# Prove union of equivalence classes is the whole set

Prove union of equivalence classes is the whole set:

Given a set $$X$$ and let $$∀x∈X$$ , $$\left[x\right]$$ be the equivalence class of $$x$$ , then we want to show that $$\bigcup_{x∈X}\left[x\right]=X$$ or equivalently $$\bigcup_{\left[x\right]∈X/\sim }\left[x\right]=X$$

proofwiki proves this theorem but it says $$∃x∈X:x∉ \left[x\right]$$ is equivalent to $$∃x∈X: x∉\bigcup\left[x\right]$$ which is not right because it is not what union of sets states.

I've tried myself like this: From the definition of equivalence relation and using the symmetric property of $$\sim$$ we know $$∀x∈X:x∈ \left[x\right]$$ if and only if $$¬(∃x∈X:x∉ \left[x\right])$$ holds, then from the definition of intersection it follows :$$¬(x∉ \bigcap_{x∈X}\left[x\right])$$ This is true if and only if: $$x∈\bigcap_{x∈X}\left[x\right]$$

But this is not what I wanted, so how can I prove that?

The proof that $$\bigcup_{x\in X}[x]=X$$ is simpler than what proofwiki does and simpler than what you're trying to do.
• By definition, $$[x] \subseteq X$$ for every $$x\in X$$, and therefore $$\bigcup_{x\in X}[x]\subseteq X$$.
• Let $$y\in X$$ be arbitrary. It is a theorem (and easy to show) that $$y\in[y]$$, and therefore $$y\in \bigcup_{x\in X}[x]$$ (since $$y\in X$$ is one of the indices in the union). Since $$y\in X$$ was arbitrary, this proves that $$X\subseteq \bigcup_{x\in X}[x]$$.
You are correct that there is an error in proofwiki. The def'n of "big union" $$\bigcup$$ is $$\forall w\,\forall y\,(y\in \bigcup w\iff \exists z\,(y\in z \in w)).$$ In particular, for any set $$x$$ we have $$\bigcup \{x\}=x.$$ For example if $$x\not \in x$$ and $$[x]=\{x\}$$ then $$x\not \in x=\bigcup \{x\}=\bigcup [x].$$
An easier way than proofwiki's (attempted) approach is that $$X=\cup_{x\in X}\{x\}\subset \cup_{x\in X}[x]\subset \cup_{x\in X}X=X$$ because for all $$x\in X$$ we have $$x\in \{y:x\sim y\in X\}\subset X$$ and $$[x]=\{y: x\sim y\}=\{y:x\sim y\in X\}\subset X.$$