Subsuccessions

Subsucessions arise from extracting new sequences, whose terms are from the original sequence and in the same order

That is, we take infinite terms, skipping some, but without going back

For example, given the succession:

s_1, s_2, s_3, s_4, s_5, s_6, s_7, s_8, s_9, \cdots, s_n, \cdots

And now we are left with the terms that occupy the odd position:

s_1, s_3, s_5, s_7, s_9, \cdots, s_{2 \cdot n + 1}, \cdots

And now we are left with the terms that occupy the even position:

s_2, s_4, s_6, s_8, s_{10}, \cdots, s_{2 \cdot n}, \cdots

Both the odd and even sequence are subsuccessions of our initial succession

Many different ways can be devised to extract successions from the initial sequence with this procedure.

Dividing the initial sequence into subsucessions, allows us to demonstrate properties of the theory of real functions of real variables, in a simpler way than if we would do it directly on the function

Definition of subsuccession

Given a succession (s_n), it is said that another succession (t_n) is a subsuccession of (s_n) if there is a function \varphi | \mathbb{N} \to \mathbb{N} strictly increasing, that is:

\varphi(1) < \varphi(2) < \varphi(3) < \cdots < \varphi(n) < \varphi(n + 1) < \cdots, \forall n \in \mathbb{N} | t_n = s_{\varphi(n)}

From the definition of limit, it is easy to verify that if a succession is convergent, any subsucession of its will be convergent and will have the same limit

Examples

• It n_0 \in \mathbb{N} | \varphi(n) = n + n_0 and continuing with the initial succession of the example, the subsuccession is obtained:

  s_{n_0 + 1}, s_{n_0 + 2}, s_{n_0 + 3} + s_{n_0 + 4} + s_{n_0 + 5} + s_{n_0 + 6} + s_{n_0 + 7} + s_{n_0 + 8} + s_{n_0 + 9} + \cdots + s_{n_0 + n}, \cdots

  What is obtained from the initial by suppressing the n_0 terms

• The nth term succession t_n = 4 \cdot n^{2} is a subsuccession of the nth term s_n = (-1)^n \cdot n^2, how can we check if we take \varphi(n) = 2\cdot n

• The succession (1, \frac{1}{3}, \frac{1}{2}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}, \frac{1}{8}, \frac{1}{9}, \cdots, \frac{1}{n}, \cdots) is not a subsuccession of \left ( \frac{1}{n} \right )^\infty_{n=1}, since although they have the same terms, they do not have the same order

  The succession (1, 0, \frac{1}{3}, 0, \frac{1}{5}, 0, \frac{1}{7}, 0, , \cdots, \frac{1 + (-1)^{n + 1})}{2\cdot n}, \cdots) nor is it a subsuccession of \left ( \frac{1}{n} \right )^\infty_{n=1}, since although they have the same terms, it alternates between values ​​with 0, breaking the order

• Every subsuccession is a subsuccession of itself (reflective property)

  It also complies with the transitive property: if (u_n) is a subsuccession of (t_n) and (t_n) is a subsuccession of (s_n), which in turn is a subsuccession of (s_n)

• In C) we have seen that \left ( (-1)^n \right ) is not a convergent succession

  However, the subsuccession of its even terms converges to 1 and the subsuccession of its odd terms converges to -1

• The succession (x^n) for x \in [0, 1) converges to 0, since it is convergent and \underset {n \to \infty} {\lim} x^n = a, we see that \underset {n \to \infty} {\lim} x^{n+1} = a\cdot x

  Even though (x^{n+1}) is a subsuccession of (x^n) (the one that corresponds to \varphi(n) = n + 1 in the definition), then your limit will be \underset {n \to \infty} {\lim} x^{n + 1} = a, with a \cdot x = a and as x\neq 1 then a = 0

• The diagonal numbering of all rational numbers form a succession (s_n) which is not convergent

  But it has a surprising property, it has convergent subsucessions to any real number

  Given \alpha \in \mathbb{R}, we will build a subsuccession (s_{n_k}) such that \left \| s_{n_k} - \alpha \right \| < \frac{1}{k}, k\geq 1 and therefore convergent to \alpha

  To find the subsuccession we will proceed by induction over k

  We select n_1 such that \left \| s_{n_1} - \alpha \right \| < 1, this is possible since in the interval (\alpha - 1, \alpha + 1) there are infinite rational numbers

