# Thread: Partition of a set

1. ## Partition of a set

The question is to determine whether $\mathcal{A}$ is a partition of the set $A$.

$A = \mathbb{R}$ and $\mathcal{A} = \{S_y : y \epsilon \mathbb{R}\}$, where $S_y = \{x \epsilon \mathbb{R} : |x| = |y|\}$.

I understand what it means to be a partition, but I'm not sure what $\mathcal{A}$, or particularly, $S_y$ means.

Does this mean $\mathcal{A} = \{ \{1, 1\}, \{1.1, 1.1\}, \{1.2, 1.2\}, ..., \{-1, 1\}, \{-1.1, 1.1\}, \{-1.2, 1.2\}, ... \{1, -1\}, \{1.1, -1.1\}, ... \{-1, -1\}, ... \{-5.4, 5.4\} \}$? Where I just find any real number pair so that their absolute values are equal?

2. ## Re: Partition of a set

$S_x = \{ x, -x \} \text{ for } x \in \mathbb{R}, x \ne 0. \ S_0 = \{ 0 \}. \ \mathcal{A} = \{ S_y : y \in \mathbb{R} \}.$

$\text{Since } S_x = S_{-x} \text{ for } x \ne 0, \text{ have that }$

$\mathcal{A} = \{ S_y : y \in \mathbb{R} \} = \{ S_y : y \in \mathbb{R}, y \ge 0 \}.$

Are those sets in $\mathcal{A}$ disjoint? Non-empty? Does their union equal the whole set (in this case, $\mathbb{R}$)? If so, then $\mathcal{A}$ is a partition of $\mathbb{R}.$

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The only potentially confusing thing here is that the original definition of $\mathcal{A}$ had redundancies - two different names for the same set - and so that might seem to make it not a partition. However, a set with redundant (even if differently named) elements is the same as the set with unique elements.

i.e. If a = -2, b = -2, then U = {a, b} = { -2, -2 } = { -2 } = { a }.

Likewise: If A = {5, -2}, B = {5, -2}, then V = {A, B} = { {5,-2}, {5,-2} } = { {5,-2} } = { A }.