# Changing the order of quantifiers in continuity $\epsilon$-$\delta$ definition

I refer to the 2 alleged definitions in the remarks here.

Let $$p \in I$$ arbitrarily chosen. A function $$f:I \rightarrow \mathbb{R}$$ is continuous in p if ($$y \in I)$$:

1. $$\forall \epsilon(\epsilon >0 \rightarrow \forall y\exists \delta(\delta > 0 \land (d(x,y) < \delta \rightarrow d(f(x),f(y)) < \epsilon)))$$ i.e. $$\forall \epsilon \forall y(\epsilon >0 \rightarrow \exists \delta(\delta > 0 \land (d(x,y) < \delta \rightarrow d(f(x),f(y)) < \epsilon)))$$

2. $$\forall \epsilon(\epsilon >0 \rightarrow \exists \delta(\delta > 0 \land \forall y(d(x,y) < \delta \rightarrow d(f(x),f(y)) < \epsilon)))$$

The website says the first definition is true, which I doubt. But what is the exact difference when switching the order of $$\exists \delta > 0$$ and $$\forall y$$

I am having problems with even understading (i). My idea is that (1) is more loose because I can even pick $$x,y$$ somewhere in $$I$$ such that $$f(x),f(y)$$ are $$\epsilon$$-close and choose a suitable $$\delta$$. For example the Dirichlet-function fullfills (i).

If those $$x,y$$ exist in a $$\delta$$-Ball around $$p$$ (i.e f is in fact continuous in p), one chooses $$\delta$$ suitable small and for all $$y$$ with $$d(x,y) \geq \delta$$ (i) is not interesting because it yields no result. I am quite unsure, though.

EDIT: My question is what is the effect of changing the order of $$\exists \delta > 0$$ and $$\forall y$$. How does (ii) relate to (i)

• It's not clear what your actual question is. – Derek Elkins Mar 7 at 9:28

After reading through a few articles about using "$$\exists$$" along with "$$\rightarrow$$" (implication) [ 1 2 3], I realized that my question is quite similar.

Consider the two first-order sentences stated in the OP, simplified:

(Let $$f$$ a function, $$p$$ some point in $$domain(f)$$)

1. $$\forall \epsilon \ \exists \delta \ \forall y \ (|p-y| < \delta \to |f(p)-f(y)|< \epsilon)$$
2. $$\forall \epsilon \ \forall y \ \exists \delta \ (|p-y| < \delta \to |f(p)-f(y)|< \epsilon)$$

The quantifier $$\forall y$$ in front of the implication $$(|p-y| < \delta \to |f(p)-f(y)|< \epsilon)$$ is swapped with $$\exists \delta$$.

In logic, a statement of the form $$\exists x: P \rightarrow Q$$ is uncommon. Besides being true if $$P$$ and $$Q$$ is true, the implication also holds if $$P$$ is false. Thus, if I can find an $$x$$ that $$\lnot P$$ holds, the statement $$\exists x: P \rightarrow Q$$ is vacuously true. But this is often not what we want because we are looking for statements about elements which satisfy the condition $$P$$, otherwise there is no "information" in the statement $$\exists x: P \rightarrow Q$$.

Practically, one uses statements like $$\exists x: P \land Q$$, demanding that $$P$$ and $$Q$$ need to hold for the statement to be true.

So what is the case with the "swapped statement" $$\forall \epsilon \ \forall y \ \exists \delta \ (|p-y| < \delta \to |f(p)-f(y)|< \epsilon)$$ ?

I am not completely sure about this, but this is my opinion so far:

This statement has no information, it is a tautology. Take an arbitrary $$y$$. Choose $$\delta$$ s.t. $$\delta \leq |p-y|$$. The implication is true and thus, the statement is either.