Significance of the sequence (1/n)

Significance of the sequence (1/n)

M. Velrajan

 

We know that the sequence 1/n → 0 and it follows from Archimedean Property of Real Numbers : 

If x, y are real numbers with x > 0 then there is a positive integer n 

such that nx > y.

In fact, it follows as the particular case x = ϵ > 0 and y = 1 of the

Archimedean Property.

The sequence 1/n → 0 and some of the results that follow from it are to derive many results not only in real sequences and series but also in metric spaces and topological spaces. 

We highlight some of the significance of 1/n → 0. 

First we recall some of the important results in real sequences that follow from 1/n → 0 and then their uses in metric spaces and topological spaces.    

 

1. For all real k, the sequence k/n → 0 and hence, for any real number x, 

the sequence (x + k/n) → x . ( since | (x + k/n) - x | = |k/n| )

In general, if an → a then for all real k, the sequence (an + k/n) → a.

2. limn → ∞ (1 + 1/n)n = e = ∑ 1/n!, n varies from 0 to ∞.

Think and recall many such results in real sequences and series.   

3. If |an - a| < 1/n, for all n ∈ ℕ, then an → a.

For, 

Since for all k ≥ n, 1/k ≤ 1/n we have |ak - a| < 1/k ≤ 1/n.

i.e. |ak - a| < 1/n, for all k ≥ n

Let ϵ > 0. Since the sequence 1/n → 0, there exists n0 such that 

1/n < ϵ, for all n ≥ n0. 

Hence |ak - a| < 1/n0 < ϵ, for all k ≥ n0.

Hence an → a.

3.1. Is the converse of 3. true ?

No. For example, 

1/n → 0, but |1/n - 0| = 1/n, which is not less than 1/n, for all n ∈ ℕ. 

Interestingly 3. is obtained using 1/n → 0.

As seen in an earlier post, velrajanm.blogspot.com, Real Space to Metric Space and Normed Linear Space , replacing |an - a| by d(an, a) in the case of metric space and replacing |an - a| by ||an - a|| in the case of normed linear space, we get : 

3.2. In a metric space if for each n ∈ ℕ, d(an, a) < 1/n then an → a.

3.2.1. In a normed linear space if for each n ∈ ℕ, ||an - a|| < 1/n then an → a.

3.3. Learning : 

Recall : 

Suppose (an) is a sequence in the real space or in a metric space or in a

normed linear space. Then an → a if for every ϵ > 0 there exists a natural number m such that

an ∈ B(a, ϵ), for all n ≥ m, where B(a, ϵ) is the open ball with centre a and radius ϵ. 

i.e. an → a if for every ϵ > 0, B(a, ϵ) contains all terms of

(an) except a finite number of terms.

From 3. and 3.2. we get that, if an ∈ B(a, 1/n), the open ball with centre a and

radius 1/n, for all n ∈ ℕ then an → a.

Observation : 

‘The uncountable set {B(a, ϵ) / ϵ > 0}’ is replaced by ‘its countable subset 

{B(a, 1/n) / n ∈ ℕ}’ and ‘B(a, ϵ) contains all terms of

(an) except a finite number of terms’ by ‘an ∈ B(a, 1/n)’.

3.3.1. Note that if an ∈ B(a, 1/n) then ak ∈ B(a, 1/n), for all k ≥ n, since B(a, 1/n) ⊇ B(a, 1/(n + 1)), for all n ≥ 1.  

3.3.2. In a metric space X as a topological space, {B(a, ϵ) / ϵ > 0} is a basis at a and {B(a, 1/n) / n ∈ ℕ} is a countable basis at a and B(a, 1/n) ⊇ B(a, 1/(n + 1)), for all n ≥ 1.

Think, how 3.2. can be extended to a topological space.

Recall

4. Suppose (an) is a sequence in a topological space X, a ∈ X and 𝔅 is a basis at a. Then an → a in X if and only if for each B ∈ 𝔅 there exists m ∈ ℕ such that

an ∈ B, for all n ≥ m.

5. In a first countable topological space X, 

every a ∈ X has a countable basis {Bn / n ∈ ℕ} at a. 

But Bn need not contain Bn+1, for all n ≥ 1.

Think how to get a decreasing countable basis at a.

