The epsilon–delta definition of continuity is:

"A function $f(x)$ from $\mathbf{R}$ to $\mathbf{R}$ is continuous at point $x_0 \in \mathbf{R}$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that whenever $|x–x_0| < \delta$ then $|f(x)–f(x_0)| < \epsilon$"

$\text{ }$

The intuitive informal statement of continuity is:

"A function $f(x)$ from $\mathbf{R}$ to $\mathbf{R}$ is continuous at point $x_0 \in \mathbf{R}$ if there is no sudden jump in $f(x)$ in the immediate neighbourhood of point $x_0$"

After wondering for quite some time how on earth does the definition statement imply the intuitive statement, I came up with the following interpretation:

Let us first take $\epsilon=\epsilon_1$.

Therefore there exists a $\delta_1 > 0$ such that whenever $|x–x_0| < \delta_1$ then $|f(x)–f(x_0)| < \epsilon_1$

Therefore $|f(\text{immediate neighbourhood of } x)–f(x_0)| < \epsilon_1$

This guarantees that the discontinuity at $x_0$ is less than $\epsilon_1$

Since $\epsilon_1$ could be "any small number greater than zero", the discontinuity at $x_0$ is less than "every small number greater than zero". That is, the discontinuity at $x_0$ is zero. That is, there is continuity at $x_0$.

  • 2
    $\begingroup$ Yes, this seems like a valid way of interpreting the definition $\endgroup$
    – n7kvz
    Jun 10, 2019 at 14:56
  • $\begingroup$ Does anyone have anything else to say about this interpretation? $\endgroup$
    – Joe
    Jun 10, 2019 at 15:14

1 Answer 1


Therefore $|f(\text{immediate neighbourhood of } x)–f(x_0)|< \epsilon_1$

I think you meant neighbourhood of $x_0$. But yes, your understanding of the definition seems right, especially at the end when you talk about the discontinuity at $x_0$ being less than $\epsilon_1$, and then concluding the discontinuity at $x_0$ has to be $0$.

If you're interested, you should read up about the oscillation of a bounded function at a point. This is defined as follows: let $A \subset \Bbb{R}$ be non-empty, (or more generally, a non-empty subset of $\mathbb{R}^n$), fix a point $a \in A$, and let $f: A \to \Bbb{R}$ be a bounded function. For $\delta > 0$, define the following quantities: \begin{align} M(a,f,\delta) := \sup \{f(x)|\, x \in A \text{ and } |x-a|< \delta \} \\ m(a,f,\delta) := \inf \{f(x)|\, x \in A \text{ and } |x-a|< \delta \} \end{align}

Since we assumed $f$ is bounded, these quantities exist. Now, define the oscillation of $f$ at $a$, denoted $o(f,a)$ by \begin{equation} o(f,a) := \lim_{\delta \to 0} \left[M(a,f,\delta) - m(a,f,\delta) \right] \end{equation} Then, one can prove that $f$ is continuous at $a$ if and only if $o(f,a) = 0$.

Thus, the oscillation of a function at a point is a precise way of seeing how much a bounded function fails to be continuous; in a sense, this is the precise formulation of your last paragraph.

For more information about oscillation, see Spivak's Calculus on Manifolds, page $13$, on the section about functions and continuity.

  • $\begingroup$ May I know the reason for downvote? $\endgroup$
    – Joe
    Jun 11, 2019 at 6:21
  • $\begingroup$ @Joe I'd like to know the reason as well... my best guess is that my answer defines the concept of "oscillation at a point", which requires the use of supremum, infimum, and limits to answer a question about the $\varepsilon$-$\delta$ definition of continuity. As such the answer may be viewed as "too much machinery" for a "simple" question. Although, even if one ignores the technical details in the middle, my first and last paragraphs answer your question in a simple manner. $\endgroup$
    – peek-a-boo
    Jun 11, 2019 at 6:44

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