how to show the following problem on combinatorics $Y_1,Y_2,\ldots,Y_{n+1}$ be non-empty subsets of $\{1,2,3\ldots,n\}$. Prove that there exists non-empty disjoint subsets $A_1$ and $A_2$ of $\{1,2,3\ldots,n+1\}$ such that $$\bigcup\limits_{i\in A_1} Y_{i}=\bigcup\limits_{j\in A_2} Y_{j}.$$
Please give a hint for this problem. I am trying but could not proceed.
 A: Previously this was a hint. However, as @mindlack pointed out it does not seem to be straightforward to deduce from here there are two disjoint subsets.
Attempt. Define $[n] = \{1, 2, \ldots, n\}$ and let $\mathcal{P}([n])$ be the power set of $[n]$, i.e., the set of all subsets of $[n]$.
Consider the function $f: \mathcal{P}([n+1])\setminus\{\varnothing\} \to \mathcal{P}([n])$ such that $A\mapsto \cup_{i \in A} X_i$. The pidgeonhole principle implies that there must be $2$ different $A$s with the same image.
A: Let $e_i$ for $i = 1, \ldots, n$ be the standard basis for $\Bbb{R}^n$. That is, $e_i$ is the $n$-tuple where the $i$'th component is $1$ and all the other components are $0$. To each $j \in \{1, \ldots, n+1\}$ we define a non-zero vector $v_j \in \Bbb{R}^n$ by
$$v_j = \sum_{i \in X_j}e_i.$$
We then have $n+1$ non-zero vectors in the $n$-dimensional vector space $\Bbb{R}^n$. So there is a non-trivial linear relation among them. I.e there exists numbers $c_j \in \Bbb{R}$ with not all $c_j = 0$ such that
$$\sum_{j=1}^{n+1}c_j v_j = 0.$$
Now define $A = \{j | c_j > 0\}$ and $B = \{j | c_j < 0\}$. And define two vectors $a,b \in \Bbb{R}^{n}$ as $a = \sum_{j\in A}c_j v_j$ and $b = \sum_{j \in B}-c_j v_j$. Then we have
$$0 = \sum_{j=1}^{n+1} c_j v_j = \sum_{c_j > 0}c_j v_j + \sum_{c_j <0} c_j v_j = a - b.$$
In other words
$$a = b.$$
This shows that both $A$ and $B$ are non-empty, since at least one of them is non-empty.
Let $a_i$ for $i=1, \ldots, n$ be the coordinates of $a$ in the standard basis. That is,
$$a = \sum_{i=1}^{n}a_i e_i.$$
Since $a$ is the linear combination with positive coefficients of the $v_j$s for $j \in A$, which are themselves linear combination of basis vectors with positive coefficients, we have that $a_i>0$ if and only if there exists a $j \in A$ such that the coefficient of $e_i$ in the expression for $v_j$ is positive. And by the definition of $v_j$ this is the case if and only if $i \in X_j$.
In other words we have shown that $a_i > 0 \iff i \in \cup_{j\in A}X_j$.
Similarly we write
$$b = \sum_{i=1}^{n}b_i e_i.$$
And similarly we have $b_i > 0 \iff i \in \cup_{j \in B}X_j$.
But since $a = b$ we have $a_i = b_i$ for all $i$. Hence $\cup_{j\in A}X_j = \cup_{j\in B}X_j$. And this is what we needed to show since $A$ and $B$ are non-empty and  disjoint by construction.
