# How does this property differ from the actual $\varepsilon - \delta$ definition of continuity?

I have written the question below verbatim:

Suppose a function $$f : \mathbb { R } \rightarrow \mathbb { R }$$ satisfies the following property: $$\forall \varepsilon > 0 , \exists \delta > 0 , \forall x , y \in \mathbb { R } , \quad | x - y | < \delta \Rightarrow | f ( x ) - f ( y ) | < \varepsilon$$

(in words: for any positive number $$\varepsilon$$, there exists a positive number $$\delta$$ such that for any pair of real numbers $$x, y$$ the inequality $$| x - y | < \delta$$ implies $$f ( x ) - f ( y ) | < \varepsilon$$)

Does it then follow that $$f$$ is continuous on $$\mathbb { R }$$?

However, this simply looks to be the $$\varepsilon - \delta$$ definition of continuity to me. Is there any subtle difference that I am missing here? If so, what would the answer be?

• Note the ordering of the quantifiers. This notion is called uniform continuity. – jgon Nov 4 '18 at 17:31
• This is uniform continuity and is a property defined on the entire set, whereas continuity is usually defined with regards to a single point $c$. – twnly Nov 4 '18 at 17:32
• This defines uniform continuity. Notice that the same $\delta$ applies to all points. Compare this to the definition of continuity where the $x$ is picked first. – John Douma Nov 4 '18 at 17:33
• So would giving an example of a uniform continuous function that is not continuous be enough to show that $f$ is not necessarily continuous? – Mohammed Shahid Nov 4 '18 at 21:22