Definition:Continuity

= Real Function =

Continuity at a Point
Let $$f$$ be a real function defined on an open interval $$\left({a \, . \, . \, b}\right)$$.

Let $$c \in \left({a \, . \, . \, b}\right)$$.

Then $$f$$ is continuous at (the point) $$c$$ iff $$\lim_{x \to c} f \left({x}\right) = f \left({c}\right)$$.

Loosely speaking, this means that continuity at a point is defined as the graph of $$f$$ not having a "break" at $$c$$.

Note that from the definition of a limit of a function, it is necessary that $$f \left({c}\right)$$ is both:
 * the limit from the left, and
 * the limit from the right

of $$f \left({x}\right)$$ as $$x$$ tends to $$c$$ from both above $$c$$ and below $$c$$.

Continuous on the Left
Let $$f$$ be a real function defined on a half open interval $$\left({a \, . \, . \, b}\right]$$.

Let $$\lim_{x \to b^-} f \left({x}\right) = f \left({b}\right)$$.

Then $$f$$ is continuous on the left at the point $$b$$

Continuous on the Right
Let $$f$$ be a real function defined on a half open interval $$\left[{a \,. \, . \, b}\right)$$.

Let $$\lim_{x \to a^+} f \left({x}\right) = f \left({a}\right)$$.

Then $$f$$ is continuous on the right at the point $$a$$

Open Interval
Let $$f$$ be a real function defined on an open interval $$\left({a \, . \, . \, b}\right)$$.

Then $$f$$ is continuous on $$\left({a \, . \, . \, b}\right)$$ iff it is continuous at every point of $$\left({a \, . \, . \, b}\right)$$.

Closed Interval
Let $$f$$ be a real function defined on a closed interval $$\left[{a \,. \, . \, b}\right]$$.

Then $$f$$ is ''continuous on $$\left[{a \,. \, . \, b}\right]$$'' 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.

Half Open Intervals
Similar definitions apply to half open intervals.

Let $$f$$ be a real function defined on a half open interval $$\left({a \, . \, . \, b}\right]$$.

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

Let $$f$$ be a real function defined on a half open interval $$\left[{a \,. \, . \, b}\right)$$.

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

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 $$A_1$$.

Epsilon-Delta Definition
Then $$f$$ is continuous at (the point) $$a$$ (with respect to the metrics $$d_1$$ and $$d_2$$) iff:

$$\forall \epsilon > 0: \exists \delta > 0: d_1 \left({x, a_1}\right) < \delta \Longrightarrow d_2 \left({f \left({x}\right), f \left({a_1}\right)}\right) < \epsilon$$.

Epsilon-Neighborhood Definition
An alternative statement of this is as follows:

$$f$$ is continuous at (the point) $$a$$ (with respect to the metrics $$d_1$$ and $$d_2$$) iff:

$$\forall N_\epsilon \left({f \left({a}\right)}\right): \exists N_\delta \left({a}\right): f \left({ N_\delta \left({a}\right)}\right) \subseteq N_\epsilon \left({f \left({a}\right)}\right)$$.

where $$N_\epsilon \left({a}\right)$$ is the $\epsilon$-neighborhood of $$a$$ in $$M_1$$.

That is, for every $$\epsilon$$-neighborhood of $$f \left({a}\right)$$ in $$M_2$$, there exists a $$\delta$$-neighborhood of $$a$$ in $$M_1$$ whose image is a subset of that $$\epsilon$$-neighborhood.

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

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 $$\forall \epsilon > 0: \exists \delta > 0: d_Y \left({x, a_1}\right) < \delta \Longrightarrow d_2 \left({f \left({x}\right), f \left({a_1}\right)}\right) < \epsilon$$.

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

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

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

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

= Topological Space =

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 $$U \in \vartheta_2 \Longrightarrow f^{-1} \left({U}\right) \in \vartheta_1$$.

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