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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?

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

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