The epsilon-delta definition of continuity As we know the epsilon-delta definition of continuity is:
For given $$\varepsilon > 0\  \exists \delta > 0\ \text{s.t. } 0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon $$
My question: Why wouldn't this work if the implication would be:
For given $$\varepsilon > 0\ \exists \delta > 0\ \text{s.t. } |f(x) - f(x_0)| < \varepsilon  \implies  0 < |x - x_0| < \delta ?$$

Consider the implications of using this definition for any constant function (which should all be continuous, if any function is to be continuous).


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*In particular, for $c \in \mathbb{R}$ consider the constant function $f(x) = c$. Given $x_0 \in \mathbb{R}$, taking $\varepsilon = 1$, note that for any $\delta > 0$ if $x = x_0 + \delta$ we have that 


*

*$| f(x) - f(x_0) | = | c - c | = 0 < 1 = \varepsilon$;  but 

*$| x - x_0 | = |(x_0 + \delta ) - x_0 | = \delta \not< \delta$.


Therefore the implication $| f(x) - f(x_0) | < \varepsilon \rightarrow | x - x_0 | < \delta$ does not hold. It follows that the function $f$ does not satisfy the given property.
