Logic question in Real Analysis

Question

I thought about posting some minor insecurities and some main doubts i have on a particular topic.

Let R with the usual metric be a metric space, $A \subseteq R$ , $f:A\to R$ and p is a limit point of A.

Consider the limit of f as its dependent variable on the A-domain tends to the point p being defined with the common $\delta-\varepsilon$ definition.

I have the following sequence characterization .

$P_1$ iff $P_2$

Where
$P_1$ = $\exists L \in R ( lim_{x\to p}f(x) = L)$ .
$P_2$ = $\exists L \in R $ ( for every sequence eventually non-constant sequence $x_n \in A$ converging to p, $f(x_n)$ converges to L ).

Now, i'm atempting to negate the previous equivalence in order to get a criterium for the non-existence of the functional limit .

I guess that $\neg P_1$ would become "$\forall L \in R ( lim_{x\to p}f(x) \neq L) $. In words, it would become "$lim_{x\to p}f(x)$ doesn't exist ".
Is that correct ? (Insecurity #1)

But my problem lies on $P_2$.
The natural negation that comes to me would be :
"$\forall L\in Y$ (there exists a eventually non-constant sequence $x_n \in A$ converging to p such that $f(x_n)\neq L$ ) .
In words, it would become equivalent to "there exists an eventually non-constant sequence $x_n \in A$ converging to p such that $f(x_n)$ doesn't converge".
Is that reasoning correct ? (Insecurity #2)

Now, i have read that there is some other proposition $P_3$ that is is also equivalent to $\neg P_2$, namely, :
"there exists eventually non-constant sequences $x_n$ and $y_n$ $\in A$ converging to p such that $lim_{n\to \infty} f(x_n) \neq lim_{n\to \infty} f(y_n)$ ".

I'm curious on this alternative equivalence. I can see how $\neg P_2$ iff $P_3$ on the context of sequences :
If $\neg P_2$ is true, then letting the sequence that satisfies it be $x_n$, then $P_3$ will be true for any eventually non constant sequence $y_n$ converging to p.
If $P_3$ is true, then we can build a sequence $z_n$ such that $f(z_n)$ won't converge by defining $z_{2n} = x_n , z_{2n+1}=y_n$, and which will satisfy $\neg P_2$ .
Is that correct ? (#Doubt 1)

But i'm curious on the generality of that equivalence. Is it certain that the $\neg P_2$ and $P_3$ propositions can not be expressed formally with first order logic, but i guess that the formal system of higher order logic would be able to formally express them.
So, would there be a logical axiom of higher order logic representing generally the equivalence that was presented on the context of sequences between $\neg P_2$ and $P_3$ Or is the generality of that equivalence much more simple than i'm making ? (#Doubt 2) .

Thanks a lot in advance.

Answer 1

A few points, in the order I think of them:

• For $P_2$, you should have $\lim_{n \to \infty }f(x_n) = L$
• Your negation of $P_1$ is correct
• The verbal of $P_2$ should be "for every $L$, there exists a sequence $x_n$ such that $f(x_n)$ does not converge to $L$"
• $P_3$ assumes that we can construct some sequence that converges to two non-equal things. That is, we have to assume that there are an $x_n$ and $y_n$ such that $f(x_n)$ and $f(y_n)$ are convergent and have non-equal limits.
• However, assuming we have such an $x_n$ and $y_n$ exist, we can indeed make a $z_n$ as you suggest, so that $P_1 \iff P_2 \implies \neg P_3$

For an example of that convergence only existing in the one direction, consider $$ f(x) = 1/x $$ $f$ has no limit at $0$, but $P_3$ is not true.