I was reading a proof about cardinal numbers, but I do not understand one step. The proof goes as follows:
"Let $\beta$ be any ordinal, and for each ordinal $\alpha \lt \beta$, let $\kappa_{\alpha}, \lambda_{\alpha}$ be cardinals, where $\kappa_{\alpha} \lt \lambda_{\alpha}$. Then, alpha
$$\sum_{\alpha\lt\beta}\kappa_{} \lt \prod_{\alpha \lt \beta}\lambda_{\alpha}$$
The basic idea of the proof is to prove there is an injection from the cardinal sum to the cardinal product, to show that the cardinal sum is less than or equal to the cardinal product. Then, the author assumes that they are equal to each other and finds a contradiction. I do not understand the injection he creates
"Proof:
Define
$$f\colon \bigcup_{\alpha \lt \beta}(\kappa_{\alpha} \times \langle \alpha\rangle) \rightarrow \prod_{\alpha \lt \beta}\lambda_\alpha$$
by taking $f(\xi, \gamma)$ to be that element of $\prod_{\alpha \lt \beta}\lambda_{\alpha}$ that takes the value $\xi \in \lambda_\gamma$, in the $\gamma$'th place and the value $0$ elsewhere. That is,
$f(\xi,\gamma)(\nu) = \xi$ (if $\nu = \gamma$) and $0$ otherwise."
I do not understand this function, or what it does. I do not understand what "that element of $\prod_{\alpha \lt \beta}\lambda_\alpha$ that takes the value $\xi \in \lambda_{\gamma}$ in the $\gamma$'th" place means. I also don't understand the notation $f(\xi, \gamma)(\nu)$ means as well. Could someone help me?
An element of the product is given by specifyig for each index an element of the corresponding factor. In other words, it is a function $\phi$ from the index set (here $\beta$) to $\bigcup \lambda_\alpha$ such that for each $\alpha\in \beta$ we have $\phi(\alpha)\in\lambda_\alpha$. Now the map $\phi=f(\xi,\gamma)$ that $f$ yields for a given $(\xi,\gamma)$ (with $\xi\in\kappa_\gamma$!) is chosen such that $\phi(\gamma)=\xi$ (which is allowed because $\xi\in\kappa_\gamma\subseteq \lambda_\gamma$); since we cannot safely guess anything else, we also let $\phi(\nu)=0\in\lambda_\nu$ for any $\nu\ne\gamma$. In short (and replacing $\phi$ by $f(\xi,\gamma)$ again) $$f(\xi,\gamma)(\nu)=\begin{cases}\xi&\text{if $\nu=\gamma$}\\0&\text{else}\end{cases} $$