Yes, 

6. If {Bn / n ∈ ℕ} is a countable basis at a and 

Cn = B1 ∩ B2 ∩ . . . ∩ Bn, for all n ≥ 1 then 

{Cn / n ∈ ℕ} is also a countable basis at a and 

Cn ⊇ Cn+1, for all n ≥ 1. (Verify !)

i.e. In a first countable topological space X, 

every a ∈ X has a countable basis {Bn / n ∈ ℕ} at a such that 

Bn ⊇ Bn+1, for all n ≥ 1.

7. Suppose (an) is a sequence in a first countable topological space X, a ∈ X and {Bn / n ∈ ℕ} is a countable basis at a and Cn = B1 ∩ B2 ∩ . . . ∩ Bn, for all n ≥ 1.

If an ∈ Cn, for all n ∈ ℕ then an → a in X. 

For,

By 6., {Cn / n ∈ ℕ} is also a countable basis at a and 

Cn ⊇ Cn+1, for all n ≥ 1. 

Let n ∈ ℕ. Since Cn ⊇ Cn+1, for all n ≥ 1 and ak ∈ Ck, for all k ∈ ℕ we have

ak ∈ Cn, for all k ≥ n. Hence, By 4., an → a in X. 

8. For any real number x, there exist a countable sequence of rational numbers and 

an uncountable sequence of irrational numbers that converges to x.  

For,

Let x be a real number.

Then [nx] ≤ nx < [nx] + 1, for all n ∈ ℕ, 

where [x] is the largest integer less than or equal to x.

Hence [nx]/n ≤ x < [nx]/n + 1/n, for all n ∈ ℕ.

i.e. |[nx]/n - x| < 1/n, for all n ∈ ℕ.

By 3., [nx]/n → x, and by 1., ([nx] + a)/n → x, for any real number a.

Note that, if a is rational (or irrational) then 

([nx] + a)/n is also rational (or irrational).

Hence, for any rational(or irrational) a, the sequence ([nx] + a)/n of rational

(or irrational) numbers converge to  x. 

In particular, for any rational (or irrational) a, the sequence 

([10nx] + a)/10n of rational (or irrational) numbers converges to  x, 

since ([10nx] + a)/10n is a subsequence of ([nx] + a)/n.  

Also since ([n(-x)] + a)/n → - x, - ([n(-x)] + a)/n → x and 

in particular, - ([10n(-x)] + a)/10n → x.

Thus for any real number x, we have a countable sequence of rational numbers

and an uncountable sequence of irrational numbers that converges to x.

9. If S is a nonempty subset of the real space ℝ,

and a = Sup S then there exists a sequence (an) in S such that an → a.

For,

Suppose S is bounded above. 

Then a ∈ ℝ and for any ϵ > 0, a - ϵ < a, 

hence a - ϵ is not an upper bound of S and hence 

there exists s ∈ S such that a - ϵ < s ≤ a.

In particular, there exists an ∈ S such that a - 1/n < an ≤ a 

i.e.  such that |an - a| < 1/n, for all n ∈ ℕ.

By 3., an → a.

Suppose S is not bounded above. 

Then a = ∞. Hence for each n ∈ ℕ, there exists an ∈ S such that an > n 

and hence an → ∞ = a.

9.1. If S is a nonempty subset of the real space ℝ,

and b = Inf S then there exists a sequence (bn) in S such that bn → b.

(Can be proved by replacing above by below, a - ϵ by a + ϵ, < by >,

upper by lower, etc.) 

10. Suppose A is a nonempty subset of a metric space X.

If x ∈ Cl A, the closure of A then there exists a sequence (an) in A 

such that an → x in X.

For,

Since x ∈ Cl A, for every ϵ > 0 the open ball B(x, ϵ) intersects A.

In particular, for each n ∈ ℕ, the open ball B(x, 1/n) intersects A

i.e. there exists an ∈ A with d(an, x) < 1/n.

By 3.2., an → x in X.

10.1. Suppose A is a nonempty subset of a topological space X.

If x ∈ Cl A, the closure of A then there need not exists a sequence (an) in A 

such that an → x in X. (Think to get an example)

But the converse is true, 

i.e. Suppose A is a nonempty subset of a topological space X.

If there exists a sequences (an) in A such that an → x in X 

then x ∈ Cl A, the closure of A.

10.2. Suppose A is a nonempty subset of a first countable topological space X.

If x ∈ Cl A, the closure of A then there exists a sequence (an) in A 

such that an → x in X.

