Matric space full bsc notes for third year in easy language in English.
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Course
121212
Institution
Kalyani
The BSc third year 6th-semester syllabus for metric spaces typically covers concepts such as distance functions, open and closed sets, convergence, continuity, compactness, completeness, and topological properties in Euclidean and abstract spaces. It also includes the study of metric space properti...
, SEMESTER-VI
Course: MATH-H-CC-T-13
Metric space
Unit 1. Metric spaces: sequences in metric spaces, Cauchy sequences.
Complete metric spaces, Cantor’s theorem.
Convergence and Completeness:-
Sequence: Let (X,d) be a metric space. If for each 𝑛 ∈ 𝑁 we have an element
𝑥𝑛 (say) in X then the set {𝑥𝑛 } is called a sequence in X.
An element 𝑙 ∈ 𝑋 is said to be limit of a sequence {𝑥𝑛 } if for each 𝜀 > 0 we have an
integer 𝑚 > 0 such that 𝑥𝑛 ∈ 𝑆𝜀 (𝑙) ∀𝑛 ≥ 𝑚 i.e. 𝑑(𝑥𝑛 , 𝑙) < 𝜀 ∀𝑛 ≥ 𝑚.
A sequence {𝑥𝑛 } in a metric space is said to be convergent if it has a limit.
Q. Prove that a convergence sequence in a metric space cannot have more than one limit.
Let, {𝑥𝑛 } be a convergent sequence in the metric space (X,d).
If possible let 𝑙1 , 𝑙2 be two different limits of the convergent sequence {𝑥𝑛 }.
Since 𝑥𝑛 → 𝑙1 as 𝑛 → ∞, ∴ for a given 𝜀 > 0 ∃ a positive integer 𝑛0′ such that
𝜀
𝑑(𝑥𝑛 , 𝑙1 ) < ∀𝑛 ≥ 𝑛0′
2
Again since 𝑥𝑛 → 𝑙2 as 𝑛 → ∞, ∴ for a given 𝜀 > 0 ∃ a positive integer 𝑛0′′ such that
𝜀
𝑑(𝑥𝑛 , 𝑙2 ) < ∀𝑛 ≥ 𝑛0′′
2
′ ′′
Let 𝑛0 = max(𝑛0 , 𝑛0 ) then,
𝜀 𝜀
𝑑(𝑥𝑛 , 𝑙1 ) < 𝑎𝑛𝑑 𝑑(𝑥𝑛 , 𝑙2 ) < ∀𝑛 ≥ 𝑛0
2 2
𝜀 𝜀
Now, 𝑑(𝑙1 , 𝑙2 ) ≤ 𝑑(𝑙1 , 𝑥𝑛 ) + 𝑑(𝑥𝑛 , 𝑙2 ) = 𝑑(𝑥𝑛 , 𝑙1 ) + 𝑑(𝑥𝑛 , 𝑙2 ) < + = 𝜀 ∀𝑛 ≥ 𝑛0
2 2
∴ 𝑑(𝑙1 , 𝑙2 ) < 𝜀 ∀𝑛 ≥ 𝑛0
Since 𝑙1 and 𝑙2 are two fixed numbers and 𝜀 being an arbitrary positive number
then the only possibility is 𝑑(𝑙1 , 𝑙2 ) = 0 𝑖. 𝑒. 𝑙1 = 𝑙2 .
Thus the convergent sequence cannot have more than one limit.
Cauchy Sequence:- A sequence in a metric space is said to be a Cauchy sequence if
for each 𝜀(> 0), ∃ a positive integer 𝑛0 such that,
𝑑(𝑥𝑚 , 𝑥𝑛 ) < 𝜀 ∀𝑚, 𝑛 ≥ 𝑛0
Q. Prove that every convergent sequence in a metric space is Cauchy but the converse
may not be true.
Let {𝑥𝑛 } be a convergent sequence in a metric space (X,d) and let it converges to
x.
Thus for a given 𝜀 > 0 ∃ a positive integer 𝑛0 such that
1
, 𝜀
𝑑(𝑥𝑛 , 𝑥) <∀𝑛 ≥ 𝑛0
2
Let m,n be two positive integers such that 𝑚, 𝑛 ≥ 𝑛0
𝜀 𝜀
∴ 𝑑(𝑥𝑚 , 𝑥𝑛 ) ≤ 𝑑(𝑥𝑚 , 𝑥) + 𝑑(𝑥, 𝑥𝑛 ) = 𝑑(𝑥𝑚 , 𝑥) + 𝑑(𝑥𝑛 , 𝑥) < + =𝜀
2 2
Thus
𝑑(𝑥𝑚 , 𝑥𝑛 ) < 𝜀 ∀𝑚, 𝑛 ≥ 𝑛0
Hence {𝑥𝑛 } is a Cauchy Sequence.
The converse of the above theorem is not always true. For example let us
consider the subspace 𝑋 = (0,1] with distance function 𝑑(𝑥, 𝑦) = |𝑥 − 𝑦|.
1
Let us consider a sequence {𝑥𝑛 } = { : 𝑛 ∈ 𝑁} of x which is Cauchy sequence in
𝑛
the space but it is not convergent in X, since the point 0 is not a point of the space.
