I have a question that's been bothering me lately.
Theorem. Let $f:X\rightarrow Y$ be a map between metric spaces. Then $f$ is continuous at $x_{0}$ if and only if $x_{n}\rightarrow x_{0}$ always implies $f(x_{n})\rightarrow f(x_{0})$.
Here is the standard proof that I'm aware of.
Proof. $(\implies).$ Let $(x_{n})$ be a sequence in $X$ that converges to $x_{0}$. Let $\epsilon > 0$. By continuity of $f$, there exists $\delta > 0$ such that
$$ d(x, x_{0}) < \delta \quad \implies\quad d(f(x), f(x_{0})) < \epsilon. $$
By convergence of our sequence, choose $N\in\mathbb{N}$ such that $n\ge N$ implies $d(x_{n}, x_{0}) < \delta$.
Then the above implies
$$ n\ge N \quad \implies \quad d(f(x_{n}), f(x_{0})) < \epsilon, $$
and so $f(x_{n})\rightarrow f(x_{0})$.$(\impliedby).$ For this, we will prove the contrapositive. Assume $f$ is not continuous at $x_{0}$. Then there exists $\epsilon > 0$ such that for all $\delta > 0$, there exists some $d(\tilde{x}, x_{0}) < \delta$ that has $d(f(\tilde{x}), f(x_{0}))\ge \epsilon$.
We use this to construct the following sequence. For $\delta = \tfrac{1}{n}$ we choose $x_{n}$ to be the element such that $d(x_{n}, x_{0}) < \tfrac{1}{n}$ but $d(f(x_{n}), f(x_{0}))\ge\epsilon$.
Then we have $x_{n}\rightarrow x_{0}$ but $f(x_{n})\not\rightarrow f(x_{0})$. $\;\square$
For the backwards direction, we used the contrapositive, which boils down to proof by contradiction, as far as I'm aware. Usually I'm okay with these kinds of proofs, but I find it strange that we need to use proof by contradiction. I can't seem to find a proof that doesn't rely on contradiction here.
My Attempt
Suppose $x_{n}\rightarrow x_{0}$ always implies $f(x_{n})\rightarrow f(x_{0})$. Let $\epsilon > 0$.
I want to show that there exists $\delta > 0$ such that $d(x, x_{0})<\delta$ implies $d(f(x), f(x_{0}))<\epsilon$.
Perhaps I can consider the set of all possible sequences $S = (x_{n})$ for which $d(x_{n}, x_{0}) < \tfrac{1}{n}$. By hypothesis, each sequence $S$ has an index $N_{S}$ for which $n\ge N_{S} \implies d(f(x_{n}),f(x_{0}))<\epsilon$. By the well-ordering principle, we may take $N_{S}$ to be the least such natural number for $S$.
Let's say I take $\delta = \inf \tfrac{1}{N_{S}}$ (where the infimum is over all those $S$). What exactly prevents us from having $\delta = 0$?
I see that every sequence $S$ has $N_{S}$, but if there are uncountably many possible $S$, it feels as though you can choose some contrived sequence that gets closer and closer to $x_{0}$ but has nonetheless an arbitrarily large $N_{S}$.
My Questions
My main question is, is the use of the law of non-contradiction unavoidable? Is there a convincing argument one way or the other?
Best Answer
Notice that the contrapositive is not the same as a 'proof by contradiction'.
Say we wish to prove $P$. In a proof by contradiction, we first assume $\neg P$, and then deduce $\neg Q$. However, we know that $Q$ holds from before our assumption of $\neg P$. This is the contradiction, and therefore, our assumption of $\neg P$ must be incorrect; in other words, $P$ must hold.
The contrapositive shows up when proving implications. Its essence is the observation that
$$(P \implies Q) \iff (\neg Q\implies\neg P),$$
that is, proving that '$P$ implies $Q$' is logically equivalent to proving that 'the negation of $Q$ implies the negation of $P$'.