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Suppose that a set X is expressed as a union of disjoint subsets. For $j, \, k \in X$ define $j \sim k$ that $j$ and $k$ lie in the same subset. Prove equivalency.

How do I start?

My attempt:

I would write $X=\bigcup_{i}A_i$ [where $A_j \neq A_k$] [fault]

Then, I would let $B \subset X$. Therefore, if $k$ or $j$ $\in B$ then they would also have to lie within $X$. But hereafter, I am stuck...

Does anyone have any suggestions?

1 Answers1

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Here is how to start. First, write down the properties you need to show about $\sim$:

  1. $\sim$ is reflexive: That is, $x \sim x$ for all $x \in X$.
  2. $\sim$ is symmetric: That is, whenever $x_1,x_2 \in X$ satisfy $x_1 \sim x_2$, then also $x_2 \sim x_1$.
  3. $\sim$ is transitive: That is, whenever $x_1,x_2,x_3 \in X$ satisfy $x_1 \sim x_2$ and $x_2 \sim x_3$, then also $x_1 \sim x_3$.

Now prove them. The first two are really just translations of the definition of $\sim$. The hardest is transitivity, and for it you need to use the fact that the sets in the partition given are not just disjoint but pairwise disjoint.

Matthew Leingang
  • 26,112
  • Okey, here is my attempt: Reflexitivity. Let $x \in X$ and therefore $x \in A_1 \cup A_2 \cdots$. Because all $A_i$ are disjoint, that means that $(x \in A_1)\vee(x \in A_2)\cdots$. –  Feb 26 '19 at 18:08
  • Symmetry Let $x_1$ and $x_2$ $\in X$. Because all $A_i$ are disjoint, that means that $[(x_1\in A_1)\vee (x_1 \in A_2)\cdots]\wedge[(x_2 \in A_1)\vee(x_2\in A_2)\cdots]$ @Matthew here I am stuck, I don't know if I am on the right track or not –  Feb 26 '19 at 18:11
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    You're going too fast. Don't overload on the notation. For reflexivity: Let $x \in X$; you want to show that $x \sim x$. In other words, you want to show that $x$ and $x$ belong to the same part of the same subset. Not much more to say than that, is there? – Matthew Leingang Feb 26 '19 at 18:16
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    For symmetry: Let $x$ and $y$ be elements of $X$ such that $x \sim y$. That is, $x$ and $y$ both belong to the same subset $A$. Do $y$ and $x$ belong to the same subset? – Matthew Leingang Feb 26 '19 at 18:17
  • Okey, I have got it now. Thank you for your answer –  Feb 26 '19 at 18:21