s ⊆ ℤxℤ

s={(z,z+1) | z ∈ ℤ}

How would I represent the following in set builders notation:



My understanding is s^+ is the union of all positive powers of relation s so s^1 union s^2 union ... so and so on


s^* is the same as the above but includes s^0 so all powers

Edit: This is what I came up with so far:

s^* = {(a,a),(a,a+1),(a,a+2),(a,a+4)|a∈ℤ}

s^* = {(a,a+1),(a,a+2),(a,a+4)|a∈ℤ}

  • $\begingroup$ A couple of things to improve your question. (1) Use MathJax to make your formulas more readable. (2) Your title has $R$ and your question has $Z$, and you should fix this inconsistency. $\endgroup$ – Lee Mosher Aug 16 '16 at 14:23

Given a relation $R \subseteq A \times A$ we have $$ R^{0} = \{ (a,a) \mid a \in A \}, $$ $$ R^{1} = R, $$ and for $n \in \mathbb N^{+} ( = \mathbb N \setminus \{0\})$ $$ R^{n+1} := \{(a,c) \mid \exists b \in A \colon (a,b) \in R^{n} \wedge (b,c) \in R \}. $$ There is an equivalent expression for $R^{n+1}$, namely $$ \begin{align*} R^{n+1} := \{ (a,c) \mid &\exists b_1, \ldots, b_n \in A \colon (a,b_1) \in R \wedge (b_1,b_2) \in R \wedge \\ & \ldots \wedge (b_{n-1},b_{n}) \in R \wedge (b_{n},c) \in R \}. \end{align*} $$ In your specific case we have, for $n \in \mathbb N^{+}$ $$ s^{n} = \{ (z,z+n) \mid z \in \mathbb Z \} $$ and hence $$ s^+ = \{ (z, z+n) \mid z \in \mathbb Z \wedge n \in \mathbb N^{+} \} $$ as well as $$ s^* = \{ (z, z+n) \mid z \in \mathbb Z \wedge n \in \mathbb N \}. $$ (Let me stress that in my notation $0 \in \mathbb N$.)


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