# Product of discontinuous functions

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Let $$\displaystyle f:\mathbb{R} \to \mathbb{R}$$ and $$\displaystyle g:\mathbb{R} \to \mathbb{R}$$ be discontinuous at a point $$\displaystyle c$$ . Give an example of a function $$\displaystyle h(x)=f(x)g(x)$$ such that $$\displaystyle h$$ is continuous at c.
$$\displaystyle f(x) = \begin{cases} 0 & \text{if } x \in \mathbb{Q} \\ 1 & \text{if } x \in \mathbb{R}-\mathbb{Q} \end{cases}$$

$$\displaystyle g(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \in \mathbb{R}-\mathbb{Q} \end{cases}$$

$$\displaystyle f,g$$ are continous nowhere but $$\displaystyle h(x)=0 \,\,\, \, \forall \,\, x \in \mathbb{R}$$.

What other examples you might think of ?

#### Sudharaka

##### Well-known member
MHB Math Helper
Let $$\displaystyle f:\mathbb{R} \to \mathbb{R}$$ and $$\displaystyle g:\mathbb{R} \to \mathbb{R}$$ be discontinuous at a point $$\displaystyle c$$ . Give an example of a function $$\displaystyle h(x)=f(x)g(x)$$ such that $$\displaystyle h$$ is continuous at c.
$$\displaystyle f(x) = \begin{cases} 0 & \text{if } x \in \mathbb{Q} \\ 1 & \text{if } x \in \mathbb{R}-\mathbb{Q} \end{cases}$$

$$\displaystyle g(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \in \mathbb{R}-\mathbb{Q} \end{cases}$$

$$\displaystyle f,g$$ are continous nowhere but $$\displaystyle h(x)=0 \,\,\, \, \forall \,\, x \in \mathbb{R}$$.

What other examples you might think of ?
Hi Zaid,

How about the set for functions, $$\{f,\,g\}$$ such that,

$f(x)=\begin{cases}a & \text{if } x \geq c \\b & \text{if } x < c\end{cases}$

$g(x)=\begin{cases}b & \text{if } x \geq c \\a & \text{if } x < c\end{cases}$

where $$a,\,b,\,c\in \Re$$ and $$a\neq b$$.

#### Deveno

##### Well-known member
MHB Math Scholar
The most general, and all-encompassing example I can think of, off the top of my head:

Let $a \neq b$ and define:

$f(x) = a,\ x \neq c$
$f(c) = b$

$g(x) = b,\ x \neq c$
$g(c) = a$

Clearly, neither $f$ nor $g$ is continuous at $c$, as can be proved straight from the definition (use an $0 < \epsilon < |b - a|$).

Just as clearly:

$fg(x) = ab,\ \forall x \in \Bbb R$, which is clearly continuous.

One can construct more "extravagant" examples, but the important part is that $a \neq b$, and that $f$ and $g$ "complement" each other. In fact, there is nothing special about the partition of $\Bbb R$ into the two disjoint sets $\{c\}$ and $\Bbb R - \{c\}$, you can use any partition (such as the Dedekind cut example Sudharaka gives, or the partition into the rationals and irrationals).

To me, this underscores the fact that continuity (of a function) is dependent on the DOMAIN OF DEFINITION of said function, not just the "rule itself" of said function.

In other words, a "continuous function" doesn't really MEAN anything, what IS meaningful is: a function continuous at all points of a set $A$. The underlying domain is important. Context is everything: a function that is perfectly continuous on the real numbers may suddenly spectacularly fail to be so on the complex numbers, for example (as is the case with:

$f(x) = \dfrac{1}{1 + x^2}$).

#### Amer

##### Active member
Characteristic function of A
$$\chi_A : \mathbb{R} \rightarrow \{0,1\}$$
$$\chi_A =\left\{ \begin{array}{lr} 1 &,x\in A \\ 0 &,x\in A^{c} \end{array} \right.$$
$$\chi_{A^{c}} = \left\{ \begin{array}{ir} 0 & , x\in A \\ 1 & , x\in A^{c} \end{array} \right.$$
Their product is zero function which is continuous