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I would like to define several ordering relations. What is the correct way in an English sentence to phrase the sentence?

I'm currently choosing between the following two awkward formulations.

Let $\sqsubset_1$ be defined such that $A \sqsubset_1 B$ means $\forall \alpha\in A\ \exists \beta \in B \ni \alpha \subset \beta$.

Let $A \sqsubset_2 B$ denote $\forall \alpha_1\in A\ \exists \beta_1 \in B \ni \alpha_1 \subsetneq \beta_1$ and $\forall \beta_2\in B\ \exists \alpha_2 \in A \ni \alpha_2 \subsetneq \beta_2$.

Isn't there some notation similar to set notation or function notation? Let $A = \{x\in \mathbb R \mid x^2 \in \mathbb R\}$, or let $f:\mathbb R \to \mathbb R^2$ by $x \mapsto (x,-x)$.

Jim Newton
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  • English sentences can have several orderings which are all acceptable. For example "Run away John!" and "John, run away!" are both valid despite changes in order. – CyclotomicField Nov 30 '17 at 13:52
  • Maybe: $\sqsubset_1 = { (\alpha, \beta) \mid \alpha \in A, \beta \in B, \text { and } \ldots }$ – Mauro ALLEGRANZA Nov 30 '17 at 13:52
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    Are you using $\ni$ for “such that”? It makes your formulas hard to read typographically. Or you using it for "contains as an element"? That makes the formulas hard to read syntactically. Either way, I'd avoid it. – Matthew Leingang Nov 30 '17 at 14:07
  • Thus, unwinding it into English: "Let the relation $\sqsubset_1$ be defined such that $A \sqsubset_1 B$ means: for every $\alpha \in A$ there is a $\beta \in B$ such that ...". – Mauro ALLEGRANZA Nov 30 '17 at 14:08
  • Hi Matthew, a discussion about whether backwards epsilon means "has element", or whether it means "such that" can be found elsewhere. That article is perhaps a more appropriate place to continue that discussion. The issue here is notation for definition of a binary relation. – Jim Newton Nov 30 '17 at 15:10
  • Relation 1 and relation 2 are different. In 2, the sub 1 and sun 2 clutter up the definition and are better not used. The set A and the set f are different. Notation used in f does not seem right. – William Elliot Nov 30 '17 at 18:44
  • Yes, relation $\sqsubset_1$ and relation $\sqsubset_2$ are different; they are intended to be different. – Jim Newton Dec 01 '17 at 08:08

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