Proof Verification - Every sequence in $\Bbb R$ contains a monotone sub-sequence

Question

Came across the following exercise in Bartle's Elements of Real Analysis. This is the solution I came up with. Would be grateful if someone could verify it for me and maybe suggest better/alternate solutions.

I also looked up these related questions - (1), (2), (3) - but was not happy with proofs given there. I seem to need some help understanding these. Any such help is appreciated.

Show that every sequence in $\Bbb R$ either has a monotone increasing sub-sequence or a monotone decreasing sub-sequence.

Let $(x_n)$ be a sequence in $\Bbb R$. Suppose $(x_n)$ is not bounded. Without loss of generality we may assume that $(x_n)$ is not bounded above. Therefore given any real number there is a member of the sequence which is greater. Let $x_{n_1}$ be any member of the sequence.

There is $x_{n_2} \gt \sup\{x_1, x_2, ..., x_{n_1} \}$. For $i \gt 1$ let $x_{n_i} = \{x_1, x_2, ..., x_{n_{i - 1}}\}$ then $(x_{n_k})$ forms a monotone subsequence of $(x_n)$.

Now suppose instead that $(x_n)$ is bounded. By the Bolzano-Weierstrass Theorem there is a subsequence $(y_n)$ of $(x_n)$ which converges to a limit $y$. Without loss of generality there are infinitely many distinct values in $(y_n)$ that are unequal to $y$. Let $y_{k1}$ be the first such element. Let $y_{k2}$ be any element in $\{ y' \in (y_n) \ \ | \ \ |y' - y | \lt |y - y_{k1}| \}$.

For $i \gt 1$ let $y_{ki} \in \{ y' \in (y_n) \ \ | \ \ |y' - y | \lt |y - y_{k \ i - 1}| \}$. Such $y_{ki}$ exists for every $i \in \Bbb N$ since $ \lim (y_n) = y $. Now let $(y_{kn})$ be the sub-sequence of $(y_n)$ thus formed. At least one of the two following sets must contain infinitely many elements.

• $\{ y \in (y_{kn}) \ \ | \ \ y \gt x\}$
• $\{ y \in (y_{kn}) \ \ | \ \ y \lt x\}$

The one which does forms a monotone subsequence.

Answer 1

This is a very classic argument, I think. Let $n\in \mathbb{N}$ be called "nice" if $a_n >a_m$ for all $m> n$. So we have only two possibilities:

(1) The sequence contains infinitely many "nice" points. If $n_1<n_2<\ldots<n_i< \ldots$ are the "nice" points, we have $a_{n_1}>a_{n_2}> \ldots>a_{n_i}> \ldots$, so $(a_{n_i})$ is a decreasing subsequence.

(2) The sequence contains finitely many "nice" points. Let $n_1$ be greater than all the "nice" points. Since $n_1$ is not "nice" there is some $n_2>n_1$ such that $a_{n_2}\ge a_{n_1}$, and we continue in this fashion to get a non-decreasing subsequence $(a_{n_i})$.

More formally: Let $N$ be a natural number which is greater than all the "nice" points. We define $n_1:= N$ and $n_{i+1}:=\min\{n> n_{i}: a_n\ge a_{n_{i}}\}$. Hence $(a_{n_i})$ is non-decreasing.

Answer 2

I like to think of this in terms of Ramsey theory. We are coloring the edges of the complete graph on $\mathbb N$, using two colors, and want to ensure that there is a complete infinite subgraph that is monochromatic.

The case that concerns us is the coloring where, for $i<j$, the edge $\{i,j\}$ is colored red if $x_i\le x_j$, and blue otherwise. An infinite monochromatic subgraph gives us the indices of a monotone subsequence: If red, the subsequence is increasing while, if blue, it is strictly decreasing.

Start by noting that there is an infinite $A_0$ with all edges $\{0,i\}$, $i\in A_0$, of the same color. Let $i_0=0$ and $i_1=\min(A_0)$. There is an infinite $A_1\subset A_0$ with all edges $\{i_1,i\}$, $i\in A_1$, of the same color. Let $i_2=\min(A_1)$, and continue this way.

