0
$\begingroup$

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)

$\endgroup$
  • 1
    $\begingroup$ It's not clear what your actual question is. $\endgroup$ – Derek Elkins Mar 7 at 9:28
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.