Does $\forall x(x\in B\implies x\in A)$ imply that $((\forall a\in A)P(a)\implies (\forall b\in B)P(b) )$?

Question

I was solving a problem like $$\textrm{if} \:f\in C(A)\:\text{and}\:B\subset A,\: \text{then}\: f|_B\in C(B).$$ I transformed $f\in C(A)$ so $f$ is continuous on A and therefore $$(\forall a\in A)(\forall \epsilon>0)(\exists\delta>0)(\forall x\in A) \quad |x-a|<\delta\Rightarrow|f(x)-f(a)|<\epsilon.$$ So, this is basically $\forall a\in A, f\:\text{continuous on}\:a.$

I have solved the problem. It can be transformed into $$[\forall x\quad (x\in B\implies x\in A)]\implies [\forall a \in A \quad P(a) \implies\forall b\in B \quad P(b)].$$

Can you tell me if this extended proposition is true? If yes, why? And if no, how is that? This is just a question about the abstraction that I extracted from the context of the problem.

Answer 1

Yes, it is true that $$\Big(\forall x\;\big(x\in B\implies x\in A\big)\Big)\implies \Big(\forall a {\in} A \;P(a) \implies\forall b{\in} B \; P(b)\Big).$$ This is logically equivalent to $$\Big(\forall x\;\big(x\in B\implies x\in A\big)\;\land\;\forall x {\in} A \;P(x)\Big) \implies\forall x{\in} B \; P(x),$$ which says that if $B$ is a subset of $A$ and the propositional function $P$ holds for every element of $A,$ then $P$ holds for every element of $B.$

Answer 2

Yes, it is true that if $f$ is continuous on a set $A$ and $B\subset A$ then $f|_{B}$ is continuous on $B$. If $x\in B$ then, also $x\in A$ so for each $\epsilon>0$ there is a $\delta>0$ satisfying the condition that if $y\in(x-\delta,x+\delta)$ then $f(y)\in(f(x)-\epsilon,f(x)+\epsilon)$, now use this same choice of $\delta$ for continuity in $B$. also check out https://math.stackexchange.com/questions/1773085/subsets-of-continuous-functions-are-also-continuous#:~:text=Let%20X%20bet%20a%20subset,is%20also%20a%20continuous%20function.