$ i, j, k$ are pairwisely distinct, right?

Consider a non empty set $ A\in F$ where $ F$ is the family of non empty subsets of $ \{1,2,\dots,n\}$ and $ |F|=m\geq 2^{n-1}+1$. Let $ F'=\{B_j|B_j=A\cup A_j \quad \forall A_j\neq A \quad and \quad A_j\in F \quad and \quad j=1,2,\dots,m\}$ Since each $ B_j$ is distinct, $ |F'|\geq2^{n-1}$ $ \Rightarrow |F\cap F'|\geq1$ and we are done.

Why $ B_j$ are distinct? Take: $ A = \{1, 2\}$, $ A_1 = \{3, 4, 5\}$, $ A_2 = \{1, 2, 3, 4, 5\}$. Then $ A \cup A_1 = A \cup A_2$.

Is Tomcio's predict correct? Otherwise, it just an easy induction.

It seems like a straightforward inductive solution should work even if you require pairwise disjointness, but it turns out there's a booby trap to watch out for. To start with, here's a solution with a hole in it: The case $ n = 3$ can be handled by a quick case analysis (alternatively, we can start our induction at $ n=2$, where there is only one possible collection of $ 3$ sets). Now let us assume that the result is true for subsets of an $ n - 1$ element set, and we are given a collection of $ 2^{n - 1} + 1$ subsets of an $ n$ element set. By pigeonhole either there are $ 2^{n - 2} + 1$ sets not containing $ n$ or $ 2^{n - 2} + 1$ sets containing $ n$. In the former case, we know by the inductive hypothesis that three of the sets not containing $ n$ satisfy $ A_i \cup A_j = A_k$, so we are done. In the latter case, we just apply the hypothesis to the (at least $ 2^{n - 1} + 1$) sets formed by taking each set containing $ n$ and dropping $ n$. Why doesn't this work?

How can we fix this?

It's ok but this solution is very easy for a 3rd question of Romanian TST.

$ F$ is the family of non empty subsets of $ [n]=\{1,\dots,n\}$ and $ |F|\geq 2^{n-1}+1$.We have $ A,B\in F$ so that $ A\cup B=[n]$ and $ A\cap B=\emptyset$. If $ [n]$ is contained in $ F$ we are done, else ( we know $ \emptyset \notin F$) we have another pair $ C,D\in F$ so that $ C\cup D=[n]$ and $ C\cap D=\emptyset$. Consider the four disjoint sets $ A\cap C, A \cap D, B\cap C$ and $ B\cap D$. Among many forbidden pairs we cannot have for example $ A\cap C, A \cap D$ both present in $ F$. Hence we have another pair $ E,G\in F$ so that $ E\cup G=[n]$ and $ E\cap G=\emptyset$. By recursively employing the same process we empty out all such pairs in $ F$ which are of course finite in number and hence the contradiction.

srikanth wrote: By recursively employing the same process we empty out all such pairs in $ F$ which are of course finite in number and hence the contradiction. I think you need to go into more detail here. For example, If $ A = \{1\}, C = \{2\}$, then $ A \cap C = \emptyset$, so $ (A \cap C, A \cap D)$ doesn't lead to a new forbidden pair. Of course, in this particular example we're not stuck yet (we can look at $ (B \cap C, B \cap D)$ instead). However, it may be possible that what eventually will happen is that all the forbidden pairs you generate in your new step will overlap with previously generated pairs in such a way, and your process will halt before you actually forbid all the pairs from $ F$.