Let $f:ℝ→ℝ$ be a real analytic function. Assume that $f$ has infinitely may zeros. Let $D$ be the set values of the function $f$. I want to construct a partition of $D$ as follow: $D=D₁∪D₂$ where $D₁= \{ x∈D:f(x)=0 \} $ and $D₂= \{ x∈D:f(x)≠0 \} $. One way to do this is to define an equivalence relation. My question is: How I can define this relation?
Asked
Active
Viewed 119 times
2
-
1Why can't you just take $D_1={x\in D:f(x)=0}$ and $D_2={x\in D:f(x)\neq 0}$? – walcher Aug 30 '13 at 15:59
-
1Let $x \sim y$ iff $1_{{0}}(f(x)) = 1_{{0}}(f(y))$. By $1_{{0}}$ I mean the indicator function of the set ${0}$. – copper.hat Aug 30 '13 at 16:05
1 Answers
3
A partition is the same as an equivalence relation: Given any partition $P=\{D_i:i\in I\}$ of any set $D$, you can define the equivalence relation $\sim_P$ by $$x\sim_P y \Longleftrightarrow \exists i\in I: x,y\in D_i.$$ Conversely, any equivalence relation gives you a partition of $D$ into equivalence classes.
walcher
- 3,435