prove that there exist two disjoint subsets $S_1$ and $S_2$ of $S$ each of cardinality $c$ such that $x \in S_1, y \in S_2$

Question

Suppose $S$ is a subset of cardinality $c$. Given two elements $x,y \in S$, prove that there exist two disjoint subsets $S_1$ and $S_2$ of $S$ each of cardinality $c$ such that $x \in S_1, y \in S_2$.

Answer 1

I will be assuming that $c=\mathfrak{c}=|\mathbb{R}|$.

Since $S$ has cardinality equal to $\mathbb{R}$, there exists some bijection $f:S\to \mathbb{R}$.

Now, let $A = \mathbb{R}\setminus\{f(x),f(y)\}$. Then there are two intervals, $[a,b]$ and $[c,d]$ such that $f(x),f(y)$ are not in either interval, and $[a,b]\cap[c,d]=\emptyset$. Then we define

$$S_1 := f^{-1}([a,b])\cup\{x\};\qquad S_2:=f^{-1}([c,d])\cup\{y\}$$

Therefore, these two sets have cardinality $\mathfrak{c}$ by the bijection, are disjoint, and each contains one of $x$ and $y$.

Answer 2

Since $|S|=c$, there exists a bijection $f: S\to \Re$. Now divide $\Re$ into $T_1=(-\infty, 0)$ and $T_2=[0,+\infty)$. Clearly, $|T_1|=|T_2|=c$ while $T_1\cap T_2=\emptyset$. Then consider $S'_1=\{x\in S\,|\, f(x)\in T_1\}$ and $S_2'=\{x\in S\,|\, f(x)\in T_2\}$. Clearly, $|S_1'|=|S'_2|=c$ and $S'_1\cap S'_2=\emptyset$.

• If $x\in S_1'$ and $y\in S'_2$, then let $S_i=S_i'$ and we are done.
• If $x,y\in S_i'$, then let $S_i=S_i'\backslash\{x\}\cup\{z\}$ and $S_j=S_j'\backslash\{z\}\cup \{x\}$ for an arbitrary $z\in S_j$, $i\ne j$.