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?

  • $\begingroup$ Are you asking whether the case where $\sim$ is equality results in each equivalence class being a set with $1$ element? $\endgroup$ – J.G. Apr 23 at 16:20
  • $\begingroup$ 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. $\endgroup$ – 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 $x \leftrightarrow \{x\}$.


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