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Hellow, what is the difference? between $$\forall{a}\in{A}:\ldots$$ and $${a}\in{A}\Rightarrow{}\ldots$$

Git Gud
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konto
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  • Welcome to MSE. You should choose your tags carefully. What has this to do with [tag:set-theory]? – José Carlos Santos Feb 06 '22 at 10:19
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    @JoséCarlosSantos letter $A$ denotes a set, how can i tag it? – konto Feb 06 '22 at 10:20
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    $\forall a \in \mathbb{R} : a^2 \geq 0$ : the square of any real number is positive. $a \in \mathbb{R} \Rightarrow a^2 \geq 0$ : if $a$ is a real number then it's square is positive. – Essaidi Feb 06 '22 at 10:22
  • You could use the [tag:elementary-set-theory] tag. But I think that a better tag would be [tag:quantifiers]. – José Carlos Santos Feb 06 '22 at 10:25
  • @JoséCarlosSantos thank you, i edited. – konto Feb 06 '22 at 10:26
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    The notation is very informal, so it's hard for me to speak in absolutes, but almost certainly a possible answer is that the first one is a statement, while the second one is just an expression (due to the presence of unbound variables). I realize this is probably not a helpful answer, but it's the one I can give with the context you have provided. – Git Gud Feb 06 '22 at 10:27
  • @GitGud so does it mean that, the first one is true whence the second can be true or not? also what is the more formal way? – konto Feb 06 '22 at 10:32
  • The first can be true (or not). For the second one, you can't even try to assess whether it is true or not. The second one is meaningless, it's like asking what the color blue tastes like (let's ignore synaesthesia), it's meaningless because of the lack of quantifiers. @konto – Git Gud Feb 06 '22 at 10:38
  • The first statement can be formulated as "whenever $a\in A$ , then ..." which is not different from the second which is "if $a\in A$ , then ...". I neither see a difference nor a problem with the missing quantifiers in the second statement. If $a$ does not belong to $A$ , nothing can be said in both cases. – Peter Feb 06 '22 at 10:52
  • @GitGud thank you i think i understood, in the second case wo dont even know exact $a$ so it is a variable. – konto Feb 06 '22 at 10:53
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    @konto Not exactly. It's not because we don't know what $a$ is. It is true that $a$ is a variable, though. To make it more concrete, consider $\forall a \in \mathbb N(a = 2)$ (which is false). The other version, as per your question, is $a \in \mathbb N \rightarrow a = 2$. This is neither true, nor false because of the free variable $a$. If you bound this variable by placing either a universal or existential quantifier, then you'll be able to assess whether it is true or false, but not before that. In fact, $\exists a(a\in \mathbb N \rightarrow a = 2)$ is true. – Git Gud Feb 06 '22 at 11:03
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    This is basically a companion question to this one posted less than $24$ hours ago.. Further to drhab's answer, note that neither of your two fragments are sentences: the first (better without the colon) literally reads as "for each element $a$ in the set $A\ldots$" whereas the second one literally reads as "element $a$ being in the set $A$ implies that $\ldots$". Note that the second is typically understood to be implicitly preceded by "for each $a$" anyway and thus be equivalent to the first. – ryang Feb 06 '22 at 11:27
  • completly different; see quantifiers and connectives. – Mauro ALLEGRANZA Feb 09 '22 at 08:12
  • See Quantification for the notation. – Mauro ALLEGRANZA Feb 09 '22 at 08:25

1 Answers1

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We can write:$$\forall a\in A [P(a)]$$This in order to state that every element of set $A$ has property $P$.

Another way of expressing this is:$$\forall a[a\in A\implies P(a)]$$ In words: for every $a$ it is true that it has property $P$ if $a$ happens to be an element $A$.

Sometimes we just leave out the quantor and abbreviate this as:$$a\in A\implies P(a)$$This with in the back of our mind the knowledge that we think of all $a$.

I hope this makes things more clear for you.

drhab
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