I have always wondered about this:

A continuous function is defined thus: for any $\epsilon>0$, there exists $\delta\in\Bbb{R}$ such that $|x-y|<\delta\implies |f(x)-f(y)|<\epsilon$ for every $x$ in its domain.

However, isn't this equivalent to saying

For any $\delta>0$, there exists $\epsilon>0$ such that $|x-y|<\delta\implies |f(x)-f(y)|<\epsilon$ for every $x$ in its domain?

I have read somewhere that the two statements are not the same, and that the latter statement may sometimes not be true for a continuous function. Could someone provide such an example?

  • 1
    $\begingroup$ Both of the statements you give are not the definition of continuity. A function $f$ is continuous at $x$ if for every $\varepsilon > 0$, there is $\delta > 0$ such that $|x - y| < \delta \Rightarrow |f(x) - f(y)| < \varepsilon$. We then say $f$ is continuous if it is continuous at every $x$ in its domain. $\endgroup$ – Michael Albanese Jun 11 '14 at 9:39
  • $\begingroup$ @MichaelAlbanese- Thanks. I added the "every $x$ in its domain" part. $\endgroup$ – freebird Jun 11 '14 at 9:40
  • $\begingroup$ What Michael Albanese was referring to is the fact that you switched the order of implication in your first "definition". $\endgroup$ – JoeyBF Jun 11 '14 at 9:43
  • $\begingroup$ I think that the implication arrow in your first statement has to be reversed to get the definition of continuity. $\endgroup$ – Giovanni De Gaetano Jun 11 '14 at 9:43
  • $\begingroup$ Your corrected statement is still wrong. You need to say for all $x$ before you introduce $\varepsilon$ and $\delta$. $\endgroup$ – Michael Albanese Jun 11 '14 at 9:48

You're right. Your first statement is really close to the definition of a continuous function, which should be

$\forall x \in \mathcal{D}_{f}\;,\; \forall \epsilon > 0 \;,\; \exists \delta > 0 \;,\; \forall y \in \mathcal{D}_{f} \;,\; |x - y| \leq \delta \Rightarrow |f(x) - f(y)| \leq \epsilon$

where $\mathcal{D}_{f}$ is the definition domain of $f$. On the other hand, the second one induces much weaker constraints on the function (if any...). An intuitive reason for that is that there is no constraint on $\epsilon$ to be small.

A good example is to consider the characteristic function of $\mathbb{Q}$ , $\chi_{\mathbb{Q}}$ , defined as

$$ \chi_{\mathbb{Q}} (x) = \begin{cases} 1 \;\; \text{if} \;\; x \in \mathbb{Q} \\ 0 \;\; \text{if} \;\; x \notin \mathbb{Q} \end{cases}$$

This function is known to be highly discontinuous. The maximum distance between two values is $1$, so that $\forall \delta$, if we choose $\epsilon = 2$, your second statement is always satisfied.

  • $\begingroup$ Thank you. I remember thinking along these lines a long time back. $\endgroup$ – freebird Jun 11 '14 at 9:44
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    $\begingroup$ What you wrote is correct, except that the first statement is not the definition of a continuous function over the reals. $\endgroup$ – JoeyBF Jun 11 '14 at 9:45
  • $\begingroup$ @JoeyBf I hope now it's okay. I was wondering what kind of constraint the second statement could imply. My feeling would be that it only forbids $f$ to diverge at the edge of the open interval of definition ? $\endgroup$ – jibe Jun 11 '14 at 9:51
  • $\begingroup$ Your new definition is correct. I think that this definition is equivalent to being bounded, but probably weaker. For example, let $f(x) = 1/x$ if $x \neq 0$ and $f(0) = 0$. Then $f$ does not satisfy this proposition at $x=0$. $\endgroup$ – JoeyBF Jun 11 '14 at 10:00
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    $\begingroup$ @JoeyBF Your conclusion is correct only if $f$ is defined on a bounded domain. If we consider the function $f(x) = x$ defined on $\mathbb{R}$, it verifies the second statement (choose $\epsilon = \delta$), but it is not a bounded function. $\endgroup$ – jibe Jun 11 '14 at 10:05

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