Am I correctly interpreting the epsilon–delta definition of continuity?

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 $$x_0$$.

• Yes, this seems like a valid way of interpreting the definition Jun 10, 2019 at 14:56
– Joe
Jun 10, 2019 at 15:14

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 $$$$o(f,a) := \lim_{\delta \to 0} \left[M(a,f,\delta) - m(a,f,\delta) \right]$$$$ Then, one can prove that $$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.

• May I know the reason for downvote?
– Joe
Jun 11, 2019 at 6:21
• @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. Jun 11, 2019 at 6:44