Q. Prove that every Cauchy sequence in a metric space is bounded.
Let {𝑥𝑛 } be a Cauchy sequence in a metric space (X,d). Then for each 𝜀(> 0), ∃ a
positive integer 𝑛0 such that,
𝑑(𝑥𝑚 , 𝑥𝑛 ) < 𝜀 ∀𝑚, 𝑛 ≥ 𝑛0
Let us choose 𝜀 = 1. Then, 𝑑(𝑥𝑚 , 𝑥𝑛 ) < 1 ∀𝑚, 𝑛 ≥ 𝑛0
Now the range set of the sequence {𝑥𝑛 } is {𝑥1 , 𝑥2 , … , 𝑥𝑛0 } ∪ {𝑥𝑛0 +1 , 𝑥𝑛0+2 , … }.
Since the set {𝑥1 , 𝑥2 , … , 𝑥𝑛0 } is finite hence it is bounded and the set
{𝑥𝑛0+1 , 𝑥𝑛0 +2 , … } is a set whose diameter is <1. Hence it is bounded.
Thus the range set {𝑥𝑛 } i.e. the set {𝑥1 , 𝑥2 , … } is the union of two bounded set.
Hence it is bounded.
Thus the Cauchy sequence is bounded.
Q. Prove that a Cauchy sequence in a metric space (X,d) is convergent if and only if it has
a convergent subsequence(the limit being the same in both cases).
Clearly, if a sequence (whether Cauchy or not) converges to a point x, then every
subsequence of it also converges to the same point x.
Conversely, Since, {𝑥𝑛𝑘 } converges to 𝑥 ∈ 𝑋, then for each 𝜀 > 0, ∃ a positive
𝑘
integer 𝑛𝜀1 such that
𝜀
𝑑(𝑥𝑛𝑘 , 𝑥) < ∀𝑛 ≥ 𝑛𝜀1 … … … … … … … (1)
2
Also {𝑥𝑛 }𝑛 being a Cauchy sequence, for the same 𝜀 > 0, ∃ a positive integer 𝑛𝜀2
such that
𝜀
𝑑(𝑥𝑚 , 𝑥𝑛 ) < ∀𝑚, 𝑛 ≥ 𝑛𝜀2 … … … … … … … (2)
2
Now if 𝑛𝜀 = max{𝑛𝜀1 , 𝑛𝜀2 }, then for all 𝑛 > 𝑛𝜀 , from (1) and (2) we can write,
𝜀 𝜀
𝑑(𝑥𝑛 , 𝑥) ≤ 𝑑(𝑥𝑛 , 𝑥𝑛𝑘 ) + 𝑑(𝑥𝑛𝑘 , 𝑥) < + = 𝜀
2 2
2
, ∴ lim 𝑥𝑛 = 𝑥
𝑛→∞
Observations: There should not be any confusion with the two terms – the limit of a
sequence to which a sequence converges; and a limit point of the range set of a
sequence. A sequence may have limit but its range set may not have a limit point. For
example, the constant sequence {𝑥𝑛 }, where 𝑥𝑛 = 2 ∀𝑛, i.e., the sequence consists of
{2,2,2,2, … … }. This sequence has the limit 2 the range set {2} being finite set of real line
does not have a limit point.
Q. prove that if a convergent sequence in a metric space has infinitely many distinct
points, the limit of the convergent sequence is a limit point of the range set.
Let {𝑥𝑛 } be a convergent sequence with the limit x in a metric space (X,d) having
a range set consisting of infinitely many distinct points of X.
We have to prove that x is a limit point of the range set of {𝑥𝑛 }. If possible let x is
not a limit point of the range set, then there exists an open ball 𝐵(𝑥, 𝜀) centred at x,
which contains no point of the range set, other than possibly the point x.
But lim 𝑥𝑛 = 𝑥, hence, the open ball 𝐵(𝑥, 𝜀) must contain all the points 𝑥𝑛 of the
𝑛→∞
sequence ∀𝑛 ≥ 𝑛𝜀 and thus all such points must coincide with x. From this we are led to
the conclusion that there are only finitely many distinct points in the range set;
contradicting the hypothesis that the range set is infinite.
Hence, x is a limit point of the range set.
Q. What are the Cauchy sequence in a discrete metric space?
Let {𝑥𝑛 } be a Cauchy sequence. Then for every 𝜀 > 0 ∃𝑛0 ∈ 𝑁 such that if 𝑚, 𝑛 ≥
𝑛0 then we have 𝑑(𝑥𝑚 , 𝑥𝑛 ) < 𝜀.
1
Now take 𝜀 = then there exist some 𝑛0 ∈ 𝑁 such that
2
1
𝑑(𝑥𝑚 , 𝑥𝑛 ) < 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑚, 𝑛 ≥ 𝑛0
2
Now, since d is the discrete metric, the only possibility is {𝑥𝑛 } is a constant
sequence.
Thus only constant sequences are the Cauchy sequence in a discrete metric space.
Complete metric space:- A metric space (X,d) is said to be a complete metric space
if every Cauchy sequence in it is convergent to a point in X.
As for example the real line with usual metric is a complete metric space.
Any metric space which is not complete can be made complete metric space by
suitably adjoining additional points. As for example if we choose 𝑋 = (0,1] and
1
{𝑥𝑛 } = { } for all 𝑛 ∈ 𝑁 then {𝑥𝑛 } is a Cauchy sequence in X but {𝑥𝑛 } converges to
𝑛
0 which is not a point of X. hence X is not a complete metric space. If we add the
point 0 to X then X will become a complete metric space.
3
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