25 Subsequences and the Bolzano-Weierstrass Theorem Practice

#todo/school : 2.5.1, 2.5.2, 2.5.5, 2.5.6 📅 2024-10-24 ✅ 2024-10-27

2.5.1

Question

Give an example of each of the following, or argue that such a request is impossible.
a. A sequence that has a subsequence that is bounded but contains no subsequence that converges.
b. A sequence that does not contain 0 or 1 as a term but contains sub-sequences converging to each of these values.
c. A sequence that contains subsequences converging to every point in the infinite set ${1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \dots}$
d. A sequence that contains subsequences converging to every point in the infinite set ${1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \dots}$ and no subsequences converging to points outside of this set.

Proof

a. Impossible. The Balzano-Weierstrass Theorem says that any bounded sequence has a convergent subsequence. If the sequence we found $(a_{n})$ has a subsequence that is bounded, then it must also have it's own subsequence that is convergent. This is itself a subsequence of $(a_{n})$ that converges, barring the second condition.

b. Define the sequence:

(a_{n}) = {\begin{cases} \frac{1}{n} & x is even \\ 1 + \frac{1}{n} & x is odd \end{cases}

Here clearly $(a_{2 n}) \to 0$ and $(a_{2 n + 1}) \to 1$ are convergent subsequences but none of these value are 0,1.

c. The construction is that we want some subsequence like:

(a_{m, n}) = \frac{1}{m} \cdot \frac{n}{n} = \frac{n}{n m}

here if $m$ is constant then as $n \to \infty$ then $(a_{m, n}) \to \frac{1}{m}$ as desired. The idea is that we could plot this out onto a table:

$m$ on row, $n$ on column	1	2	3	4	...
1	1/1	2/2	3/3	4/4	...
2	1/2	2/4	3/6	4/8	...
3	1/3	2/6	3/9	4/12	...
4	1/4	2/8	3/12	4/16	...
...	...	...	...	...	...
Then just make $(a_{n})$ be the sequence of entries going sweeping around the diagonals:

(a_{n}) = {\frac{1}{1}, \frac{2}{2}, \frac{1}{2}, \frac{1}{3}, \frac{2}{4}, \frac{3}{3}, \dots}

Then choosing the subsequence of each row gives the limit we wanted.

d. Impossible. Say such a subsequence exists. Then a subsequence exists that must converge to 0 (take terms that approach 1, then 1/2, then 1/3, ..., towards 0) which is not in the set.

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2.5.2

Question

Decide whether the following propositions are true or false, providing a short justification for each conclusion.
a. If every proper subsequence of $(x_{n})$ converges, then $(x_{n})$ converges as well.
b. If $(x_{n})$ contains a divergent subsequence, then $(x_{n})$ diverges
c. If $(x_{n})$ is bounded and diverges, then there exist two subsequence of $(x_{n})$ that converge to different limits.
d. If $(x_{n})$ is monotone and contains a convergent subsequence, then $(x_{n})$ converges.

Proof
a. True. See that $(a_{n + 1})$ must converge. We showed in 24 Monotone Sequence, MCT, and Cauchy Condensation Test#2.4.1 (b) that these must both converge and to the same limit, so clearly $(a_{n})$ must converge.

b. True. By the contrapositive of the Balzano-Weierstrass Theorem if $(x_{n})$ contains a divergent subsequence, then the original sequence $(x_{n})$ must be unbounded. As a result then $(x_{n})$ cannot be convergent (see Bounded For Sequences and the contrapositive of the theorem).

c. True. Since $(x_{n})$ is bounded and diverges, then any $| x_{i} | < M$ for the bound $M > 0$ . Then notice from 24 Monotone Sequence, MCT, and Cauchy Condensation Test#2.4.7 then we can consider the subsequences $inf x_{n}$ and $sup x_{n}$ whose limits exist. Notice that since $x_{n}$ diverges then these limits cannot equal each other.

d. True. Suppose WLOG that $(x_{n})$ is increasing and that $(x_{n_{k}})$ is our convergent subsequence. Then the subsequence is bounded: $| x_{n_{k}} | < M$ for some $M > 0$ . Then we must have it that $| x_{n} | < M$ eventually since we can pick our $n \leq n_{k}$ where $x_{n} \leq x_{n_{k}} < M$ . Since ( $x_{n}$ ) is increasing and bounded then it must be convergent.
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2.5.3

