The proof that we often see for the equivalence of $\epsilon$-$\delta$ definition of a limit and the definition of a limit in terms of sequences often use proof by contradiction. As an exercise I wanted to try a direct proof.

I'll only prove one direction as this is the part I had a question on. We consider function $f: \mathbb{R} \rightarrow \mathbb{R}$ and assume that it is continuous at $(x,y)$ in the sequential sense.

Choose any $\epsilon > 0$. Further, choose any sequence $\{x_n\}$ where $\lim x_n = x$ and $x_n \neq x$ for all $n$. We know that there must exist some $n_1$ where $n > n_1$ implies $|f(x_n) - y| < \epsilon$. Also, there must be some $n_2$ for which $n > n_2$ implies $|x_n - x| < \epsilon$. Now, we let

$$\delta = \sup \{|x_n - x| : n > \max(n_1, n_2)\}.$$

Clearly, $\delta$ is finite and larger than zero. Further if we have $x'$ such that $|x' - x| < \delta$, then there must exist some $x_{n'}$ for which $|x' - x| < |x_{n'} - x|$.

I want to conclude that $|f(x') - y| < \epsilon$, but I'm not sure how to proceed from here.


1 Answer 1


Suppose that $f$ is given by

$$f(x)=\begin{cases} 0,&\text{if }x=\frac1n\text{ for some }n\in\Bbb Z^+\\ 2x,&\text{otherwise;} \end{cases}$$

this is continuous at $x=0$. We might happen to choose the sequence $\left\langle\frac1n:n\in\Bbb Z^+\right\rangle$, so that $x_n=\frac1n$ for each $n\in\Bbb Z^+$. Now take $\epsilon=\frac12$. We can take $n_1=1$ and $n_2=2$, so that $\delta=\frac13$. Let $x'=\frac3{10}$; then $|x'|<\delta$, and $x'<\frac13=x_3$, so we can take $n'=3$. But then


Thus, on the basis of what you’ve done so far you cannot hope to prove in general that $|f(x')-y|<\epsilon$. I am not at all sure that a direct proof is possible, because somehow you really do need to use the fact that $f$ behaves at $x$ for all sequences converging to $x$.


You must log in to answer this question.

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