# Definition of continuity with epsilon and delta

I have been told that a function $f$ is continuous at $c$ if for every $\epsilon >0$ there exists a $\delta >0$ such that:

$|x-c|<\delta \Rightarrow |f(x)-f(c)|< \epsilon$

This may be a stupid question, but is this the same as saying $f$ is continuous at $c$ if for every $\delta >0$ there exists an $\epsilon >0$ such that:

$|f(x)-f(c)|< \epsilon \Rightarrow |x-c|<\delta$ ?

• It's not $\iff$ but only $\implies$. – Xam Nov 22 '16 at 21:49
• $$\boldsymbol{f(x) = \sin x}$$ is an example of a function that is continuous, but does not satisfy your alternate condition anywhere. Why? Consider for example $c = 0$. Pick any $\delta > 0$ and there must be an $\epsilon > 0$ such that whenever $|\sin x| < \epsilon$, $|x| < \delta$. But taking $x = k \pi$, since $|\sin (k \pi)| = 0 < \epsilon$, that means that $|k \pi| < \delta$ for all $k$, which never happens.
• $$\boldsymbol{f(x) = \begin{cases} x + 1 &\text{if } x > 0 \\ 0 &\text{if } x = 0 \\ x - 1 &\text{if } x < 0 \end{cases}}$$ is an example of a function that is not continuous (at $0$), but satisfies your alternate condition everywhere.
• thank you, could you help me furthermore? Is the definition the same if you say "for every $\delta > 0$ there exists $\epsilon > 0$"? (instead of for every epsilon there exists a delta). – Desmoz Nov 22 '16 at 23:29
• @ConnorGaughan No, that would not be the same either. That condition is the same as the function being bounded. If it is bounded, then for any $\delta$ we can just take $\epsilon$ to be bigger than the bound on the function. – 6005 Nov 23 '16 at 1:59