Logical problem in analysis

Question

I'm holding a logical problem in real analysis.

Let R with the usual metric be the metric space in consideration. Define a function $f : A \to R$ where $A \subseteq R$ and let p be a limit point of A.

By what i have learned so far, it would be okay to say that the proposition

$P_1$ = "(finite) limit of $f$ as $x$ tends to $p$ exists"can be formally expressed as

$P_2$ = "$\exists L \in R, \lim_{x \to p} f(x) = L $"

where "$\lim_{x \to p} f(x) $= L" is an abbreviation for some formula ( also depending on $f, L$ and $p$ ) , which can be based on $\varepsilon -\delta $ or based in sequences for example.

Now, my problem is that if $P_1$ is equivalent to $P_2$, then $\neg P_1$ should also be equivalent to $\neg P_2$.
Which would be saying that "(finite) limit of $f$ as $x$ tends to $p$ doesn't exist" should be equivalent to $\forall L \in R, \lim_{x\to p}f(x) \neq L$.

But the problem is that $\neg P_2$ only makes sense if we know that $ \lim_{x\to p}f(x) $ exists.

So, how can be two propositions $P_1$ and $P_2$ be equivalent, but whenever we negate them there are sittuations that make $\neg P_1$ is well-defined but make $P_2$ unambiguous/not making any sense?

I'm thinking the problem lies in stating the equivalence between $\exists L \in R, \lim_{x\to p}f(x) = L$ and
$\exists L \in R, \forall \varepsilon>0, \exists \delta>0, ( \forall x \in A (|x-p|<\delta ) \to |f(x)-L|<\varepsilon ) $, for the $\varepsilon -\delta $, for example ( but also for the sequence case ) .

I think they can't be equivalent because while the latter makes sense whether $f$ has a limit at $p$ or doesn't, the former only makes sense when f has a limit at $p$.

So, the statements might have some relation ( what is it ? ) but it is not a relation of equivalence.

Is it exactly the actual problem related to my initial doubt ?
Thanks in advance

Answer 1

Writing $\lim_{x\to p}f(x)=L$ is a symbolic shorthand for two statements:

1. The existence of a limit at $p$
2. the value of that limit.

There is no shorthand for:

• saying that $f$ has a limit at $p$ without specifying that value.
• There is also no shorthand to say $f$ may not have a limit at $p$ but if it does, then its value is not $L$.

The symbolic negation of $\lim_{x\to p}f(x)=L$ is $\lnot(\lim_{x\to p}f(x)=L)$, which is a different statement from $\lim_{x\to p}f(x)\neq L$, which is shorthand for there is a limit, but is not $L$.

There is way to talk about convergence without limits in $\mathbb{R}$, since it is a complete metric space: Cauchy sequences are the same as convergent sequences. If you use sequences to define the limit of $f$ at $p$, then you can write your propositions without the need to introduce temporary symbols for limits that may or may not exist.

Answer 2

I believe your second guess to be a safe way to be sure. $\exists L \in \mathbb{R}: R=\lim_{x\to p}f(x)$ would actually translate to $\exists L\in\mathbb{R}: \forall \varepsilon>0...$, which in turn would, negated, imply equivalence between your statements.

Where I suppose you are mistaken is that your negation does not actually imply the existence of the limit. Take the statement $L\neq\lim_{x\to p}f(x)$.

This statement means no more than $\exists \varepsilon>0: \forall \delta>0 \quad \exists x \in U_\delta(x): f(x)\notin U_\varepsilon(L)$.

This is notably different from the statement "$f$ has a limit, and this limit is not $L$."

If you look at that statement, in no way does it mean that there is an actual limit. I think what you have here is a problem caused by the notation - once we use $\lim$ to describe a property of $L$, namely the $\varepsilon-\delta$ criterion or so, another time you use $\lim$ to describe an object, which does not always exist.

Answer 3

This is the notorious issue of "undefined"" or "non-denoting" terms in logic. For a simpler example, compare your $P_1$ (about the function $f$ and the real number $p$) and $P_2$ with the following statements about numbers $m$ and $n$:

$Q_1$: $\exists d \in {\mathbb{N}} \cdot dm = n$

$Q_2$: $\frac{n}{m} \in {\mathbb{N}}$

Intuitively these ought to be equivalent, but it is unclear what $Q_2$ means if $m = 0$.

Russell's seminal contribution to this topic asks us to replace all propositions containing potentially undefined terms by propositions that do not suffer from this problem. There has been a lot of debate about this and it is an important issue for computer science (where the difference between $Q_1$ and $Q_2$ might be the difference between a safe program and a program with a divide-by-zero error). Many people find it intuitive just to read equations between potentially undefined terms as false, but this can be counter-intuitive when the equations occur inside complex propositions (as in your example). In pure mathematics, I would advise you to be aware of the issue and to avoid using potentially undefined terms if in any doubt.

Answer 4

Your statement $$P_1: \textrm{There exists a limit of} f(x)\textrm{ as }x\textrm{ tends to }p$$ is logically equivalent to $$P_2: \exists L, \lim_{x\to p} f(x)=L.$$ By logically equivalent here I mean that those two statements are true for precisely the same pairs of $f,p$. This is different from $P_1$ being linked to $P_2$ by a sequence of truth-preserving transformations. But that is also possible: You can consider the statement $$P_1':\textrm{There exists a limit of }f(x)\textrm{ as }x\textrm{ tends to }p\textrm{, lets call it }L$$ This statement is equivalent to $P_1$, it introduces a new name, but that does not change truth values. The statements $P_1'$ and $P_2$ are logically equivalent, because they express the same statement once as text and once as symbols. Now for the negations: $\lnot P_1$ and $\lnot P_2$ are logically equivalent because $P_1$ and $P_2$ are equivalent: $\lnot P_2$ is true precisely where $P_2$ is false, which is where $P_1$ is false, which is where $\lnot P_1$ is true.

We can easily negate $P_1$: $$\lnot P_1:\textrm{There does not exist a limit of }f(x)\textrm{ as }x\textrm{ tends to }p$$ Note how the second part of statement $P_1'$ doesn't make much sense anymore if $P_1$ is negated. Now for $P_2$: $$\lnot P_2:\lnot(\exists L,\lim_{x\to p}f(x)=L)$$ The negation of $\exists L, Q(L)$ is logically equivalent to $\forall L,\lnot Q(L)$ for a proposition $Q$. Lets apply that to $\lnot P_2$: $$\lnot P_2:\forall L, \lnot(\lim_{x\to p}f(x)=L)$$ The symbol $\lim_{x\to p} f(x)$ denotes the limit of $f$ at $p$ and using it implies the existence of the limit. For that reason writing $\lnot(\lim f(x)=L)$ is not the same as writing $\lim f(x)\neq L$. We therefore do not know immediately whether $\lnot P_2$ and the statement $$P_3: \forall L,\lim_{x\to p}f(x)\neq L$$ are logically equivalent. However, as writing $\lim f(x)\neq L$ implies the existence of the limit, we know that $L$ can have the value of this limit. Which means $\lim f(x) \neq L$ is false when $L$ has this value, i.e. for $L=\lim f(x)$. Which means $$\forall L, \lim_{x\to p} f(x)\neq L$$ is false, regardless of $f$ and $p$, and therefore not logically equivalent to $\lnot P_2$. The most natural way to write the negation of $P_2$ would probably be $$\forall L, L\textrm{ is not the limit of }f\textrm{ at }p$$ which is of course just $\lnot(\lim f(x)=L)$ written as text instead of using mathematical and logical symbols.