According to this answer:
https://math.stackexchange.com/a/2582334/1098426
We know $\forall x \ \exists y \ \forall z$ differs from $\forall x \forall z\ \exists y$ insofar as $y$ depends on $x$ versus $y$ depends on $x$ and $z$.
I traditionally translate the definition of uniform continuity on some set $A\subseteq\mathbb{R}$ as $\forall \epsilon > 0 \ \exists d > 0 \ \forall x,y \in A \ : |x-y|<d \rightarrow |f(x)-f(y)| < \epsilon$.
Edit:
The flaw in my original statement is this: $d$ is independent of $x$ and $y$. Including the $\forall$ as first quantifier suggests different $d$ for different $x$ and $y$.
Could this be written with the third quantifier as the first?
$\forall x,y \in A \ \forall \epsilon > 0 \ \exists d > 0 \ : |x-y|<d > \rightarrow |f(x)-f(y)| < \epsilon$
Typical continuity then would be something like:
$\exists c \in A \ \forall \epsilon > 0 \ \exists d > 0 \ : |x-c|<d > \rightarrow |f(x)-f(c)| < \epsilon$
Is this a reasonable interpretation?
The revised question:
If the definition of uniformy continuous is $\forall \epsilon > 0 \ \exists d > 0 \ \forall x,y \in A \ : |x-y|<d \rightarrow |f(x)-f(y)| < \epsilon$ then does this imply the formal definition of continuity is $\forall \epsilon > 0 \ \exists d > 0 \ \exists c \in A \ : |x-c|<d \rightarrow |f(x)-f(c)| < \epsilon$ as such?
Or, must we state $\exists c \in A$ as first quantifier, mirroring the typical ordinary language definition?
Per MSE guidelines, my motive is studying for an exam and trying to prove uniform continuity implies continuity. But, that is just my motive.
The formal definition of "$f$ is continuous on $A$" is $$ \forall x\in A,\ \forall\epsilon>0, \exists \delta>0,\ \forall y\in A,: |x-y|<\delta\implies |f(x)-f(y)|<\epsilon.\tag{1} $$ Note that this is just saying "$f$ is continuous at every point $x\in A$," where for fixed $x\in A$, "$f$ is continuous at $x\in A$" means $$ \forall\epsilon>0,\ \exists\delta>0,\ \forall y\in A:|x-y|<\delta\implies |f(x)-f(y)|<\epsilon. $$ When you compare $(1)$ with the definition of "$f$ is uniformly continuous on $A$": $$ \forall\epsilon>0, \exists\delta>0, \forall x,y\in A:|x-y|<\delta\implies |f(x)-f(y)|<\epsilon, $$ we see that "$f$ is uniformly continuous on $A$" implies "$f$ is continuous on $A$" in a formal way.