$\aleph(X)\geq \aleph_0$, $A_1,A_2\subset X$ and $\aleph(A_i)<\aleph(X) \implies \aleph(A_1\cup A_2)<\aleph(X)$?

Question

I'm struggling with the following (is it true?):

Let $X$ be a set and denote $\aleph(X)$ the cardinality of $X$. Suppose that $\aleph(X)\geq \aleph_0$, the cardinality $\aleph_0:=\aleph(\Bbb N)$, with $\Bbb N$ the set of natural numbers.

Consider two subsets $A_1,A_2\subset X$ with $\aleph(A_i)<\aleph(X)$, $i=1,2$. Show that $\aleph(\bigcup_1^2 A_i)<\aleph(X)$.

From this, it would follow by induction that $\aleph(\bigcup_1^n A_i)<\aleph(X)$, for any sets $A_1,\dots, A_n\subset X$ with $\aleph(A_i)<\aleph(X)$, $i=1,\dots,n$.

With this result, I would be able to solve Dugundji's Exercise 1-b (Chapter III - Topological Spaces), which asserts:

1.b) If $\aleph(X)\geq \aleph_0$, then $\scr{A}_1$$=\{\emptyset\}\cup \{A\,|\, \aleph(X-A)<\aleph(X)\}$ is a topology on $X$.

The fact that $\emptyset$ and $X$ are in $\scr A_1$ is easy. Arbitrary unions of elements of $\scr A_1$ is in $\scr A_1$, since the complement in $X$ of such an union is an intersection of complements, which have, each of of them, cardinality strictly smaller than $\aleph(X)$ (intersections decrease cardinality). But to show that finite intersections of elements of $\scr A_1$ are in $\scr A_1$, I need the fact above...

It is easy to see that it is true if $\aleph(X)=\aleph_0$, since in this case $A_1$ and $A_2$ must be finite. The problem arises if $\aleph_0\leq \aleph(A_i)<\aleph(X)$...

Answer 1

If the cardinality of a finite union is infinite, then it is the cardinality of one of those sets. Sum of two infinite cardinals is simply the largest one.

Answer 2

It is well known that for infinite cardinals $\kappa$ and $\lambda$ :

$$\kappa + \lambda= \kappa \times \lambda = \max(\kappa,\lambda)$$

And by induction this holds for all finite sums and products as well.