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?

  • $\begingroup$ Your proof uses the well-ordering principle. I wonder its use can be justified. $\endgroup$
    – Hanul Jeon
    Nov 5, 2017 at 5:28
  • 3
    $\begingroup$ According to an article by Ishihara, even the equivalence of two concepts over separable metric spaces requires a non-trivial assumption, which is not basically assumed in the constructive mathematics. $\endgroup$
    – Hanul Jeon
    Nov 5, 2017 at 5:34
  • $\begingroup$ @HanulJeon That's really interesting. $\endgroup$ Nov 5, 2017 at 7:16
  • $\begingroup$ Possible duplicate of math.stackexchange.com/questions/1312873/… $\endgroup$ Nov 5, 2017 at 7:42

1 Answer 1


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 $P$'.

  • $\begingroup$ I see; I admit that I brushed over this detail a little too fast. However, doesn't the contrapositive require proof by contradiction to be proven? I don't see how this changes things. $\endgroup$ Nov 5, 2017 at 5:47
  • 1
    $\begingroup$ @SpiralRain You are correct. The $\impliedby$ direction is equivalent to double negation elimination. Simply pick $P\equiv\top$ and assume $\neg Q\equiv(Q\Rightarrow \bot)$. $\endgroup$ Nov 5, 2017 at 9:07
  • $\begingroup$ The proof of the contrapositive does not rely on contradiction. It is a mere symbolic manipulation/rewriting of $P\implies Q$ after noting that it is logically equivalent to $\neg P \lor Q$. See this link. $\endgroup$ Nov 5, 2017 at 14:40
  • 3
    $\begingroup$ @Fimpellizieri Showing that $P\Rightarrow Q$ is logically equivalent to $\neg P \lor Q$ relies on contradiction. Simply choose $Q=P$ and you get $P\Rightarrow P$ which is surely true even constructively, and $\neg P\lor P$ which is the law of excluded middle. $\endgroup$ Nov 5, 2017 at 17:37
  • 2
    $\begingroup$ @Fimpellizieri My assumption is the question isn't: "How can I superficially rewrite my proof to not use proof by contradiction but an equivalent principle?", but is more like: "Can this be proven constructively? if so, how?" If my assumption is correct, no amount of manipulation of generic classical propositional formulas is going to help. $\endgroup$ Nov 5, 2017 at 18:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .