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If $a \implies b$ can we then generally say that $a \subseteq b$ ?
For example:
if $a: x > 15$ and $b: x >10$ then clearly $a \implies b$ and if we look at the sets represented by a $\lbrace16,17,18..\rbrace$ and b $\lbrace 11,12,13... \rbrace$ it is also obvious that $a \subset b$.

Numair Mansur
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    The connective $\to$ is used between sentences and not always a sentence express a condition regarding a set. What is linked is the formula $\forall x (A(x) \to B(x))$ with the inclusion $A \subseteq B$ where $A = { x \mid A(x) }$ and the same for $B$. – Mauro ALLEGRANZA Feb 20 '17 at 14:58
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    The reason is that in set theory $\subseteq$ is defined in terms of $\to$ (and the universal quantifier) : $A \subseteq B \Leftrightarrow \forall x (x \in A \to x \in B)$. – Mauro ALLEGRANZA Feb 20 '17 at 14:59
  • You cannot use $a\Rightarrow b$ and $a\subseteq b$ with the same $a, b$. This doesn't make sense. For $a\Rightarrow b$ to work both $a, b$ have to be sentences. On the other hand for $a\subseteq b$ to work both $a, b$ have to be sets. Even in your example you clearly distinguish between sentences $a, b$ and underlying set conditions $x > 15$, $x>10$. You even say yourself "sets represented". Now if you can define how this "representation" works in general case then we can help. So if you have a way to generate a set from a sentence then perhaps yes. Who knows? – freakish Feb 20 '17 at 15:40

1 Answers1

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No but you can say:

Let $P$ be $(x ∈ A)$
Let $Q$ be $(x ∈ B)$

$A ⊆ B ≡ ∀x, P → Q $


$⊆$ is for sets
$→$ is for propositions


Note: You cannot say the same for $A ⊂ B$
Why?

Let $A$ be ${1}$
Let $B$ be ${1}$
Let $P$ be $(x ∈ A)$
Let $Q$ be $(x ∈ B)$

$A ⊂ B$ is false.
$∀x, P → Q$ is true.
Therefore $A ⊂ B$ cannot be equivalent to $∀x, P → Q$

isaac
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Jeff Hykin
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