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The question says: $S$ is a set of cardinality $c$. Given two distinct elements $x, y ∈ S$, prove that there exist two disjoint subsets $S_1$ and $S_2$ of $S$ and each of cardinality $c$ such that $x ∈ S_1$ and $y ∈ S_2$.

Here is what I am done so far:

Since we know that $|[0,1]|=|\mathbb{R}|=c$. Let $S=[0,1]$, then pick $x=1/2, y=7/10$. Define $S_1=[0,1/2]$ and $S_2=[3/5,1]$, where $x∈S_1$, $ S_1 \subset S$ and $y∈S_2$, $ S_2 \subset S$.

Since $|[0,1/2]|=c$ and $|[3/5,1]|=c$; therefore, we can conclude that there exist two disjoint subsets $S_1$ and $S_2$ of $S$ and each of cardinality $c$ such that $x ∈ S_1$ and $y ∈ S_2$.

Can someone give me some suggestions on my proof? Thanks!

Christopher S.
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1 Answers1

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You already have almost-almost the proof: let $\;x,y\in[0,1]\;$, with $\;x<y\;$, say $\;\epsilon:=y-x>0\;$ , and define

$$A:=[0,x+\epsilon)\;,\;\;B:=(y-\epsilon, 1]\;$$

Trivially $\;x\in A\,,\,\,y\in B\;$ , and now prove all the rest: $\;|A|=|B|=|[0,1]|\;$ , and $\;A\cap B=\emptyset\;$

DonAntonio
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