Here is my attempt at a proof for the Sequential Criterion.

Definition: Sequential Criterion for Continuity

A function $f: A \to \Bbb R $ is continuous at the point $c \in A$ if and only if $\forall$ sequence $(x_n) \in A$ that converges to $c$, the sequence $(f(x)n))$ converges to $f(c)$.


i) "$\rightarrow$"

Let $f: A \to \Bbb R$ be continuous at a point $c \in A$. Since $f$ is continuous at $c$ then $\forall$ $\epsilon > 0$, $\exists$ $\delta > 0 $ such that for $x \in A$ satisfying $|x-c| < \delta$ $\implies$ $|f(x) - f(c)| < \epsilon$.

ii) "$\leftarrow$"

Let $(x_n) \in A$ be a sequence converging to some point $c$.

If $(x_n) = $ constant function, then $f(x_n) = c$ and thus $f(x_n)$ converges to $f(c)$

If $(x_n)$ is $\neq c$, then an infinite subsequence $(x_n)_k, k\in \Bbb N$, can be formed in which the subsequence converges to $c$.

Therefore if $f$ is continuous at $c$, $f(x_n)_k) \to f(c)$, thus $f(x_n) \to f(c)$.

Any advice pointing out holes in my argument is appreciated!

  • $\begingroup$ In one direction you would be expected to use your definition of $\lim_{x\to c} f(x)$ rather than this loose argument. [Probably an $\epsilon$, $\delta$ definition in your text?] In the other direction ... well you didn't consider the other direction. $\endgroup$ Jan 18, 2016 at 19:19
  • 1
    $\begingroup$ General comment: you're trying to prove a very basic result; it therefore helps to know what you can assume known, e.g., what is proved in your textbook before this. In most proofs you can freely throw out general knowledge, but not here. To me it looks like, in the next-to-last line, you are assuming (because it's general knowledge) the result you're trying to prove. $\endgroup$
    – ForgotALot
    Jan 18, 2016 at 19:30
  • $\begingroup$ Thank you, I appreciate your points. I separated the implications, and tried to clean up the argument. $\endgroup$
    – Iff
    Jan 18, 2016 at 20:16

1 Answer 1


I don't understand your forward direction, since it is essentially stating that $f$ is continuous twice: once explicitly, and then again with $\epsilon-\delta$ definition. I will try to "fill in" a little bit of the argument (albeit with some verbiage) for this direction. The problem is, as has been stated, that the result is very much intuitive and so its almost unclear what exactly you are supposed to show.

A function $f: A \to \Bbb R $ is continuous at the point $c \in A$ if and only if $\forall$ sequence $(x_n) \in A$ that converges to $c$, the sequence $f(x_n)$ converges to $f(c)$.

Well suppose that $f:A \rightarrow \mathbb{R}$ is continuous. Then for each $\epsilon$, there exists $\delta$ such that $|x-c|<\delta \implies |f(x)-f(c)|<\epsilon.$

Let $\{x_n\}$ be a sequence in $A$ that converges to $c$. We will show that for each $\epsilon$, there exists some $N$ such that $n>N \implies |f(x_n)-f(c)|<\epsilon$.

Let $\epsilon>0$. Then choose $\delta$ so that $|x-c|<\delta \implies |f(x)-f(c)|<\epsilon.$ We have that for each $\delta$ there exists $N \in \mathbb{N}$ and $n>N$ where $|x_n-c|<\delta$, implying that $|f(x_n)-f(c)|<\epsilon$, as desired.

If you have any background in general topology, you can see the standard munkres proof of "the sequence lemma" http://www.westga.edu/~faucette/teaching/Fall_15/MATH4613/Handouts/Munkres.pdf page 44


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