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Given a partition $\mathcal{P}$ on a set S, show how to define a relation $\sim$ on $S$ such that $\mathcal{P}$ is the corresponding partition.

So far, all I have down is:

Let $X$ be a subset of partition $\mathcal{P}$. Then $$a \sim b \iff a \in X \ \wedge \ b \in X $$ defines an equivalence relation on S such that $\mathcal{P}_\sim = \mathcal{P}$.

I understand my next step is to show $\mathcal{P}_\sim \subseteq \mathcal{P}$ and $\mathcal{P} \subseteq \mathcal{P}_\sim$, but I'm unsure how.

Could someone nudge me in the right direction?

No complete answers please!

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    If $x$, $y \in S$, define $x \sim y$ if and only if $x$ and $y$ belong to the same member of the partition. Then the members of the partition are precisely the equivalence classes defined by $\sim$. – 2017-01-20

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hint

Let $[a]$ be the equivalence class of the element $a \in X$ with regards to the new relation $P_{\sim}$ you have created. This consists of all elements $b$ such that both $a,b \in X$. This shows $[a] \subseteq X$. Now show the other containment.

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    Just what I needed, thank you :) – 2017-01-20
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    Quick question, is it sufficient to say "Let $X$ be a set containing some element $a \in [a]_\sim$. Then $X$ consists of all elements $b $ such that $b \sim a $, hence $X \subseteq [a]_\sim $? – 2017-01-20
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    @AndrewTawfeek it is a bit different. Let $a \in X$, then we know $a \in [a]$ (due to reflexivity, this will be true regardless of the particular equivalence relation used). Thus $X \subseteq [a]$. – 2017-01-20
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    Ohh! I feel silly - I only had to show it was a subset and not try to account for every element! Thank you again! – 2017-01-20