic elements of a base of size $n$ of $X$ in a power of $m$ spaces has size $nm$ and when $m \ge n$ we have that $w(X^m) = m$). In particular $w([0,1]^m) = m$ for all infinite cardinals $m$, as $w([0,1]) = \aleph_0$.

Answer 2

Both are right.

If $x=y$, then $x\leq y$, but also since $\leq$ is antisymmetric, and clearly $m\leq mn$,¹ we get that $mn\leq m$ implies $mn=m$.

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_{(1) In this case, at least, where $n$ is non-zero.}