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There is couple of things that i don't understand in the proof that Every bounded sequence has a convergent subsequence

here is the textbook proof (first part)

proof: let $(s_n)$ be a sequence whose range $T = \{s_n:n \in N\}$ is bounded. Suppose first that T is finite.Then there is some number x in T that is equal to $s_n$ for infinitely many values of n.That is there exist indices $n_1 < n_2 < ..... < n_k < .....$ such that $s_{n_k}=x$ for all $k \in N$.It follows that the subsequence $(s_{n_k})$ converges to x

so i don't understand what does it mean when the proof says there is some number x on T that is equal to $s_n$ for inifintely many values of n ? does it mean that there is a maximum ? which is x ?

and also why is there must exists indices in that way and why did he conclude that snk = x for all k belongs to N ?

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Just a guess. I think "Then there is some number x in T that is equal to $s_n$ for infinitely many values of n" is the same thing as "Then there is some number m in N s.t. $s_m$ = $s_n$ for infinitely many values of n and $s_m$ is in T"

More guessing: If T is finite then that means the sequence is constant after some index like -1,5,7,8,8,8,8,... or oscillates between a finite number of numbers like -1,1,-1,1,-1,1,... or 5,6,7,8,-4,1,2,4,1,2,4,1,2,4,.... If I am not mistaken those sequences have finite range.

For the first sequence the x or $s_m$ is 8. The indices in question are 4, 5, 6, 7, .... So if that sequence was $a_n$, the subsequence $a_4, a_5, a_6, a_7, ...$ converges to 8

Second: x = -1 or 1. The indices for x = -1 are odd natural numbers. So if that sequence was $a_n$, the subsequence $a_1, a_3, a_5, a_7, ...$ converges to -1. The indices for x = 1 are even natural numbers (except zero, if zero is a natural number). So if that sequence was $a_n$, the subsequence $a_2, a_4, a_6, a_8, ...$ converges to 1.

third: x = 1, 2 or 4.

I'll leave this as an exercise :))

Anyway, that's my attempt at the explanation. I am not quite sure how to prove that the x exists though. Maybe you can try to assume it does not exist and the range is finite and then arrive at a contradiction, or assume it does not exist and then show that the range is infinite.

Likely, it was proven in an earlier page. Worst case scenario it is left as an exercise.

My attempt on that:

Assume x does not exist. Then, for each value in the range, there exists only finitely many indices s.t. the corresponding numbers in the sequence equal it (that's the opposite of the definition of x, right?). Then the sequence is finite, which contradicts the definition of a sequence which is infinite.

Example: Range is {1,2,4} and x does not exist => the sequence could look something like 1,2,4,4,1,2,4 since there are only finitely many indices for each number in the range (2 for 1, 2 for 2 and 3 for 4). If the sequence goes on forever, then x exists.

Edit: Again, these are just guesses and are clearly not very precise. I am hoping that someone can state these precisely if they are right or correct them if they are wrong.

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Yes. If $T$ is finite then as a subset of the reals,it has a maximum. Now because $s_n$ is infinite and for infinitely many terms it is equal to $x$ then you can choose the first term of $s_n$ that is equal to $x$ ,and let's say we find it at $n_{1}$ position of $s_n$. Then we can find the next term at $n_{2}>n_1$ position and keep going infinitely many times because $s_n=x$ infinitely many times.

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Suppose for all $x\in T, s_n=x$ for only finitely many $n\in \mathbb{N}$. Let $T_x:=\{s_n:s_n=x\}$. Since $T$ is finite, so is $\bigcup_{x\in T}T_x=\{s_n:n\in \mathbb{N}\}$. This is a contradiction.

Now choose $x$ such that $T_x=\{s_n:s_n=x\}$ is infinite. You can define indices inductively. Let $n_0 :=\min\{n\in \mathbb{N}:s_n\in T_x\}$. Now suppose we have defined $n_0,\cdots,n_k$. Then define $n_{k+1}:=\min\{n\in \mathbb{N}\setminus\{n_0,\cdots,n_k\}:s_n\in T_x \}$. From the construction it follows that $n_k<n_{k+1}$ for all $k\in \mathbb{N}$ and $(s_{n_k})$ is a sequence with $s_{n_k}=x$ for all $k\in \mathbb{N}$.

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