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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)

Asaf Karagila
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Quantaurix
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1 Answers1

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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.

Quantaurix
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