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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.

isaac
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    It looks like you've confused some symbols in your last expression of continuity ($c$ is not used and $x,y$ are not quantified). Also note that uniform continuity is defined for a set $A$ whereas continuity is defined at a point. What you should be proving is that uniform continuity on $A$ (i.e. $\forall\epsilon\exists\delta\forall x\forall y\dots$) implies continuity at every point in $A$, i.e. $\forall x\in A,\forall\epsilon\exists\delta\forall y\dots$. – Karl Mar 19 '24 at 02:06

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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.

Alex Ortiz
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  • Thank you. I see I was missing the arbitrary $y$ in $A$ for continuity. For $f$ continuous a fixed single point $x$ in $A$, (your 2nd equation) could we say $\exists x\in A,\ \forall\epsilon>0, \exists \delta>0,\ \forall y\in A,: |x-y|<\delta\implies |f(x)-f(y)|<\epsilon$? – isaac Mar 19 '24 at 03:02
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    @isaac: When we want to say $f$ is continuous at $x$, I would leave off the first "$\exists x$" because we already froze $x$ in place and now we are stating what it means for $f$ to be continuous at $x$. This is a definition in terms of numbers $\epsilon,\delta$, and other points $y$ of $A$. – Alex Ortiz Mar 19 '24 at 13:08
  • Thank you for clarifying. I can see how fixing $x$ is a property independent of the definition of continuity, and how adding the quantifier would be unneeded or even confusing. In retrospect, I can make sense of this setting $A={x}$ for your equation $(1)$, but that is a bit of a stretch. I appreciate you taking the time. – isaac Mar 19 '24 at 19:08