We introduce a weaker notion of continuity, similar to continuity at a point, that might be useful pedalogically.

Continuity

Over \(\mathbb{R}\), a function \(f: \mathbb{R} \to \mathbb{R}\) is defined to be continuous if

\[\forall x \, \forall \epsilon \, \exists \delta\, \forall y \text{ s.t. } \vert y-x\vert < \delta \implies \vert f(y)-f(x)\vert < \epsilon\]

where \(x, y \in \mathbb{R}, \epsilon, \delta \in (0, \infty)\)

Letting \(R\) be the statement that \(\forall y \text{ s.t. } \vert y-x\vert < \delta \implies \vert f(y)-f(x)\vert < \epsilon\), we could see \(f\) is continuous if and only if

\[\forall x \, \forall \epsilon \, \exists \delta\, R\]

Weaker notions of continuity

The most common weaker notion of continuity is that of continuity at a point. We say that \(f\) is continuous at \(x\) if

\[\forall \epsilon \, \exists \delta\, R\]

and as such it’s apparent that \(f\) is continuous if and only if it is continuous at all \(x\). Inspired by this, we define that \(f\) is \(\epsilon\)-continuous for some \(\epsilon>0\) if

\[\forall x \, \exists \delta R\]

Example: The floor function is \(2\)-continuous but not \(\epsilon\)-continuous for any \(\epsilon \leq 1\).

Example: If a function is bounded then it is \(\epsilon\)-continuous for some \(\epsilon\).

It is clear that \(f\) is continuous if and only if it is continuous at all \(\epsilon\). However, \(\epsilon\)-continuity has rather unusual properties.

Claim: If \(f\) is \(\epsilon_1\)-continuous and \(g\) is \(\epsilon_2\)-continuous then \(f+g\) is \(\epsilon_1 + \epsilon_2\)-continuous.

Example: The exponential function is continuous but \(\exp(x)\text{floor}(x)\) is not \(\epsilon\)-continuous for any \(\epsilon\).

Example: The exponential function is continuous but \(\exp(\text{floor}(x))\) is not \(\epsilon\)-continuous for any \(\epsilon\).

Much of the ideas driving the above claims and examples could be traced back to the proofs of preservation of continuity under addition and multiplication.

Uniform continuity

Similarly we can say that \(f\) is uniformly \(\epsilon\)-continuous if

\[\exists \,\delta\, \forall x R\]

Again \(f\) is uniformly continuous if and only if it is uniformly \(\epsilon\)-continuous for all \(\epsilon\). Uniform \(\epsilon\)-continuity also clearly implies \(\epsilon\)-continuity.

Example: \(\text{floor}(\exp(x))\) is 2-continuous but not uniformly 2-continuous.

Acknowledgements

Thanks to Isaac Li for discussions about this topic, especially on the use of floor and exponential functions to generate counterexamples.