Let $\kappa$ be a cardinal number and let $$\tau_{\kappa}=\{U \in \mathcal P(X) : X \setminus U \space \text{has cardinal at most} \space \kappa\} \cup \{\emptyset\}$$ Determine the necessary and sufficient conditions on $\kappa$ for $\tau_{\kappa}$ to be a topology on $X$.

I had problems solving the exercise. In order for $\tau_{\kappa}$ to be a topology, the following conditions must be satisfied:

i)$X, \emptyset \in \tau_{\kappa}$. For now this doesn't imply any condition on $\kappa$.

ii) Arbitrary union of elements in $\tau_{\kappa}$, so if $U_j \in \tau_{\kappa}$, then $\cup_{j \in J} U_j \in \tau_{\kappa}$. But $\cup_{j \in J} U_j \in \tau_{\kappa}$ iff $X \setminus \cup_{j \in J} U_j \in \tau_{\kappa}$ has cardinal at most $\kappa$. So $|X \setminus \cup_{j \in J} U_j \in \tau_{\kappa}|=|\bigcap_{j \in J} (X \cap {U_j}^c)|\leq \kappa$

iii) Finite intersection of elements in $\tau_{\kappa}$ must be contained in $\tau_{\kappa}$ which means if $U_i \in \tau_{\kappa}$ for a finite index set $I$, then $\cap_{i \in I} U_i \in \tau_{\kappa}$. But $\cap_{i \in I} U_i \in \tau_{\kappa}$ iff $X \setminus \cap_{i \in I} U_i \in \tau_{\kappa}$. So $|X \setminus \cap_{i \in I} U_i \in \tau_{\kappa}|=|\bigcup_{i \in I} (X \cap {U_i}^c)| \leq \kappa$

From the conditions above, I have no idea how to deduce which are the necessary and sufficient conditions on $\kappa$, I would appreciate some suggestions/help.


If you look closely, your proof of the 2nd condition shows that this is always true. Because taking union of sets with "small complement" can only decrease the size of the complement (of the union).

The third condition is where the cake hides. It suffice to show that the intersection of two sets still have a small complement. And as we know, $X\setminus(A\cap B)=(X\setminus A)\cup(X\setminus B)$.

Therefore you should ask yourself, for which $\kappa$ the above equality holds: $\kappa=\kappa+\kappa$?

  • $\begingroup$ I think that $\aleph_0$ and $c$ satisfy the equality but I don't know if there is any other cardinal with that property. I have another doubt: if I have a finite intersection of sets, say $U_1 \cap U_2 \cap ... \cap U_n$, as $U_1 \cap U_2 \cap ... \cap U_n$ is a subset of, say $U_1 \cap U_2$, then $X \setminus (U_1 \cap U_2) \subset X \setminus (U_1 \cap U_2 \cap ... \cap U_n$ so I don't see why it is sufficient to show the intersection of two sets still has a small complement. Is there an induction proof involved in that? $\endgroup$ – user100106 Aug 13 '14 at 3:52
  • 1
    $\begingroup$ (1) If a collection of sets is closed under intersection of two sets, then it is closed under intersection of any finite number of sets. Similarly if $\kappa+\kappa=\kappa$ then for every finite $n$, $\kappa+\ldots+\kappa=\kappa$ ($n$ times). So yes, induction is involved. (2) If you don't know any other cardinal which satisfies this property, then you should go to where this question came from and protest that you are insufficiently prepared for this sort of question. There is also a finite cardinal with this property, by the way, which you seem to have missed. :-) $\endgroup$ – Asaf Karagila Aug 13 '14 at 3:56
  • $\begingroup$ Oh, sure, the cardinal of the empty set :-). $\endgroup$ – user100106 Aug 13 '14 at 4:00
  • 1
    $\begingroup$ Yes, but there are more. Many infinite cardinals satisfy this property. In fact, all of them do. $\endgroup$ – Asaf Karagila Aug 13 '14 at 4:11

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.