# Definition:Continuous Real-Valued Vector Function

## Definition

Let $\R^n$ be the cartesian $n$-space.

Let $f: \R^n \to \R$ be a real-valued function on $\R^n$.

Then $f$ is continuous on $\R^n$ if and only if:

$\forall a \in \R^n: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R^n: \map d {x, a} < \delta \implies \size {\map f x - \map f a} < \epsilon$

where $\map d {x, a}$ is the distance function on $\R^n$:

$\ds d: \R^n \to \R: \map d {x, y} := \sqrt {\sum_{i \mathop = 1}^n \paren {x_i - y_i}^2}$

where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n}$ are general elements of $\R^n$.

## Examples

### Non-Continuous Example 1

Let $f: \R^2 \to \R$ be the real $2$-variable function defined as:

$\forall \tuple {x_1, x_2} \in \R^2: \map f {x_1, x_2} = \begin {cases} 0 & : \tuple {x_1, x_2} = \tuple {0, 0} \\ \dfrac {x_1 x_2} {x_1^2 + x_2^2} & : \text {otherwise} \end {cases}$

Then the restrictions of $f$:

$f_{\restriction \R \times \set 0}$
$f_{\restriction \set 0 \times \R}$

are both constant functions with value $0$ for all arguments.

Hence both are continuous at $\tuple {0, 0}$.

But $f$ is not continuous at $\tuple {0, 0}$.

### Non-Continuous Example 2

Let $f: \R^2 \to \R$ be the real $2$-variable function defined as:

$\forall \tuple {x, y} \in \R^2: \map f {x, y} = \begin {cases} 0 & : y = 0 \\ \dfrac {x^2} y \end {cases}$

Then the restrictions of $f$:

$f_{\restriction \tuple {x, y} \in \R^2: y = m x}$

is continuous.

But $f$ is not continuous at $\tuple {0, 0}$.