I've been facing the next problem:
$\alpha$ is an ordinal, show that there is no cardinals collection $\{\beta_i\}$, when $i\in I$. Such as: $\sum_I \beta_i=\aleph_{\alpha+1}$. When $|I|<\aleph_{\alpha+1}$ and for all $i\in I$, $\beta_i < \aleph_{\alpha+1}$.
My first question. Before I got to my current soultion (will be represented immediately) I was trying to solve it using konig inequality, and also tried to find elegant explanation using cardinals definiton, cause $\aleph_{\alpha+1}$ is also an ordinal. I couldn't make it, and I would really like to get any solution using those principles.
In my current solution I was trying generalizing the lemma about countable union of countable set, my solution arises a doubt about my usage and understanding of the axiom of choice, I would be grateful for you comments and opinion about it.
So, my solution: $|I|<\aleph_{\alpha+1} $ and for for all $i \in I$, $\beta_i <\aleph_{\alpha+1} $, so the largest result of $\sum_I\beta_i $ is when the collection $\{\beta_i\}_I $ cardinality is $\aleph_{\alpha}$ and $I$ carindality is also $\aleph_{\alpha}$. Now I want to prove that the cardinality of $\aleph_{\alpha}$ union of sets with cardinality $\aleph_{\alpha}$ is also $\aleph_{\alpha}$
Let $\{B_i\}_I $ be a collection of set with cardinality $\aleph_{\alpha}$ (representing the $\{\beta_i\}_I $ caridnals collection). Given set $A$ with cardinality $\aleph_{\alpha}$, there is a bijection $g_I : I \to A$ and a set of bijections $f_i: B_i\to A$. Now we define new collection of sets composed of $\{i\}\times B_i $ (each and every $i$ with its respective $B_i$). We define a one-to-one function $H:\bigcup(\{i\}\times B_i) \to A\times A$, when $H((i,b))=(g_I(i),f_i(b))$. the function is one to one, cause for $(i_1,b_1)\neq (i_2,b_2)\Rightarrow$ assuming $i_1 \neq i_2\Rightarrow g_I(i_1)\neq g_I(i_2)$ cause $g_i$ is a bijection, while in case $i_1 = i_2$ and $b_1 \neq b_2\Rightarrow f_i(b_1)\neq f_i(b_2)$, also cause $f_i$ is a bijection.
So now we know that $|\bigcup (\{i\}\times B_i)|\le |A\times A|$. Using the axiom of choice I want to show that $|\bigcup B_I|\le|\bigcup (\{i\}\times B_i)| $
The axiom of choice promises existence of a function $C : \bigcup B_i\to I$ such that for each $ b\in \bigcup B_i, C(b) = i$ when $b\in B_i$.
Now using $C$, we define $ F: \bigcup B_I \to \bigcup (\{i\}\times B_i), F(b) = (C(b),b) $. This, function is one-to-one cause for $b_1\neq b_2\Rightarrow (C(b_1),b_1)\neq (C(b_2),b_2)$, no matter how $C$ function operates upon $b_i$ elements.
so from transitivity $|\bigcup B_i|\le|\bigcup (\{i\}\times B_i)|\le |A\times A|$. And a known result about infinite sets which derives from ZL, $|A\times A|= |A|=\aleph_{\alpha}$. While, for a $B_0\in \{B_i\}_I, B_0 \subseteq \bigcup B_I\Rightarrow |B_0|\le|\bigcup B_I|$, so from CSB: $\aleph_{\alpha}=|B_0|\le|\bigcup B_I|\le|A\times A|=\aleph_{\alpha}$.
What conclusively means that $\sum_I\{\beta_i\} \le \aleph_{\alpha}<\aleph_{\alpha+1}$