Spivak Calculus, Ch. 5 Limits: Understanding the mathematical logic underlying conclusion of proof of uniqueness of limit of function

Question

In Spivak's Calculus, Ch. 5 on Limits, there is the following theorem about the uniqueness of a limit of a function near a point:

A function cannot approach two different limits near $a$. In other words, if $f$ approaches $l$ near $a$, and $f$ approaches $m$ near $a$, then $l=m$

Here is the proof:

"f approaches $l$ near $a$"

$$\forall \epsilon>0\ \exists\delta_1>0, |x-a|<\delta_1\implies |f(x)-l|<\epsilon$$

"f approaches $m$ near $a$"

$$\forall \epsilon>0\ \exists\delta_2>0, |x-a|<\delta_2\implies |f(x)-m|<\epsilon$$

$\delta=min(\delta_1, \delta_2)$

$$\implies \forall \epsilon>0\ \exists\delta>0, |x-a|<\delta\implies |f(x)-m|<\epsilon,|f(x)-l|<\epsilon\tag{1}$$

Assume $m\neq l$ Choose $\epsilon=\frac{|m-l|}{2}$.

$$|x-a|<\delta \implies |f(x)-m|<\frac{|m-l|}{2}, |f(x)-l|<\frac{|m-l|}{2}$$

\begin{align*} |m-l|&=|m-f(x)+f(x)-l|\\ &\leq |m-f(x)|+|f(x)-l|\\ &< \frac{|m-l|}{2}+\frac{|m-l|}{2}\\ &= |m-l| \end{align*} Therefore, with this reasoning we've reached the conclusion that $|m-l|<|m-l|$, which is a contradiction.

I am fine with the proof itself. My questions are about the actual logic that allows me to conclude that the proof was successful.

Questions:

• When we say the conclusion is a contradiction, what is it contradicting? The fact that a number is smaller than itself? So here some basic axiom of numbers is being contradicted?

• More importantly, what is the exact statement that because of our conclusion, is now shown to be false?

I believe it is the statement $(1)$ above. What is the negation of statement $(1)$ exactly?

In general terms I believe it is "we found an $\epsilon$ for which there is no $\delta$ such that $|x-a|<\delta\implies|f(x)-m|<\epsilon,|f(x)-l|<\epsilon$".

Reformulating this statement:

$$\exists\epsilon>0\ \text{such that}\ \nexists \delta>0\ \text{such that}\ |x-a|<\delta\implies |f(x)-m|<\epsilon,|f(x)-l|<\epsilon$$

Is this the correct conclusion? Ie, the negation of $(1)$, which is true?

Finally, as an extra if anyone can give their two cents: I haven't studied mathematical logic, but I think it would be a good idea, though I am not sure what to study (propositional logic?). What does one study so that questions of the type I am asking above aren't an issue anymore?

Answer 1

The part where he says "assume $m\ne l$" is the only assumption he makes. This has to be the part responsible for the contradiction. Therefore, $m\ne l$ must be incorrect and so $m=l$ is correct which is what was to be proved. Statement $(1)$ can't be the problem, because it follows directly from the given assumptions.

(As you say, the contradiction is that a number is smaller than itself.)

Answer 2

Therefore, with this reasoning we've reached the conclusion that $|m-l|<|m-l|$, which is a contradiction.

• When we say the conclusion is a contradiction, what is it contradicting?

The contradiction is this: $$|m-l|<|m-l|\quad\text{and}\quad|m-l|\not<|m-l|.$$

Assume $m\neq l$

• More importantly, what is the exact statement that because of our conclusion, is now shown to be false?

The assumption $$m\neq l$$ has now been disproven.

$$ \forall \epsilon>0\ \exists\delta>0, |x-a|<\delta\implies |f(x)-m|<\epsilon,|f(x)-l|<\epsilon\tag{1}$$

is the negation of statement $(1)$ $$\exists\epsilon>0\ \text{such that}\ \nexists \delta>0\ \text{such that}\ |x-a|<\delta\implies |f(x)-m|<\epsilon,|f(x)-l|<\epsilon\quad?$$

Yes, and this negation is equivalent to $$ \exists \epsilon{>}0\; \forall\delta{>}0\; \Big(|x-a|<\delta\;\text{and}\;\big(|f(x)-m|\geq\epsilon\;\text{or}\;|f(x)-l|\geq\epsilon\big)\Big).$$

What does one study so that questions of the type I am asking above aren't an issue anymore?

A quick read is sections 2 and 5—save the last two chapters of each—of this freely-downloadable book.