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.