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Product of discontinuous functions

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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
Feb 5, 2012
1,621
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
Feb 15, 2012
1,967
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
Mar 1, 2012
275
Characteristic function of A
[tex]\chi_A : \mathbb{R} \rightarrow \{0,1\} [/tex]
[tex]\chi_A =\left\{ \begin{array}{lr} 1 &,x\in A \\ 0 &,x\in A^{c} \end{array} \right.[/tex]
[tex]\chi_{A^{c}} = \left\{ \begin{array}{ir} 0 & , x\in A \\ 1 & , x\in A^{c} \end{array} \right. [/tex]
Their product is zero function which is continuous