The interactive applications linked to this page are intended to demonstrate three definitions of a continuous function on a metric space.

If viewing this on a mobile device we suggest that you use 'landscape' mode.

## Definition using inverse images

The function \(f:A \rightarrow B\) is continuous if and only if for every open set, \(C\) in the codomain
the inverse image of \(c, f^{-1}(c)\) is an open set in the domain.

Definition of continuity using inverse image (Desktop)

Definition of continuity using inverse image (Mobile Device)

As calculation of inverse images can take a considerable time on a mobile device the 'mobile' option below has a fixed size and uses pre-calculated images.

Definition of continuity using inverse image (Desktop)

Definition of continuity using inverse image (Mobile Device)

## Definition using sequences

The function \(f:A \rightarrow B\) is continuous if and only for every convergent sequence \(\{x_n\}\)
in the domain with limit \(a\), the image sequence \(\{f(x_n)\}\) in the codomain is convergent with limit \(f(a)\).

Definition of continuity using sequences

Definition of continuity using sequences

## Definition using '\(\epsilon-\delta\)'

The function \(f:A \rightarrow B\) is continuous at a point \(z\) in its domain if and only for every \(\epsilon\) there is
a \(\delta\) such that the image of the open ball \(B(z,\delta)\) is contained in the
open ball \(B(f(z),\epsilon)\).

The \(\epsilon-\delta\) definition of continuity.

The \(\epsilon-\delta\) definition of continuity.

The functions and sequences used are complex valued. This is simply a convenient way to
illustrate the definitions. No axes or details of position are shown as these are not relevant.
The fifth function is not continuous on the whole domain and the points selected are chosen to be
points where it is not continuous.

- \(2z+1\)
- \(4z^2+z\)
- \(z+1/z\) -- the Joukowski function
- \(1/\bar{z}\) -- inversion
- \(\sqrt{z}\)

- \(\{(0.9)^ne^{\frac{2\pi i}{n}}-1\}\)
- \(\{(-1)^n\times ((0.9)^ni)-1\}\)
- \(\{(-1)^n\times ((0.9)^ne^{\frac{n\pi i}{20}})-1\}\)
- \(\{\frac{n^3-i}{n^3+i}\}\)