trying to recall the $\epsilon, \delta$ definition of continuity, I came up with the following:

A function is continuous at $x$ if $\forall \epsilon > 0 \; \exists \; \delta > 0: |f(x-\delta) - f(x+\delta)| < \epsilon$.

This is very likely not equivalent to the Weierstrass' definition of continuity at $c$:

$\forall \epsilon > 0 \; \exists \; \delta > 0: |x-c| < \delta \Rightarrow |f(x) - f(c)| < \epsilon $.

Could you please point out where the first statement fails to be equivalent to Weierstrass'?

Many thanks!

  • 3
    $\begingroup$ First of all, in the first definition $f$ need not to be defined at $x$. $\endgroup$ – enzotib Jul 11 '14 at 11:08
  • $\begingroup$ A definition like this cannot fail per se as you are free to define any concept you like as long as it is self-consistent. However, it is not equivalent to the standard definition of continuous. $\endgroup$ – Winther Jul 11 '14 at 11:12
  • $\begingroup$ @Winther: thank you, edited. $\endgroup$ – vojta havlíček Jul 11 '14 at 11:18

The Weierstrass' definition considers all values of $x$ within distance $\delta$ of $c$.

Your definition only says that there are two equidistant points from $x$, of which the difference in function values is less than $\epsilon$.

Consider the function $f$ which is constantly 0 except for $f(0) = 1$. This is clearly not continous, but following your definition, for any $\epsilon > 0$ I can certainly pick any positive $\delta$ so that $$|f(-\delta) - f(\delta)| = 0 < \epsilon$$


Hint For both definition, consider the continuity in $0$ of

$$f(x) =\left\{ \begin{array}{ll} 132 & \text{if } x = 0 \\ 0 & \text{else} \end{array}\right. $$


This may seem natural to you, but it has some serious problems. For example, it says that every even function is continuous at $0$--that's not really acceptable, is it?

  • $\begingroup$ Original: Does it say so even for $f(x) = 1 (x > 0)$, $f(x) = $ for $(x \leq 0)$ though? Edit: Missed the "even". Thank you! $\endgroup$ – vojta havlíček Jul 11 '14 at 11:21
  • $\begingroup$ No, but that's not an even function. An even function satisfies $f(-x)=f(x)$, so you always have |$f(0-\delta)-f(0+\delta)|=0<\epsilon$ for every choice of $\delta$ and $\epsilon$ $\endgroup$ – MPW Jul 11 '14 at 11:25
  • $\begingroup$ Please see the edit above. Thank you! $\endgroup$ – vojta havlíček Jul 11 '14 at 11:26
  • $\begingroup$ I understand, thanks $\endgroup$ – MPW Jul 11 '14 at 11:27

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