We get $i_0<i_1<\dots$ with the property that, for any $n$, all pairs $\{i_n,i_m\}$ with $m>n$, have the same color. Call it $c_n$. Now, the sequence $c_0,c_1,c_2,\dots$ is a sequence that only takes two values, so it has a constant subsequence. The corresponding $i_n$ form the monochromatic set we were looking for.

Answer 3

I think the other proofs are superior, but this is a sketch of the method I came up with when trying to solve it on my own:

If $a_n$ is unbounded, we are done.

Suppose $a_n$ is bounded and let $R_N = \{a_n| n \ge N \}$ (i.e. the range of $a_n$ after $N$). Since $R_N$ is nonempty and bounded, it has a supremum. Note that for every $N$, $\sup R_{N+1} \le \sup R_N$. Thus if $\forall N \sup R_N$ is a member of the sequence, then we can construct a decreasing subsequence.

Now to finish up, suppose that for some $N$, $x = \sup R_N$ is not a member of the sequence. From the definition of supremum, we know that $\forall \epsilon >0$ there exists an element of the sequence, $a_n$, with $x- \epsilon <a_n$, and we can build an increasing subsequence based on that.

Answer 4

Here is my version of the proof, which I came up with as an alternative to the much better argument which is peak points:

Case 1: Unbounded

Suppose that $(x_n)$ is unbounded, we have that: $\forall \ \epsilon \in \mathbb{R} \ \exists \ r \in \mathbb{N} \ \text{such that} \ x_r > \epsilon$

choose $\epsilon_1=\epsilon_f$, we have that: $\exists \ r_1 \in \mathbb{N} \ \text{such that} \ x_{r_1} > \epsilon_f$

Consider: $S_1(n)=\{ x_n \}\backslash\{x_1,x_2,......,x_{r_1}\}$, This is also unbounded, for if it were bounded, the union of $S_1(n)$ with $\{x_1,x_2.....,x_{r_1}\}$ will give me a bounded set, but this contradicts the initial statement that $(x_n)$ is unbounded. Now for this set, choose $ \epsilon_2=x_{r_1}$. Then we have, $\exists \ r_2 \in \mathbb{N} \ \text{such that} \ x_{r_2} > x_{r_1}>\epsilon_f $ and $r_2>r_1$

Consider: $\{ x_n \}\backslash\{x_1,x_2,......,x_{r_1},....,x_{r_2}\}$, This is also unbounded. Now for this set, choose $ \epsilon_3=x_{r_2}$. Then we have, $\exists \ r_3 \in \mathbb{N} \ \text{such that} \ x_{r_3} > x_{r_2}> x_{r_1} > \epsilon_f $ and $r_3>r_2>r_1$. As such, this can be carried on forever to construct a monotone increasing sequence of $x_{r_i}$

Case 2: Bounded

Suppose that $(x_n)$ is bounded. Start by defining the set $S_k =\{ x_n : k \leq n \}$. $S_1=\{x_1,x_2,......\}$, $S_2=\{x_2,x_3,......\}$. We can see that $S_{k+1} \subseteq S_{k}\subseteq....\subseteq S_1$. These sets $S_i$ are all bounded, and hence they all have supremums, which we can call $Sup(S_{i})=U_{i}$

Subcase 1: finitely many i $\in \mathbb{N}$ such that $U_i \in S_i$

Suppose the sets $\{ S_{r_1},S_{r_2}....S_{r_p}\}$ contain their own supremum, and the remaining $S_i$ dont contain their supremums. This means that, for $i>r_p$, $U_i=U_{i+1}=U_{i+2}....=U$ to see this, consider $S_i=\{ x_i,x_{i+1},x_{i+2}.....\}$ and $S_{i+1}=\{x_{i+1},x_{i+2}.....\}$ and since none of $x_{j} : j\geq i$ is the supremum of $S_i$, it is clear that none of $x_{j} : j\geq i+1$ is the supremum of $S_{i+1}$

In this case, consider the (infinite) bounded set $S_{r_p+1}=\{x_{r_p+1},x_{r_p+2},....\}$. The supremum of this set is $U$. Hence, $\exists s_{1} \in S_{r_p+1} \ \text{such that} \ U-\frac{1}{n_1}<s_{1}$

