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Prove that if $\left\{ x_n \right\}$ is an infinite sequence of real numbers, $x \in \mathbb{R}$, and every subsequence $\left\{ x_{n_k} \right\}$ has a subsequence $\left\{ x_{n_{k_j}} \right\}$ with $x_{n_{k_j}} \rightarrow x$, then $x_n \rightarrow x$.

I know that if every subsequence of a sequence converges to the same number, then the sequence converges to that same number. But I don't know if the same can be applied to subsubsequences. So for this problem, can I safely state that because $x_{n_{k_j}} \rightarrow x$, it is also true that $x_{n_k} \rightarrow x$? If this is true, then does that mean every subsequence $\left\{ x_{n_k} \right\}$ also converges to $x$?

mr eyeglasses
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  • http://math.stackexchange.com/q/1005363/, http://math.stackexchange.com/q/451764/, http://math.stackexchange.com/q/397978/ – Jonas Meyer Dec 08 '14 at 00:08
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    Your reasoning isn't correct, because of the difference between "every subsequence" and "has a subsequence". – Jonas Meyer Dec 08 '14 at 00:12

3 Answers3

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HINT: Suppose that $x_n\not\to x$, and show that $\langle x_n:n\in\Bbb N\rangle$ has a subsequence that is bounded away from $x$.

