The extended real number system
the extended real number system is obtained from the real number system
$\mathbb{R}$ by adding two elements $+\infty$ and $-\infty$ ,
we must note that this elements are not real numbers but rather symbols for describing
various limiting behaviours
.$\mathbb{R}$ by adding two elements $+\infty$ and $-\infty$ ,
we must note that this elements are not real numbers but rather symbols for describing
various limiting behaviours
the extended real number system is described by $\mathbb{R}$ or $[-\infty,+\infty]$ or
$\mathbb{R}\cup\{-\infty,+\infty\}$.
Examples
(1)the Cauchy Sequence:The cauchy sequence definition of real number allows
$(-\infty)$ or $(+\infty)$ as the set of all sequences of Rational numbers which
for $k>0$.
(2)Lebesgue Measure:When assigning measures to $\mathbb{R}$ that agrees with the usual
ength of intervals, this measure must be larger than any finite real number.
(3)Infinite integral: when considering infinite integrals such as $\int^\infty_{0}\frac{dx}{x}$,
here we can see that the value "infinity" arises.
$(-\infty)$ or $(+\infty)$ as the set of all sequences of Rational numbers which
for $k>0$.
(2)Lebesgue Measure:When assigning measures to $\mathbb{R}$ that agrees with the usual
ength of intervals, this measure must be larger than any finite real number.
(3)Infinite integral: when considering infinite integrals such as $\int^\infty_{0}\frac{dx}{x}$,
here we can see that the value "infinity" arises.
Counter examples
(1)the expression $\frac{1}{0}$ is not defined either as $+\infty$ or $-\infty$,
because although it is true that whenever $f(x)\rightarrow0$ for a continuous function
$f(x)$,then it must be the case that $\frac{1}{f(x)}$ is eventually contained in every neighbourhood
of the set $\{-\infty,+\infty\}$ it is not true that $\frac{1}{f(x)}$
must tend to one of these points.
hence it is not in the extended real line.
(2)Another example is the function $g(x)=\frac{1}{sin\frac{1}{x}}$
which is of the form $\frac{1}{f(x)}$ but does not tend to neither $(+\infty)$ nor $(-\infty)$
when $sin(\frac{1}{x})\rightarrow0$ for instance
$\lim_{x\rightarrow\frac{1}{\pi}}\frac{1}{sin(\frac{1}{x})}=0$
but $\lim_{x\rightarrow\frac{1}{\pi}}\frac{1}{sin\frac{1}{x}}$
does not exists because
$\lim_{x\rightarrow\frac{1}{\pi^+}}\frac{1}{sin\frac{1}{x}}=+\infty$ but
$\lim_{x\rightarrow\frac{1}{\pi^-}}\frac{1}{sin\frac{1}{x}}=-\infty$ and the modulus
$|\frac{1}{f(x)}|$ nevertheless does not approach $+\infty$.
hence the function is not on the real line.
because although it is true that whenever $f(x)\rightarrow0$ for a continuous function
$f(x)$,then it must be the case that $\frac{1}{f(x)}$ is eventually contained in every neighbourhood
of the set $\{-\infty,+\infty\}$ it is not true that $\frac{1}{f(x)}$
must tend to one of these points.
hence it is not in the extended real line.
(2)Another example is the function $g(x)=\frac{1}{sin\frac{1}{x}}$
which is of the form $\frac{1}{f(x)}$ but does not tend to neither $(+\infty)$ nor $(-\infty)$
when $sin(\frac{1}{x})\rightarrow0$ for instance
$\lim_{x\rightarrow\frac{1}{\pi}}\frac{1}{sin(\frac{1}{x})}=0$
but $\lim_{x\rightarrow\frac{1}{\pi}}\frac{1}{sin\frac{1}{x}}$
does not exists because
$\lim_{x\rightarrow\frac{1}{\pi^+}}\frac{1}{sin\frac{1}{x}}=+\infty$ but
$\lim_{x\rightarrow\frac{1}{\pi^-}}\frac{1}{sin\frac{1}{x}}=-\infty$ and the modulus
$|\frac{1}{f(x)}|$ nevertheless does not approach $+\infty$.
hence the function is not on the real line.
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