For, 

Suppose {Bn / n ∈ ℕ} is a countable basis at x and 

Cn = B1 ∩ B2 ∩ . . . ∩ Bn, for all n ≥ 1.

Since x ∈ Cl A, Cn intersects A.

Let an ∈ Cn ∩ A, for all n ∈ ℕ.

Then (an) is a sequence in A and an ∈ Cn, for all n ∈ ℕ.

Hence, By 7., an → x in X.  

11. Suppose X and Y are metric spaces and f : X → Y be a function.

If a ∈ X and for all sequences an → a in X,  f(an) → f(a) in Y then 

f is continuous at a.

For,

(We cannot proceed from the given hypothesis, so we prove contrapositively.)

Suppose f is not continuous at a.

Then there exists ϵ > 0 such that, for all δ > 0, there exists 

x ∈ X with d(x, a) < δ but d(f(x), f(a)) ≥ ϵ.

(Note how to write negation, by replacing ‘for all’ by ‘there exists’,

‘there exists’ by ‘for all’, ‘<’ by ‘≥’, etc)  

In particular, for each n ∈ ℕ, there exists 

xn ∈ X with d(xn, a) < 1/n but d(f(xn), f(a)) ≥ ϵ. 

Since for each n ∈ ℕ, d(xn, a) < 1/n,

By 3.2., xn → a,

but since for each n ∈ ℕ, d(f(xn), f(a)) ≥ ϵ, f(xn) ⇸ f(a).

Which contradicts the hypothesis, 

for all sequences an → a in X,  f(an) → f(a) in Y.

Hence f is continuous at a. 

11.1. Note that, if X and Y are topological spaces and f : X → Y be a function.

If a ∈ X and for all sequences an → a in X,  f(an) → f(a) in Y then 

f need not be continuous at a. (Think to get an example)

But the converse is true, 

i.e. if X and Y are topological spaces and f : X → Y is continuous at a 

then for all sequences an → a in X,  f(an) → f(a) in Y.

11.2.  If X and Y are topological spaces, X is first countable, a ∈ X 

and f : X → Y be a function.

If for all sequences an → a in X,  f(an) → f(a) in Y then f is continuous at a.

For,

Suppose {Bn / n ∈ ℕ} is a countable basis at a and 

Cn = B1 ∩ B2 ∩ . . . ∩ Bn, for all n ≥ 1.

(We cannot proceed from the given hypothesis, so we prove contrapositively.)

Suppose f is not continuous at a.

Then there exists an open set V in Y such that f(a) ∈ V and f - 1(V) is not open in X.

(Note how negation is written, by replacing ‘for all’ by ‘there exists’,

‘open in X’ by ‘not open in X’)  

Since a ∈ f - 1(V) and {Cn / n ∈ ℕ} is a basis at a, Cn ⊈ f - 1(V) for all n ≥ 1.

(if Cn ⊆ f - 1(V) for some n ≥ 1 then f - 1(V) is open in X)

Hence for each n ≥ 1, there exists xn ∈ Cn such that xn ∉ f - 1(V) 

Since xn ∈ Cn, for all n ≥ 1,  By 7., xn → a in X.

Since f(xn) ∉ V, for all n ≥ 1 and V is a neighbourhood of f(a), f(xn) ⇸ f(a).

Which contradicts the hypothesis, 

for all sequences an → a in X,  f(an) → f(a) in Y.

Hence f is continuous at a. 

12. Arbitrary intersections of open sets need not be open.

For example, for each n ∈ ℕ, (- 1/n, 1/n) is open in ℝ, but their intersection is {0}

(Verify and see how 1/n → 0 helps), which is not open in in ℝ.

And for each n ∈ ℕ, (- n, n) is open in ℝ,

and their intersection is (-1, 1)., which is also open in in ℝ.

Think and recall many such results in metric spaces and topological spaces

that follow from 1/n → 0 and its consequences.

Note that 3.2. is a consequence of 1/n → 0 and how the learning and mathematical

thinking lead to 7., an extension of 3.2. to a topological space.

Also lead to 10.2. and 11.2. as consequences of 7., extending 10. and 11.,

which are consequences of 3.2., to a topological space.

This is a way of learning Mathematics.

If you learn in such ways you can Enjoy Mathematics and Crack any Competitive Exam. 

So, to succeed, change your way of learning.

To download PDF Click the link

Significance of the sequence (1/n)