Question

a. Prove that if an infinite series converges, then the associative property holds. Namely, if some $\sum a_{n}$ converges to limit $L$ , then regrouping the terms:

(a_{1} + a_{2} + \dots + a_{n_{1}}) + (a_{n_{1} + 1} + \dots + a_{n_{2}}) + (a_{n_{2} + 1} + \dots + a_{n_{3}}) + \dots

leads to a series that also converges to $L$ .
b. Compare this result to the alternating harmonic series where infinite addition was shown not to be associative. Why doesn't our proof in (a) apply to this example?

We won't prove this, but it's a good exercise left to the reader (HAHAHAHAHAHA).

2.5.5

Question

Suppose $(a_{n})$ is a bounded sequence with the property that every convergent subsequence of $(a_{n})$ converges to the same limit $a \in R$ . Show that $(a_{n})$ must converge to $a$ .

Proof

Assume for contradiction that $(a_{n}) ↛ a$ . Then $\exists ϵ > 0$ where for any $N \in N$ then there is some $n > N$ where:

| a_{n} - a | \geq ϵ

Now construct the subsequence $b_{n}$ for each $N \in N$ we choose the first $n > N$ (guaranteed by the finding above) such that the inequality holds (each $b_{k}$ is associated with some $a_{n_{k}}$ ). This is clearly a subsequence of $(a_{n})$ so by our given above, then there is some $N_{0} \in N$ where for all $n > N_{0}$ then:

| b_{n} - a | < ϵ \leq | a_{n} - a | \Rightarrow | b_{n} | < | a_{n} |

But since $a_{n}$ is bounded then $| a_{n} | < M$ for some $M > 0$ . That implies $| b_{n} | < M$ so then $b_{n}$ is bounded. But look, since $n > N_{0}$ then we must also have that:

| b_{n} - a | \geq ϵ

which is a contradiction.

2.5.6

Question

Use a similar strategy to the one in 25 Subsequences and the Bolzano-Weierstrass Theorem Practice#2.5.3 to show that $lim_{n \to \infty} b^{\frac{1}{n}}$ exists for all $b \geq 0$ and find the value of the limit. (The results in Exercise 23 Algebraic and Order Limit Theorems Practice#2.3.1 (Finish) may be assumed).

Proof

We'll want to show that $b^{\frac{1}{n}} = \sqrt[n]{b}$ is monotonic and bounded to use the MCT. Notice first that if $b = 0$ then it clearly converges to the value $0$ so assume $b \neq 0$ , so then $b > 0$ . We'll consider the cases of $0 < b < 1$ and $b > 1$ separately:

( $0 < b < 1$ ): Notice that $\sqrt[n]{b} \leq 1$ (bounded above). This is because $b < 1^{n} ⟺ \sqrt[n]{b} < 1$ . Now for increasing notice that:

\begin{aligned} \sqrt[n + 1]{b} \geq \sqrt[n]{b} & ⟺ b \geq b^{2} & Raise to n + 1 power \\ ⟺ 1 \geq b \end{aligned}

Thus we know this is true, so then our sequence increases, showing that $\sqrt[n]{b}$ converges.

( $b > 1$ ): Notice that $\sqrt[n]{b} \geq 1$ (bounded below) because $b > 1^{n} \Rightarrow \sqrt[n]{b} > 1$ . Now for decreasing the same argument above works.

Consider the subsequence $b_{n}$ given as:

b^{\frac{1}{2 n}} = \sqrt{b^{\frac{1}{n}}}

Since $b^{\frac{1}{n}}$ converges then $b^{\frac{1}{2 n}}$ converges and to the same limit $L$ , so then:

\begin{aligned} lim_{n \to \infty} b^{\frac{1}{2 n}} & = lim_{n \to \infty} \sqrt{b^{\frac{1}{n}}} \\ L & = \sqrt{L} \\ L^{2} - L & = 0 \\ L (L - 1) & = 0 \\ L & = 0, 1 \end{aligned}

We know it can't be 0 since of our bound so then $L = 1$ for $b \neq 0$ and $L = 0$ for the $b = 0$ case.
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