The extended real number system
the extended real number system is obtained from the real number system
R by adding two elements +∞ and −∞ ,
we must note that this elements are not real numbers but rather symbols for describing
various limiting behaviours
.R by adding two elements +∞ and −∞ ,
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 R or [−∞,+∞] or
R∪{−∞,+∞}.
Examples
(1)the Cauchy Sequence:The cauchy sequence definition of real number allows
(−∞) or (+∞) as the set of all sequences of Rational numbers which
for k>0.
(2)Lebesgue Measure:When assigning measures to 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 ∫∞0dxx,
here we can see that the value "infinity" arises.
(−∞) or (+∞) as the set of all sequences of Rational numbers which
for k>0.
(2)Lebesgue Measure:When assigning measures to 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 ∫∞0dxx,
here we can see that the value "infinity" arises.
Counter examples
(1)the expression 10 is not defined either as +∞ or −∞,
because although it is true that whenever f(x)→0 for a continuous function
f(x),then it must be the case that 1f(x) is eventually contained in every neighbourhood
of the set {−∞,+∞} it is not true that 1f(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)=1sin1x
which is of the form 1f(x) but does not tend to neither (+∞) nor (−∞)
when sin(1x)→0 for instance
limx→1π1sin(1x)=0
but limx→1π1sin1x
does not exists because
limx→1π+1sin1x=+∞ but
limx→1π−1sin1x=−∞ and the modulus
|1f(x)| nevertheless does not approach +∞.
hence the function is not on the real line.
because although it is true that whenever f(x)→0 for a continuous function
f(x),then it must be the case that 1f(x) is eventually contained in every neighbourhood
of the set {−∞,+∞} it is not true that 1f(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)=1sin1x
which is of the form 1f(x) but does not tend to neither (+∞) nor (−∞)
when sin(1x)→0 for instance
limx→1π1sin(1x)=0
but limx→1π1sin1x
does not exists because
limx→1π+1sin1x=+∞ but
limx→1π−1sin1x=−∞ and the modulus
|1f(x)| nevertheless does not approach +∞.
hence the function is not on the real line.
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