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I've got something like this snippet for something I'm working on:

$$ S = \lbrace (a, (b_0, \dots, b_n), c) : a \in A, \lbrace b_0, \dots, b_n \rbrace \subseteq B, c \in C, n \in {\mathbb N} \rbrace $$

My goal here is to specify in the set constraint that the tuple $(b_0, \dots, b_n)$'s elements should all be members of set B, even though the tuple itself could contain duplicates. Is this even valid, and is there a better way to express this that I'm missing?

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It is not including when the b's are not unique.
S = $\cup${ A × $B^n$× C : n in N } includes nonunique b's.

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  • $\begingroup$ Hm? Yes it is. ${1,2,1}\subseteq{1,2,3}$. And the $n\in N$ should be below the $\bigcup$ at the beginning: otherwise, if you write $\cup X$, you mean $\bigcup_{S\in X} S$, which is a very different thing. $\endgroup$ – Deusovi May 24 '18 at 1:54
  • $\begingroup$ What does "1,2,1 subset 1,2,3" mean quibbling @Deusovi? $\endgroup$ – William Elliot May 24 '18 at 9:53
  • $\begingroup$ ...It means I forgot to escape my braces. My bad. Should've been $\{1,2,1\}\subseteq\{1,2,3\}$. The point was that his original notation seems to me to allow some of the $b_i$s to be equal, since sets can be written with repeated elements - they're just the same as if those elements weren't repeated. $\endgroup$ – Deusovi May 24 '18 at 19:10
  • $\begingroup$ It did not @Deusovi. As {1,2,1} has only two elements, it is rendered (1,2) and not (1,2,1). $\endgroup$ – William Elliot May 25 '18 at 1:53
  • $\begingroup$ That is one way to write the set {1,2}, but it is not the only way. {1,2,1} is just as valid of an expression, no? It's not used as often, sure, but it is equally valid. $\endgroup$ – Deusovi May 25 '18 at 2:53

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