Equivalent continuity definition

[JanEnClaesen]

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]

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".

 

[economicsnerd]

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).

 

[JanEnClaesen]

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.

 

[economicsnerd]

... 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).

 

[JanEnClaesen]

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]

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).

 

[economicsnerd]

But the given definition with closed neighbourhoods is still equivalent.