Brian M. Scott
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  • What does "bounded away" mean? – mr eyeglasses Dec 08 '14 at 00:23
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    @user130018: It means that there is an $\epsilon>0$ such that $|x_{n_k}-x|\ge\epsilon$ for all $k\in\Bbb N$, where $\langle x_{n_k}:k\in\Bbb N\rangle$ is the subsequence in question. The distance of the terms from $x$ is bounded away from $0$, meaning that it has a positive lower bound. – Brian M. Scott Dec 08 '14 at 00:24
  • @BrianM.Scott Can I ask, what special meaning has here $<>$? What's the different to the notation from question. Any that I know, doesn't make sense. – Tacet Dec 08 '14 at 00:36
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    @Silencer: I prefer angle brackets to parentheses or curly braces for ordered tuples and sequences; parentheses already have more than enough different uses, and curly braces already have a very different meaning. In particular, it’s important to distinguish the sequence $\langle x_n:n\in\Bbb N\rangle$ from the set ${x_n:n\in\Bbb N}$. This angle bracket notation is one of the standard notations, though it’s probably most common among those with a lot of set-theoretic background. – Brian M. Scott Dec 08 '14 at 00:41
  • @BrianM.Scott Thank you for your explanation. I see, I have to learn worldwide standards of notation. I have use for this purpose normal parentheses. I met angel brackets around sequences of natural numbers and ranges. Thank you again for answer. – Tacet Dec 08 '14 at 00:51
  • @Silencer: You’re very welcome. – Brian M. Scott Dec 08 '14 at 00:52
  • @BrianM.Scott So if $x_n \nrightarrow x$, then there exists an $\epsilon > 0$ such that for all $N \in \mathbb{N}$, there exists an $n \geq N$ such that $\lvert x_n - x \rvert \geq \epsilon$ (I'm not sure if I negated the implication correctly). Now I have to show there is a subsequence of $\langle x_n:n\in\Bbb N\rangle$ that does not converge, so can I say that because a sequence is a subsequence of itself, a diverging subsequence exists? And then also how do I obtain from the subsequence, a diverging subsubsequence? – mr eyeglasses Dec 08 '14 at 01:46
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    @user130018: Your first sentence is correct. it shows that if $M={n\in\Bbb N:|x_n-x|\ge\epsilon}$, then $M$ is infinite, so we can enumerate it as $M={m_k:k\in\Bbb N}$, with $m_k<m_{k+1}$ for each $k\in\Bbb N$. Then $\langle X_{m_k}:k\in\Bbb N\rangle$ is a subsequence of the original sequence, and it has the property that $|x_{m_k}-x|\ge\epsilon$ for all $k\in\Bbb N$. Now use your hypothesis to get a contradiction. – Brian M. Scott Dec 08 '14 at 01:49
  • @BrianM.Scott What does it mean that "$M$ is infinite?" That the set $M$ has an infinite number of elements? I thought there was always a countably infinite number of terms in a sequence, so there's an infinite number of subscripts that determines the position of each term? Also what does "enumerate" mean here? I don't see what you did to transform the set $M$ into a different form. – mr eyeglasses Dec 08 '14 at 01:56
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    @user130018: Yes, it means that the set $M$ has an infinite number of elements. What you’re missing, I think, is that while a sequence $\langle y_n:n\in\Bbb N\rangle$ has an infinite number of terms, some (or even all) of those terms could be the same number, so the set ${y_n:n\in\Bbb N}$ might be finite. Consider, for instance, the sequence $\langle (-1)^n:n\in\Bbb N\rangle$; it has a term for each natural number $n$, but the set ${(-1)^n:n\in\Bbb N}={-1,1}$ has only two elements. – Brian M. Scott Dec 08 '14 at 02:00
  • @BrianM.Scott Your explanation makes sense, but now what I don't understand is why the set is infinite. I am having a difficult time reading the meaning of $x_n \nrightarrow x$ that I stated, because reading it doesn't make logical sense to me. The part that says "for all natural numbers, there is a natural number greater than or equal to it" doesn't make sense to me because how can a natural number be greater than every single natural number? – mr eyeglasses Dec 08 '14 at 02:08
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    @user130018: That’s why I prefer to use the word each where you have all: for each natural number $N$ there is an $n_N\ge N$ such that ... . The numbers $n_N$ might all be different, or some of them might be the same, but there are certain to be infinitely many different ones. It definitely does not mean that there is a single $n$ that is greater than or equal to all natural numbers: that is indeed impossible. \ It’s often less confusing to read $\forall x$ as ‘for each $x$’ rather than as ‘for all $x$’. – Brian M. Scott Dec 08 '14 at 02:12
  • @BrianM.Scott How come you write $n_N \geq N$ (as opposed to $n \geq N$)? I don't understand the "$n_N$" part, i.e. why did you put $N$ as a subscript? I thought $n$ was just a natural number, so how can a natural number have a subscript? – mr eyeglasses Dec 08 '14 at 02:17
  • @user130018: I subscripted it to show that it depends on $N$: $n_0,n_1,n_2,\ldots$ are all natural numbers, but there’s no reason to think that any of them are the same natural number. – Brian M. Scott Dec 08 '14 at 02:18
  • @BrianM.Scott I don't see why $n$ depends on $N$. $N$ denotes one of the terms of the sequence, e.g. if the sequence is $1, 2, 3, 4$, etc. and we start at $n=1$, then $N=2$ would denote the the second term in the sequence, which here is $2$. So $n \geq 2$ would mean the second term onwards. So then what would $n_2 \geq 2$ mean for this example? i.e. what is the "meaning" of $n_2$ here and how does it depend on the second position of this sequence? – mr eyeglasses Dec 08 '14 at 02:24
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    @user130018: No, $N$ does not denote a term of the sequence. $N$ denotes an arbitrary natural number. The fact that the sequence does not converge tells us that there is an $\epsilon>0$ such that no matter what natural number $N$ you pick. there is an index $n\ge N$ such that the term $x_n$ is at least $\epsilon$ units away from $x$. If you look at a different natural number, say $M$, you will very likely need to pick a different index — say $m$ — to get one that is $\ge M$ and such that the term $x_m$ is at least $\epsilon$ units away from $x$. Instead of running through the alphabet, ... – Brian M. Scott Dec 08 '14 at 02:33
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    ... I simply let $n_N$ be the index of a term satisfying the two conditions, (a) that the index is $\ge N$, and (b) that the term $x_{n_N}$ is at least $\epsilon$ units away from $x$. Change the natural number $N$, and you’ll generally have to change the index $n_N$ of a term that meets these conditions. – Brian M. Scott Dec 08 '14 at 02:34
  • @BrianM.Scott Okay, $x_n \nrightarrow x$ means that there exists an $\epsilon > 0$ such that for each $k \in \mathbb{N}$, there exists an $n_k \geq k$ such that $\lvert x_{n_k} - x \rvert \geq \epsilon$. Does this mean we've found a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ that diverges? If we want to contradict the hypothesis, then we have to find a subsequence that doesn't have any converging subsubsequences, but it is possible for a subsequence to diverge and have at least one converging subsubsequence, so how can we show a contradiction here? – mr eyeglasses Dec 08 '14 at 02:59
  • @user130018: You’re making it too hard. You have a subsequence bounded away from $x$. It doesn’t matter whether this subsequence diverges or converges to something different from $x$. Apply your hypothesis to it: it has a sub-subsequence that converges to $x$. Why is this impossible? – Brian M. Scott Dec 08 '14 at 03:01
  • @BrianM.Scott I'm not sure why. I am just guessing by process of elimination that it has something to do with $x_n \nrightarrow x$, but I don't see why (but perhaps that's because it has nothing to do with it?). – mr eyeglasses Dec 08 '14 at 03:16
  • @user130018: Every term of the subsequence is at least $\epsilon$ away from $x$; how can it possibly have a sub-subsequence that converges to $x$? That would mean that it had terms less than $\epsilon$ away from $x$. – Brian M. Scott Dec 08 '14 at 03:17
  • @BrianM.Scott It's not intuitive to me why you can't have terms of the sub-subsequence less than $\epsilon$ away from $x$, even though the parent subsequence has each term at least $\epsilon$ away from $x$. I know that in general a subsequence has some sort of dependence on the parent sequence, but the exact relationships between them are not always clear to me. – mr eyeglasses Dec 08 '14 at 03:20
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    @user130018: Every term of the sub-subsequence is a term of the subsequence, so if every term of the subsequence is at least $\epsilon$ units away from $x$, then automatically every term of the sub-subsequence is at least $\epsilon$ units away from $x$. \ When you form a subsequence of a given sequence, you’re just throwing away any number of terms of the original sequence. The terms that remain, taken in their original order, are the subsequence. If $x_n=\frac1{2^n}$ for $n\in\Bbb N$, then the subsequence $\langle x_{2n}:n\in\Bbb N\rangle$ is $\langle x_0,x_2,x_4,\ldots\rangle$, ... – Brian M. Scott Dec 08 '14 at 03:26
  • ... which is $$\left\langle\frac1{2^0},\frac1{2^2},\frac1{2^4},\ldots\right\rangle =\left\langle 1,\frac14,\frac1{16},\ldots\right\rangle;.$$ – Brian M. Scott Dec 08 '14 at 03:27
  • @BrianM.Scott Another thing that confuses me a bit is that $x_{n_k}$ is the general notation for subsequence, but in your particular example you wrote it $x_{2n}$ (which is something I see often in textbooks), but this is not in the form of double subscripts like $x_{n_k}$. Is there an equivalent way to write $x_{2n}$ in the form of $x_{n_k}$? – mr eyeglasses Dec 08 '14 at 03:31
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    @user130018: Sure: just define $n_k=2k$ for each $k\in\Bbb N$. – Brian M. Scott Dec 08 '14 at 03:39
  • @BrianM.Scott Since we supposing every subsequence converges, is it enough to say that since a sequence is itself a subsequence of itself, then the sequence converges? – user5826 Mar 12 '16 at 05:46
  • @AlJebr: No: the hypothesis says that the original subsequence has a subsequence that converges to $x$, simply because the original sequence is a subsequence of itself, but that alone does not say that the original sequence converges to $x$. – Brian M. Scott Mar 12 '16 at 05:52
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If $x_n$ doesn't converge to $x$, no tail of it converges. But then we can construct a subsequence of elements that are at least $\varepsilon$ from $x$, but that subsequence has to have a subsequence that converges to $x$. By contradiction $x_n$ converges to $x$.

