Definition:Continuous Real Function

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This page is about Continuous Mapping in the context of Real Analysis. For other uses, see Continuous Mapping.

Definition

Informal Definition

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

Loosely speaking, a real function $f$ is continuous at a point $p$ if and only if the graph of $f$ does not have a "break" at $p$.


Continuity at a Point

$f$ is continuous at $x$ if and only if the limit $\ds \lim_{y \mathop \to x} \map f y$ exists and:

$\ds \lim_{y \mathop \to x} \map f y = \map f x$


Continuous Everywhere

Let $f: \R \to \R$ be a real function.


Then $f$ is everywhere continuous if and only if $f$ is continuous at every point in $\R$.


Continuity on a Subset of Domain

Let $A \subseteq \R$ be any subset of the real numbers.

Let $f: A \to \R$ be a real function.


Then $f$ is continuous on $A$ if and only if $f$ is continuous at every point of $A$.


Class of Continuous Functions

The class of continuous real functions is often denoted $C$.

Hence:

$\map f x \in C$ at $a$

if and only if:

$\ds \lim_{x \mathop \to a} \map f x = \map f a$


Continuity from One Side

Let $A \subseteq \R$ be an open subset of the real numbers $\R$.

Let $f: A \to \R$ be a real function.


Continuity from the Left at a Point

Let $x_0 \in A$.

Then $f$ is said to be left-continuous at $x_0$ if and only if the limit from the left of $\map f x$ as $x \to x_0$ exists and:

$\ds \lim_{\substack {x \mathop \to x_0^- \\ x_0 \mathop \in A} } \map f x = \map f {x_0}$

where $\ds \lim_{x \mathop \to x_0^-}$ is a limit from the left.


Continuity from the Right at a Point

Let $x_0 \in S$.

Then $f$ is said to be right-continuous at $x_0$ if and only if the limit from the right of $\map f x$ as $x \to x_0$ exists and:

$\ds \lim_{\substack {x \mathop \to x_0^+ \\ x_0 \mathop \in A}} \map f x = \map f {x_0}$

where $\ds \lim_{x \mathop \to x_0^+}$ is a limit from the right.


Continuity on an Interval

Where $A$ is a real interval, it is considered as a specific example of continuity on a subset of the domain.

Open Interval

This is a straightforward application of continuity on a set.

Let $f$ be a real function defined on an open interval $\openint a b$.


Then $f$ is continuous on $\openint a b$ if and only if it is continuous at every point of $\openint a b$.


Closed Interval

Let $f$ be a real function defined on a closed interval $\closedint a b$.


$f$ is continuous on $\closedint a b$ if and only if it is:

$(1): \quad$ continuous at every point of the open interval $\openint a b$
$(2): \quad$ continuous on the right at $a$
$(3): \quad$ continuous on the left at $b$.

That is, if $f$ is to be continuous over the whole of a closed interval, it needs to be continuous at the end points.

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 points on which 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 $\hointl a b$.

Then $f$ is continuous on $\hointl a b$ if and only if it is:

$(1): \quad$ continuous at every point of $\openint a b$
$(2): \quad$ continuous on the left at $b$.


Let $f$ be a real function defined on a half open interval $\hointr a b$.

Then $f$ is continuous on $\hointr a b$ if and only if it is:

$(1): \quad$ continuous at every point of $\openint a b$
$(2): \quad$ continuous on the right at $a$.


Examples

Example: $\sqrt x$ at $x = 1$

Let $f: \R_{\ge 0} \to \R$ be the real function defined as:

$\map f x = \sqrt x$

Then $\map f x$ is continuous at $x = 1$.


Example: $\dfrac {\sin x} x$ with $1$ at $x = 0$

Let $f: \R_{\ge 0} \to \R$ be the real function defined as:

$\map f x = \begin {cases} \dfrac {\sin x} x & : x \ne 0 \\ 1 & : x = 0 \end {cases}$

Then $\map f x$ is continuous at $x = 0$.


Example: $\map \sin {\dfrac 1 x}$ with $0$ at $x = 0$

Let $f: \R_{\ge 0} \to \R$ be the real function defined as:

$\map f x = \begin {cases} \map \sin {\dfrac 1 x} x & : x \ne 0 \\ 0 & : x = 0 \end {cases}$

Then $\map f x$ is not continuous at $x = 0$.


Also see

  • Results about continuous real functions can be found here.


Generalizations


Historical Note

The concept of a continuous real function was pioneered by the work of Carl Friedrich Gauss, Niels Henrik Abel‎ and Augustin Louis Cauchy.


Sources