I am a bit lost on this question to the point that I don't know where to start. I am confused as to how I am supposed to show this without a defined ~ relation. any help would be greatly appreciated, thank you.
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Do you know the def of equivalence relation ? – Mauro ALLEGRANZA May 09 '19 at 10:57
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Do you know what $[a]$, the equivalence set of $a$, means? – Graham Kemp May 09 '19 at 10:58
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@MauroALLEGRANZA the relation has to be transitive, symmetric, and reflexive, right? I know the definitions of these 3 terms. – j zeiagler May 09 '19 at 10:58
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For the first, assume that $a,b \in S$ and $a \sim b$ and assume for contradiction that $[a] \ne [b]$. – Mauro ALLEGRANZA May 09 '19 at 10:59
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If $[a] \ne [b]$, this means that for some $z : z \in [a]$ and $z \notin [b]$. – Mauro ALLEGRANZA May 09 '19 at 11:00
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@GrahamKemp [a] is the set of all _____ that satisfies ~. I am a bit lost on what the _____ should be. – j zeiagler May 09 '19 at 11:01
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But from $z \in [a]$ we have that ... ? – Mauro ALLEGRANZA May 09 '19 at 11:01
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@MauroALLEGRANZA z ∈ {b ∈ S | a~b}? – j zeiagler May 09 '19 at 11:06
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$[a]$ is the set of all elements of $S$ that are $\sim$ related to $a$.$$[a]={s\in S: a\sim s}$$ So $z\in [a]\to\underline\quad$? – Graham Kemp May 09 '19 at 11:06
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Yes, but that means : $z \sim a$. And we have : $a \sim b$; thus, by transitivity... – Mauro ALLEGRANZA May 09 '19 at 11:07
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@GrahamKemp z∈[a]→ {s∈S:a∼s}, is this correct? – j zeiagler May 09 '19 at 11:08
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@MauroALLEGRANZA z~b, so then z ∈ [b], ok I see – j zeiagler May 09 '19 at 11:10
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Similar for the second : assume that $[a] \cap [b] \ne \emptyset$ that means : for some $z : z \in [a]$ and $z \in [b]$. But $a \nsim b$. – Mauro ALLEGRANZA May 09 '19 at 12:43
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If a ~ b, then for all x,
x in [a] iff x ~ a iff x ~ b iff x in [b].
Thus [a] = [b].
If not a ~ b and x in [a] $\cap$ [b], then
x ~ a and x ~ b, so a ~ b, a contradiction.
Thus [a] $\cap$ [b] is empty.
William Elliot
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