Definition:Continuity

The concept of continuity makes precise the intuitive notion that a function has no "jumps" at a given point.

Loosely speaking, in the case of a real function, continuity at a point is defined as the property that the graph of the function does not have a "break" at the point.

This concept appears throughout mathematics and correspondingly has many variations and generalizations.

The most general definition of continuity is the one in topological spaces; see below.

= Real Function =

Definition: Continuity at a point
Let $$A \subseteq \R$$ be any subset of the real numbers, and $$f: A \to \R$$ be a function.

Let $$x \in A$$ be a point of $$A$$.

We say that $$f$$ is continuous at $$x$$ when the limit of $$f(y)$$ as $$y \to x$$ exists and
 * $$\lim_{y \to x} f \left({y}\right) = f \left({x}\right)$$.

Continuity from one side
There is a related concept of continuity where one only approaches the point $$x$$ only from the right or from the left:

Definition: Continuity from the left at a point
We say that $$f$$ is continuous from the left at $$x$$ when the limit from the left of $$f(y)$$ as $$y \to x$$ exists and
 * $$\lim_{\underset{y \in A}{y \to x^-}} f(y) = f(x).$$

Definition: Continuity from the right at a point
We say that $$f$$ is continuous from the right at $$x$$ when the limit from the right of $$f(y)$$ as $$y \to x$$ exists and
 * $$\lim_{\underset{y \in A}{y \to x^+}} f(y) = f(x).$$

Definition: Continuity on a Set
Let $$A \subseteq \R$$ be any subset of the real numbers, and $$f: A \to \R$$ be a function.

We say that $$f$$ is continuous on $$A$$ if $$f$$ is continuous at every point of $$A$$.

Continuity on an Interval
This is a specific example of continuity on a set.

Let $$f$$ be a real function defined on a closed interval $$[a, b]$$.

Then $$f$$ is continuous on $$[a, b]$$ iff it is:
 * continuous at every point of $$\left({a, b}\right)$$;
 * continuous on the left at $$b$$;
 * continuous on the right at $$a$$.

That is, if $$f$$ is to be continuous over the whole of a closed interval, it needs to be continuous at the end points as well. However, because we only have "access" to the function on one side of each end point, all we can do is insist on continuity on the side of the end point that the function is defined.

Discontinuity
If a function $$f$$ is not continuous at a point $$x$$, then $$f$$ is described as being discontinuous at $$x$$.

The point $$x$$ is called a discontinuity of $$f$$.

= Metric Space =

Let $$M_1 = \left({A_1, d_1}\right)$$ and $$M_2 = \left({A_2, d_2}\right)$$ be metric spaces.

Let $$f: A_1 \to A_2$$ be a mapping from $$A_1$$ to $$A_2$$.

Let $$a \in A_1$$ be a point in $$A_1$$.

Definition using Limit
We say that $$f$$ is continuous at (the point) $$a$$ (with respect to the metrics $$d_1$$ and $$d_2$$) when the limit of $$f \left({x}\right)$$ as $$x \to a$$ exists and
 * $$\lim_{x \to a} f \left({x}\right) = f \left({a}\right)$$.

Open Set Definition
Yet another statement of this is:

$$f$$ is continuous iff: for every set $$U \subseteq M_2$$ which is open in $$M_2$$, $$f^{-1} \left({U}\right)$$ is open in $$M_1$$.

Warning
When $$f: M_1 \to M_2$$ is continuous, it does not necessarily follow that if $$U$$ is open in $$M_1$$ then $$f \left({U}\right)$$ is open in $$M_2$$.

For example, let $$f: \R^2 \to \reals$$ such that $$\forall x \in \R^2: f \left({x}\right) = 0$$.

Then $$f$$ is continuous but for any non-empty open set $$U \in M_1$$, $$f \left({U}\right) = \left\{{0}\right\}$$ which is not open in $$M_2$$.

Equivalence of Definitions
All these statements are equivalent by Equivalence of Metric Space Continuity Definitions.

If necessary, for clarity, we can say that $$f$$ is $$\left({d_1, d_2}\right)$$-continuous.

If $$f$$ is continuous in this sense for all $$a \in A_1$$, then $$f$$ is $$\left({d_1, d_2}\right)$$-continuous on $$A_1$$.

Metric Subspace
Let $$M_1 = \left({A_1, d_1}\right)$$ and $$M_2 = \left({A_2, d_2}\right)$$ be metric spaces.

Let $$f: A_1 \to A_2$$ be a mapping from $$A_1$$ to $$A_2$$.

Let $$Y \subseteq A_1$$.

By definition, $$\left({Y, d_Y}\right)$$ is a metric subspace of $$A_1$$.

Let $$a \in Y$$ be a point in $$Y$$.

Then $$f$$ is $$\left({d_Y, d_2}\right)$$-continuous iff $$\forall \epsilon > 0: \exists \delta > 0: d_Y \left({x, a_1}\right) < \delta \implies d_2 \left({f \left({x}\right), f \left({a_1}\right)}\right) < \epsilon$$.

Warning
Note that a function which is $$\left({d_Y, d_2}\right)$$-continuous might not also be $$\left({d_1, d_2}\right)$$-continuous.

For example, let $$f: \R \to \R$$ be given by:

$$f \left({x}\right) = \begin{cases} 0 & : x \in \Q \\ 1 & : x \in \R \end{cases}$$

where $$\Q$$ is the set of rational numbers.

Then $$f \restriction_{\Q}: \Q \to \R$$ is the constant function $$f_0$$ with value $$0$$, which is continuous at every point, but $$f$$ is not continuous on $$\R$$.

Complex Function
As the complex plane is a metric space, the same definition of continuity applies to complex functions as to metric spaces.

= Topological Space =

Continuous Mapping
Let $$T_1 = \left({A_1, \vartheta_1}\right)$$ and $$T_2 = \left({A_2, \vartheta_2}\right)$$ be topological spaces.

Let $$f: A_1 \to A_2$$ be a mapping from $$A_1$$ to $$A_2$$.

Then $$f$$ is continuous (with respect to the topologies $$\vartheta_1$$ and $$\vartheta_2$$) iff:
 * $$U \in \vartheta_2 \implies f^{-1} \left({U}\right) \in \vartheta_1$$.

If necessary, we can say that $$f$$ is $$\left({\vartheta_1, \vartheta_2}\right)$$-continuous.