Prove union of equivalence classes is the whole set

Question

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?

Answer 1

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 $X\subseteq \bigcup_{x\in X}[x]$.

Answer 2

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 $[x]=\{y: x\sim y\}=\{y:x\sim y\in X\}\subset X.$