I am struggling wiht this exercise regarding cardinals and cofinality.
In the framework of ZFC, let $k$ be a regular cardinal and let $\alpha < k$. Let $(k_\beta)_{\beta<\alpha}$ be a sequence of cardinals, each strictly smaller than $k$. Prove that: $\sum_{\beta<\alpha} k_{\beta} < k$.
Now, this is what I did:
$\sum_{\beta<\alpha} k_{\beta} \leq \sum_{\beta<k} k_\beta = |\bigcup_{\beta<k}\{\beta\}\times k_{\beta} | \leq k \cdot k = k $
So I proved that $\sum_{\beta<\alpha} k_{\beta} \leq k $. Now I assumed by contradiction that equality holds, by doing so I get that there exists a bijection $h:\bigcup_{\beta<\alpha}\{\beta\}\times k_{\beta} \to k$
Now I wanted to use this fact to contradict the regularity of $k$, maybe exploiting $h$ to get a cofinal map from $\alpha$ to $k$, thus getting $cf(k) = k< \alpha$ and so a contradiction, but I don't see anything smart to do in order to get this result. Any help is very much appreciated :)
If $\sum_{β<α}κ_β=κ$ then the function $f:α→κ$ defined by $f(β)=\sum_{γ<β}κ_γ$, as $\sup\{f(β)\mid β<α\}=\sum_{β<α}κ_β=κ$ it means that there exists some minimal $η≤α$, such that $\sup\{f(β)\mid β<η\}=κ$ which implies that $cof(κ)≤η$ with the cofinal function $f\restriction\eta$, which is contradiction to the fact $κ$ is regular