COMPLETE METRIC SPACE #complete #metric $space
COMPLETE METRIC SPACE :
A metric space X with the metric d is called a complete metric space
if every cauchy sequence in x converges in X. ∴
For Example : 1. consider the metric space (0,1) and {1/n} is a cauchy sequence in (0,1).
We know that the sequence {1/n} converges to 0.
since this limit point 0 does not lies in (0,1) , (0,1) is not a complete metric space.
Instead of (0,1) if we consider the metric space [0,1], the above mensioned limit
point 0 lies in [0,1] and hence [0,1] is a complete metric space.
2. The real line is a complete metric space.
3. The complex plane is also complete which is explained in the following argument .
Let { zn } , where zn = an + bn , be a Cauchy sequence of complex numbers. Then { an } and { bn } are themselves Cauchy sequences of real numbers.
Since the real line is complete and { an } and { bn } are Cauchy sequences in the real line R, the sequences { an } and { bn } and convergent to the real numbers a and b (Say ) i.e. an → a and bn → b.
Since an → a, for a given ꜫ > 0,
there is a positive integer m1
such that | an – a | < ꜫ/2 for all n ≥ m1
Since bn → b,
for a given ꜫ
> 0, there is a positive integer m2
such that | bn – b | < ꜫ/2 for all n ≥ m2
Put
m= max { m1, m2 }
∴ | an
– a | < ꜫ/2 and | bn – b | < ꜫ/2 for all n ≥ m
Put z= a+ib.
Now |zn – z | = | ( an +ibn
) – (a+ib) |
= | (an -a) + i
( bn -b) |
≤
| an – a | + |bn – b |
< ꜫ/2 + ꜫ/2 = ꜫ.
∴
| zn – z | < ꜫ for all n ≥ M
Thus { zn
} converges to z in the complex plane C.
Hence the complex plane is a complete metric space.
A convergent sequence is not convergent " on its own ; it must converge to some point in the space.
* why | -2 | = 2 ? original meaning
*4th dimensional object HYPERCUBE
#complete #metric #space #real line #complex #plane #convergence #of #a #sequence

Nice
ReplyDelete