Lemma: Given any function $f:M\to N$ where $M$ and $N$ are both metric spaces, $\lim_{x\to a}f(x)$ converges to $L$ only if given any function $\gamma:\mathbb{I}\to M$ (where $\mathbb{I}$ is the unit interval) such that $\lim_{t\to1}\gamma(t)=a$, $\lim_{t\to1}f(\gamma(t))=L$.

I've seen the above lemma used implicitly in various analysis proofs without acknowledgement several times. My attempt to prove it goes as follows:

Assume that the lemma was false. Then all possible $\lim_{t\to1}f(\gamma(t))$ converge to $L$, but $\lim_{x\to a}f(x)$ does not. Then, by the definition of limits, there is an $\epsilon>0$ such that for all $\delta$ there exists an $x$ such that $d(x, a)<\delta$ and $d(f(x), L)\ge\epsilon$. Denote the set of all such $x$ as $C_\delta$. Define $\gamma:\mathbb{I}\to M$, $\gamma(x)=$ some element in $C_{1-x}$. By the axiom of choice such a function exists. Obviously $\lim_{t\to1}\gamma(t)=a$, but $\lim_{t\to1}f(\gamma(t))\not=L$. This contradictions our assumptions. Thus, by contradiction, $\lim_{x\to a}f(x)=L. \square$

This proof uses the axiom of choice, which is quite unsatisfactory. Does anyone know if this theorem can be proven without using the axiom of choice? I'm fine with assuming that M and N are complete, although not much more than that.

  • $\begingroup$ You need more than completeness: the irrationals are completely metrizable, but the only continuous maps from the unit interval to them are constant. $\endgroup$ Commented Sep 15, 2012 at 21:51
  • $\begingroup$ Why does that matter? $\endgroup$
    – Parakee
    Commented Sep 15, 2012 at 22:04
  • $\begingroup$ Let $\Bbb P$ be the irrationals. Define $f:\Bbb P\to\Bbb P$ by $f(x)=\sqrt2$ if $x\ne\sqrt2$, and $f(\sqrt2)=\sqrt3$. If $\gamma:\Bbb I\to\Bbb P$ satisfies $\lim_{t\to 1}\gamma(t)=\sqrt2$, then $\gamma(t)=\sqrt2$ for all $t\in\Bbb I$, so $\lim_{t\to 1}f(\gamma(t))=f(\sqrt2)$ even though $f$ is not continuous. $\endgroup$ Commented Sep 15, 2012 at 22:13
  • $\begingroup$ Actually, you did: you said that it was a path, and paths are by definition continuous. If you don’t care about continuity, you should say so explicitly. But in that case there’s no point in using $\Bbb I$: you’re really just looking at convergent sequences. $\endgroup$ Commented Sep 15, 2012 at 22:25
  • $\begingroup$ Sorry. I didn't realize that all paths must be continuous. I've fixed it in my question. My choice to use the unit interval was arbitrary, but Im going to stick with it for now. $\endgroup$
    – Parakee
    Commented Sep 15, 2012 at 22:30

1 Answer 1


Your lemma is equivalent to the assertion that $f:M\to N$ is continuous iff it is sequentially continuous, i.e., iff whenever $\langle x_k:k\in\Bbb N\rangle$ is a convergent sequence in $M$ with limit $x$, then $\langle f(x_k):k\in\Bbb N\rangle$ is a convergent sequence in $N$ with limit $f(x)$. This result requires the axiom of countable choice: for each sequence $\langle X_k:k\in\Bbb N\rangle$ of non-empty sets, $\prod_{k\in\Bbb N}X_k\ne\varnothing$.

No choice is required to prove that a function from $\Bbb R$ to $\Bbb R$ is continuous iff it is sequentially continuous. However, the following two statements are equivalent to the axiom of countable choice restricted to subsets of $\Bbb R$:

  1. A function $f:\Bbb R\to\Bbb R$ is continuous at a point $x\in\Bbb R$ iff it is sequentially continuous at $x$.

  2. A function $f:X\to\Bbb R$ defined on some subspace of $\Bbb R$ is continuous iff it is sequentially continuous.

For more on this see Horst Herrlich, Axiom of Choice, Lecture Notes in Mathematics 1876, Springer, 2006.

  • $\begingroup$ You can also point to that huge thread of discussions about continuity and sequential continuity we had a while ago... $\endgroup$
    – Asaf Karagila
    Commented Sep 16, 2012 at 8:45
  • $\begingroup$ @Asaf: I’m not sure which one you have in mind, but this question and its answers are informative. $\endgroup$ Commented Sep 16, 2012 at 8:55
  • $\begingroup$ That is the one I had in mind. $\endgroup$
    – Asaf Karagila
    Commented Sep 16, 2012 at 8:58

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