# Equivalent continuity definition

For every interval [ f(a)-e, (fa)+e ] there exists an interval [ f(a-d), f(a+d) ] such that [ f(a)-e, (fa)+e ] includes [ f(a-d), f(a+d) ]

is this definition equivalent to the epsilon-delta definition?

CompuChip
Homework Helper
Hi Jan,

Yes, it is almost a literal translation. Try writing out the limit definition, then note that if ##\epsilon > 0##, "for all x such that ##|x - a| < \epsilon##" means "for all x in ##[a - \epsilon, a + \epsilon]##"; and ##x \in A \implies x \in B## means that "B is a subset of A", in other words, "A includes B".

Not quite. Where you put $[f(a-\delta),f(a+\delta)]$, you would need to put $$f([a-\delta,a+\delta])$$ instead.

Under your definition, any function $f:\mathbb R\to \mathbb R$ with the property that $f(-23)=f(0)=f(23)$ would be continuous at $a=0$ (by setting $\delta=23$).

Quite a point, but mightn't f([a-d, a+d]) also be a point equal to f(a) if f(a-d) = f(a) = f(a+d)? f(a-d) = f(a) = f(a+d) implies the existence of a d for every e, so that if we use this definition of continuity every symmetric function is continuous around its centre of symmetry a.

The problem f([a-d, a+d]) certainly fixes is that it requires all the points in the interval to be part of the range.

... so that if we use this definition of continuity every symmetric function is continuous around its centre of symmetry a.

$f([a-\delta,a+\delta])$ is only a singleton if $f$ is constant on the whole interval $[a-\delta,a+\delta]$, a much stronger restriction than just $f(a-\delta)=f(a)=f(a+\delta)$.

$f([a-\delta,a+\delta])$ is only a singleton if $f$ is constant on the whole interval $[a-\delta,a+\delta]$, a much stronger restriction than just $f(a-\delta)=f(a)=f(a+\delta)$.

You're right, I made an error of interpretation.

pasmith
Homework Helper
Hi Jan,

Yes, it is almost a literal translation. Try writing out the limit definition, then note that if ##\epsilon > 0##, "for all x such that ##|x - a| < \epsilon##" means "for all x in ##[a - \epsilon, a + \epsilon]##";

The intervals must be open: $(a - \epsilon, a + \epsilon)$.

But the given definition with closed neighbourhoods is still equivalent.