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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 $a \subset b$.

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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 $B$. – 2017-02-20
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    The reason is that in set theory $\subseteq$ is defined in terms of $\to$ (and the universal quantifier) : $A \subseteq B \Leftrightarrow \forall x (x \in A \to x \in B)$. – 2017-02-20
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    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? – 2017-02-20

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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