Henrik supports the community
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Suppose that $\{X_n\}$ does not converge to $x$. Then, there is $\varepsilon_0>0$ such that $$\forall N\in\mathbb N,\exists \hspace{.2cm}n=n(N) : n>N~~~and ~~~ |X_n -x|>\varepsilon_0 $$

For $N_1=1$ there exists $n_1$ such that $$n_1>N_1 ~~~and ~~~ |X_{n_1} -x|>\varepsilon_0 $$ Taking successively $N_{k+1}> \max\{N_k, n_k,k+1\}$ there exists $n_{k+1>N_{k+1}}$ such that,

$$ |X_{ n_{k+1}} -x|>\varepsilon_0 $$

It is easy to see that, $\{X_{ n_k}\}_k$ is a subsequence of $\{X_{ n}\}_n$ since $$ n_k< n_{k+1} \quad i.e ~~\text{the map }~~k\mapsto n_k~~~\text{Is one-to-one}$$

However, $$\forall k,~~ |X_{ n_{k}} -x|>\varepsilon_0 \qquad \text{and}~~~\{X_{ n_{k}} \}~~~\text{is bounded} $$

Therefore By Bolzano-Weierstrass Theorem's there exists $\{X_{ n_{k_p} }\}_p$ subsequence of $\{X_{ n_{k} }\}_k$ which converges to some limit $\ell_1 $ but $\{X_{ n_{k_p} }\}_p\to \ell_1$ is also a congering subsequence of $\{X_n\}_n$

By assumption, $\ell=\ell_1$ that is together with the fact $\{X_{ n_{k_p} }\}_p$ is a subsequence of $\{X_{ n_{k} }\}_k$ we have

$$0=\lim_{p\to\infty } |X_{ n_{k_p} }-\ell|>\varepsilon_0>0~~~\text{CONTRADICTION}$$

Note that $$\forall p,~~|X_{ n_{k_p} }-\ell|>\varepsilon_0~~~ Since ~~~\forall k,~~|X_{ n_{k}} -\ell|>\varepsilon_0$$

Guy Fsone
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