A relation between some ideals

For a field $k$, let $R:=k[x_1, y_1, x_2, y_2...,x_n, y_n...]$, the polynomial ring over $k$ with infinitely many variables. Now let $I_{(a_1, a_2, a_3,...)}$,  $a_i\in \{0, 1\}$ for all $i$, be the family of ideals of $R$ generated by a subset of $\{x_1, y_1, x_2, y_2...,x_n, y_n...\}$ as follow:

If $a_1=0$, then $x_1$ is in the generating set of $I_{(a_1, a_2, a_3,...)}$, and if  $a_1=1$, then $y_1$ is in the generating set of $I_{(a_1, a_2, a_3,...)}$.

If $a_2=0$, then $x_2$ is in the generating set of $I_{(a_1, a_2, a_3,...)}$, and if  $a_2=1$, then $y_2$ is in the generating set of $I_{(a_1, a_2, a_3,...)}$.

\begin{array}{c}   . \\   . \\   . \end{array}

For example $I_{(0, 0, 1, 0, 1 ,1, 0 ...)}=\langle x_1, x_2, y_3, x_4, y_5, y_6, x_7...\rangle$.

Now let $\Gamma$ be an infinite subset of $\{0, 1\}^\mathbb{N}$, how can we show that there exists $\alpha\in \Gamma$ such that $I_\alpha\subseteq\cup_{\alpha\neq\beta\in\Gamma}I_\beta$.

• WhatsUp@ When $\Gamma$ is the whole set $\{0, 1\}^{\mathbb{N}}$ the result is true. Actually, for every $\alpha\in\{0, 1\}^{\mathbb{N}}$, we have $I_\alpha\subseteq\cup_{\alpha\neq\beta\in$\{0, 1\}^{\mathbb{N}}$}I_\beta$. – Rostami Aug 14 '16 at 12:46
• Let $\gamma_i \in \{0, 1\}^{\mathbb{N}}$ be the sequence whose $i$-th term is $1$ and all other terms are $0$. Take $\Gamma$ to be the set $\{\gamma_i: i \in \mathbb{N}\}$. Then there is no $\alpha$ in $\Gamma$ satisfying the required property, since every $\gamma_i$ contains a unique member, namely $y_i$. – WhatsUp Aug 14 '16 at 12:53

If $\Gamma$ is uncountable then there is such $\alpha$. Otherwise, (as WhatsUp commented) there is a counterexample.
Assume that $\Gamma$ is a counterexample, i.e. for every $\alpha \in \Gamma$, $I_\alpha \not\subseteq \bigcup_{\alpha \neq \beta \in \Gamma} I_{\beta}$.
For a given $\alpha \in \{0, 1\}^\mathbb{N}$, $I_\alpha$ is the vector field $$Sp_k \{ \prod_{i \in s} x_i \cdot \prod_{j \in t} y_j \mid s \subseteq \alpha^{-1}(0) \text{ finite }, t \subseteq \alpha^{-1}(1) \text{ finite}\}.$$ Since $I_\alpha \not\subseteq \bigcup_{\alpha \neq \beta \in \Gamma} I_{\beta}$, then there is finite $s, t$, such that $s\subseteq \alpha^{-1}(0), t\subseteq \alpha^{-1}(1)$, and $\prod_{i \in s} x_i \cdot \prod_{j \in t} y_j \notin I_\beta$ for every $\beta \in \Gamma$, $\beta \neq \alpha$.
Thus, we can define a one to one function sending $\alpha$ to any such pair $(s, t)$. Since there are only countably many pairs of finite subsets of $\mathbb{N}$, $\Gamma$ must be countable.