# Set of equivalence classes

I'm new to equivalence classes here, so I hope someone could help me out

Mathworld (http://mathworld.wolfram.com/EquivalenceClass.html) defined equivalence classes as "a subset of the form$$(x \in X:x\sim a),$$ where a is an element of X and the notation $$"x\sim y"$$ is used to mean that there is an equivalence relation between $$x$$ and $$y$$.

I wonder if that is the case, then is the equivalence classes of the relation $$x\sim y$$ if $$x=y$$ in the real is just the set of $$x$$ itself? Then would the quotient space $$R / \sim$$ be an empty set?

• Are you asking whether the case where $\sim$ is equality results in each equivalence class being a set with $1$ element? – J.G. Apr 23 at 16:20
• Usually, we say "If ~ is an equivalence relation on a set $X$ and $a\in X,$ then the equivalence class of $a$ is the set of all $x\in X$ such that $x\sim a.$" So the equivalence class of $a$ and the relation $=$ is $\{a\}.$ The set in the Wolfram Alpha page has $a$ as a fixed element of $X,$ and $\{x\in X\mid x\sim a\}$ is asking for all values $x.$ In particular, $\{x\}$ wouldn't make sense for an equivalence class. – Thomas Andrews Apr 23 at 16:20

If the equivalence relation is the equals relation on $$\mathbb{R}$$, then the equivalence class of $$x \in \mathbb{R}$$ is $$\{x\}$$, the set containing only $$x$$.
The quotient space $$\mathbb{R}/\mathord{=}$$ is bijective to $$\mathbb{R}$$ by the mapping $$x \leftrightarrow \{x\}$$.