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 $\aleph_0\leq \aleph(A_i)<\aleph(X)$...

  • $\begingroup$ It's a good thing that you picked the right tags... rolls eyes $\endgroup$ – Asaf Karagila Sep 17 '17 at 15:47
  • $\begingroup$ @AsafKaragila ? $\endgroup$ – let's have a breakdown Sep 17 '17 at 15:50
  • $\begingroup$ There are tag excerpts which give you some hint as to what is the topic of questions which are expected to have this tag. Have you even bothered to take the few seconds it takes to read the information about the large-cardinals tag? Or understand the difference between the two set theory tags? $\endgroup$ – Asaf Karagila Sep 17 '17 at 15:51
  • $\begingroup$ Unfortunate choice of notation. You seem to get a more interesting exercise under the usual meaning of the $\aleph (\cdot) $ symbol. $\endgroup$ – Andrés E. Caicedo Sep 17 '17 at 17:17

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.