  Suppose we have already chosen n_1 < n_2 < \cdots < n_k such that \left \| s_{n_j} - \alpha \right \| < \frac{1}{j}, j = 1, 2, \cdots, k

  Since in an interval \left (\frac{\alpha - 1}{k + 1}, \frac{\alpha + 1}{k + 1} \right ) there are infinite rational numbers, we can choose n_{k + 1} > n_k such that s_{n_{k+1}} belongs to that interval and, therefore, \left \| s_{n_{k + 1}} - \alpha \right \| < \frac{1}{k + 1}

  Through this procedure we have constructed a subsuccession of (s_n) such that \underset {k \to \infty} {\lim} s_{n_k} = \alpha

An observation that can be useful in some circumstances:

If the subsuccessions (s_{2_n}) and (s_{n + 1}) are convergent to the same value, then the subsuccession (s_n) will be convergent and its limit will coincide with that of the subsuccessions

Theorem of Bolzano-Weierstrass

Every bounded sequence has a convergent subsuccession

Lemma of Theorem of Bolzano-Weierstrass

Every subsuccession has a monotonous subsuccession

Proof of the Lemma of Theorem of Bolzano–Weierstrass

We will call peak point of a succession (a_n), \forall n\in \mathbb{N}|a_m < a_n with m > n

We distinguish between the following cases:

• The sequence has infinite peak points

  If n_1 < n_2 < n_3 < \cdots are the infinite peak points of the succession, you have to a_1 > a_2 > a_3 > \cdots, so that (a_{n_k}) is a decreasing subsuccession, and therefore, it is the monotonous subsuccession we were looking for

• The succession has a finite set of summit points

  It n_1 higher than all peak points

  As n_1 it is not a peak point, \exists n_2 > n1\rightarrow a_{n_2}\geq a_{n_1}

  As n_2 is not a peak point (since it was greater than n_1, and therefore greater than all the peak points), \exists n_3 > n2\rightarrow a_{n_3}\geq a_{n_2}

  Continuing with this series we will be able to build (a_{a_k}) which is a non decreasing subsuccession, which is what we were looking for

Proof of Theorem of Bolzano-Weierstrass

It (s_n) a bounded succession

By the Lemma of Theoremof Bolzano-Weierstrass , it will possess a monotonous subsucession (s_{\varphi_(n)})

As the succession is bounded, the succession will also be bounded, and therefore also convergent. Thus being demonstrated

Cauchy successions

A succession (s_n)^\infty_{n=1} it is said to be from Cauchy if \forall \epsilon > 0, \exists N \in \mathbb{N}|n,m > N\Rightarrow \left \| s_n - s_m \right \| < \epsilon

If we take for example \epsilon=1 in the definition of Cauchy succession, we inferred that if \exists N \in \mathbb{N}|n > N\Rightarrow \left \| s_n - s_{N + 1} \right \| < 1

From this fact we can immediately deduce that a Cauchy succession is always bounded

Proposition

A succession is convergent if and only if it is Cauchy's

Demonstration of the proposition

Are s_n \to a \in \mathbb{R} and \epsilon > 0

By definition of limit \exists N \in \mathbb{N} | n > N\Rightarrow \left \| s_n - a \right \| < \frac{\epsilon }{2}

Therefore, if n, m > N \Rightarrow \left \| s_n - s_m \right \| = \left \| s_n - a + a - s_m \right \| \leq \left \| s_n - a \right \| + \left \| s_m - a \right \| < \frac{\epsilon }{2} + \frac{\epsilon }{2} = \epsilon

Consevered, either (s_n) a Cauchy succession, by the theorem of Bolzano–Weierstrass is bounded and assures us of the existence of a subsucession (s_{\varphi(n)}) convergent to a a \in \mathbb{R}

Given \forall \epsilon > 0, \exists N \in \mathbb{N}|n,m > N\Rightarrow \left \| s_n - s_m \right \| < \frac{\epsilon }{2}

In particular, we need to \left \| s_n - s_{\varphi(n)} \right \| > \frac{\epsilon }{2}

If we take the limit in m we have that if n > N\Rightarrow \left \| s_n - a \right \| \leqslant \frac{\epsilon }{2} < \epsilon

Thus it is shown that if the succession is Cauchy, it is also convergent