Is it correct to say "$a_t=\alpha:\alpha\in A_{\sigma}$" with the following meaning: $a_t$ is equal to $\alpha$ and $\alpha$ belongs to set $A_\sigma$.

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    $\begingroup$ Yes that is correct. $\endgroup$ – eeeeeeeeee Jan 20 '13 at 0:08
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    $\begingroup$ Personally I use $∋$ for such that. [related] $\endgroup$ – Alexander Gruber Jan 20 '13 at 1:27
  • $\begingroup$ It’s surprisingly popular, but I’ve seen it only here, and I find it rather unreadable. I routinely use the colon in set-builder notation in preference to the vertical bar, but it is not a general symbol for such that and ought not to be so used. I would discourage it even in $\exists\alpha\in A_\sigma:a_t=\alpha$, in favor of $\exists\alpha\in A_\sigma(a_t=\alpha)$, my preferred form, or, as a distant second choice, $(\exists\alpha\in A_\sigma) a_t=\alpha$. (The ranking of those two is a matter of taste.) $\endgroup$ – Brian M. Scott Jan 20 '13 at 8:38
  • $\begingroup$ Go here :P $\endgroup$ – Mr Pie Sep 8 '18 at 7:48

That's a reasonable way to interpret the phrase. But it would be better to write it with words. For example: "There is an $\alpha\in A_\sigma$ such that $a_t=\alpha$."

However, you could say the same thing but have a different emphasis if you wrote "$a_t=\alpha$, where $\alpha$ is an element in $A_\sigma$". This would imply that $a_t$ is the object of focus, and $\alpha$ is almost incidental. But if you said it the first way, it places a great bit of significance on $\alpha$.

Edit: And the others considered something I didn't: if you wrote "(Notice/Then) $a_t=\alpha$ and $\alpha\in A_\sigma$", that would mean something like: these are two facts which you could figure out from the previous work, but you feel a need to state both before combining them somehow.

If you read through papers with an eye to this kind of detail, you will find that (good) authors state similar concepts in subtly different ways to best serve the need of the audience. Eventually you will see words as providing their own kind of hint to how you should mentally reconstruct arguments, and will probably use some of your own verbal tricks quite frequently.


Is it correct to say "$a_t=\alpha:\alpha\in A_{\sigma}$" with the following meaning: $a_t$ is equal to $\alpha$ and $\alpha$ belongs to set $A_\sigma$?

Wikipedia's Table of mathematical symbols will verify, that the symbol "$\,:\,$" can mean "such that" when used in proofs, and in particular, like the symbol "$\,|\,$", when used in Set Builder Notation.

So what you wrote it can be taken to mean: "$a_t = \alpha\,$ such that $\,\alpha \in A_{\sigma};\,$ or alternatively: $$\exists \alpha \in A_\sigma: a_t = \alpha.$$

You can say precisely "$a_t$ is equal to $\alpha$ and $\alpha$ belongs to set $A_\sigma$" by using the following: $$a_t=\alpha\, \land\,\alpha\in A_{\sigma}$$

But: There is usually no need translate everything to symbolic notation, nor is doing always desirable to do so in mathematical exposition. Clarity and precision are desirable, but this can be accomplished best, at times, with the use "plain old words." It would be perfectly appropriate to write: $$\,a_t = \alpha,\,\text{ where}\;\alpha\, \text{ belongs to}\; A_\sigma.$$ Doing so puts the emphasis on $a_t\;$ (as Eric Stucky expresses nicely in his post: what you say and how you say it can subtly alter the emphasis of the statement).

Also note: There are generally a number of ways to accurately state propositions.

  • $\begingroup$ I suppose I should take it that imitation is the sincerest form of flattery ... $\endgroup$ – Peter Smith Jan 20 '13 at 0:22
  • $\begingroup$ I think you mean "There is no need to translate..."? $\endgroup$ – preferred_anon Jan 20 '13 at 0:25

The existential quantifier $\exists$ and universal quantifier $\forall$ can be used together with "such that" ($:$) without ambiguity if you follow this commonly accepted syntax: $$ \exists a, b \in \mathbb{Z} : a < b. $$

"There exist $a, b \in \mathbb{Z}$ such that $a < b$."

$$ \forall x \in \mathbb{N} : x*1 = x. $$

"For all $x \in \mathbb{N}$, $x*1 = x$."

Another advantage of this syntax is that the quantifiers can be nested without ambiguity since the "such that" operator is right-associative. That is,

$$ \exists s \in \mathbb{R} : s^2 = 2, \forall a \in \mathbb{Z} : \not\exists b \in \mathbb{Z} : s = a/b $$

means the same as

$$ \exists s \in \mathbb{R} : (s^2 = 2, \forall a \in \mathbb{Z} : (\not\exists b \in \mathbb{Z} : s = a/b)) $$

which says that "there exists some real $s$ whose square is 2 and for any integer $a$, there exists no integer $b$ such that $s = a/b$."

  • $\begingroup$ Oops, sorry for the misconception, but I guess the idea of using prefix notation (and right-associativity) to reduce ambiguity still holds. $\endgroup$ – Herng Yi Jan 20 '13 at 2:40

It can't be incorrect if that is how you explicitly define the notation at the outset! But if you are asking whether the notation is common, it will probably depend on the sub-discipline you are working in.

But why not write, transparently, "$a_t=\alpha$ where $\alpha\in A_{\sigma}$" if you are writing ordinary informal mathematics? While if you are writing inside a fully formalized language, you will probably want "$a_t=\alpha \land \alpha\in A_{\sigma}$".


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