Moreover, $\forall j \in \mathbb{N},\exists s(j) \in S_{r_p+1} \ \text{such that} \ U-\frac{1}{j}<s(j)$

assertion:

There exists $n_2 > n_1$ such that $s_{1}<U-\frac{1}{n_2}$

proof:

Suppose not, i.e, for a particuar $j$, no matter what $i \in \mathbb{N}$ we use, $s(j) \geq U-\frac{1}{i}$

$\iff \ \forall i \in \mathbb{N} \ s(j)+\frac{1}{i} \geq U$

$\iff \ \lim_{i \to \infty} (s(j)+\frac{1}{i} )=s(j)\geq U$ Which is absurd, hence, assumption is false.

Therefore, for all $j \ \in \mathbb{N} \ \exists i > j : i \in \mathbb{N} \ \text{and} \ s(j) \in S_{r_p+1} \ \text{such that} \ U-\frac{1}{j} < s(j) <U-\frac{1}{i}$

Choose j=1, then:

$\exists i_1 > 1 : i_1 \in \mathbb{N} \ \text{and} \ s_{1} \in S_{r_p+1} \ \text{such that} \ U-1 < s_1 <U-\frac{1}{i_1}$

Choose $j=i_1$ , then: $\exists i_2 > i_1 : i_2 \in \mathbb{N} \ \text{and} \ s_{2} \in S_{r_p+1} \ \text{such that} \ U-1 < s_1 <U-\frac{1}{i_1} < s_2 <U-\frac{1}{i_2}$

and continue this way, with $j=i_k$ to get $s_{k+1}$, hence forming an infinite sequence of $s_i$ which is monotone increasing.

Subcase 2: infinitely many $i \in \mathbb{N}$ such that $U_{i} \in S_i$

We have that, $\{S_{k_1},S_{k_2},....,S_{k_i}....\}$ for all $i \in \mathbb{N}$ is an infinite set of $S_i$ such that $U_i \in S_i$

notice then, that, for any $l$ and $r$ $\in \{k_1,k_2.....\}$, $U_l \geq U_r $ if $r>l$. Choose $U_{k+i}=s_{k_i}$, and see that $s_{k_i} \geq s_{k_{i+1}}$, for all $i \in \mathbb{N}$. This gives us a monotonic decreasing sequence of $s_i$

Answer 5

For any converging real sequence, there is a monotone sub-sequence.

Proof:

Let $(x_n)_{n}\subset\mathbb{R}$ with $x_n\rightarrow x$. We assume that $x_n\neq x$ for all $n\in\mathbb{N}$.

• As $x$ is a limit point of the sequence, either $$\{y:y > x\}\cap (x_n)_{n} \text{ or } \{y:y < x\}\cap (x_n)_{n}$$ contain infinitely many points.

• Assume wlog $X:=\{y:y > x\}\cap (x_n)_{n\in\mathbb{N}}$ contains infinitely many points. For any $\epsilon > 0$, $$ X \cap [x,x+\epsilon] $$ contains infinitely many points because the sequence converges.

• We can define our sequence $(y_n)_{n}\subset (x_n)_n$ recursively. Choose any $x_k\in X$ and set $y_1 := x_k$. For any $k > 1$ choose $$ y_{k+1} \in X\cap [x,y_{k}]. $$ The sequence will be well-defined and decreasing, which is easy to check. As $(y_n)_n\subset X\subset (x_n)_n$, we have constructed the sequence we were looking for.

For any real sequence, there is a monotone subsequence.

Proof: Let $(x_n)_n$ be some real sequence. There are two cases for $(x_n)_n$.

• $(x_n)_n$ is unbounded. In this case we are done, as mentioned in another answer.

• $(x_n)_n$ is bounded by some $K\in\mathbb{R}$. Then $(x_n)_n\cap [-K,K]$ is compact and contains the sequence. By Bolzano-Weierstrass, $(x_n)_n$ contains a converging subsequence. Then by the above proof we are done as well.