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The formulation I'm looking at goes: If $\lbrace x_n\rbrace$, $\lbrace y_n\rbrace$ and $\lbrace z_n \rbrace$ are sequences such that $x_n \le y_n \le z_n$ for all $n \in \mathbb N$, and $x_n \to l$ and $z_n \to l$ for some $l \in \mathbb R$, then $y_n \to l$ also.

So we have to use the definition of convergence to a limit for a sequence: $$\forall \varepsilon > 0, \space \exists N_\varepsilon \in \mathbb N, \space \forall n \ge N_\varepsilon, \space |a_n - l| < \varepsilon$$

I've been trying to say something like: $|y_n - l| < |x_n - l| + |z_n - l| \le \frac\varepsilon 2 + \frac\varepsilon 2 = \varepsilon$ for every $\varepsilon > 0$, but I'm not sure how to get there or if there may be a better way to prove the theorem. Any help would be greatly appreciated.

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6 Answers 6

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Let $\varepsilon > 0$. Since $x_n \to l$, there exists $N_1 = N_1(\varepsilon)$ such that $|x_n - l| < \varepsilon$ for all $n \ge N_1$. Since $z_n \to l$, there exists $N_2 = N_2(\varepsilon)$ such that $|z_n - l| < \varepsilon$ for all $n \ge N_2$. Set $N = \max\{N_1,N_2\}$. If $n \ge N$, then $$y_n - l \le z_n - l < \varepsilon$$ and $$y_n - l \ge x_n - l > -\varepsilon$$ Hence $|y_n - l| < \varepsilon$ for all $n \ge N$. Since $\varepsilon$ was arbitrary, $y_n \to l$.

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    $\begingroup$ Clear and concise. +1. $\endgroup$
    – Mark Viola
    Sep 22, 2016 at 2:01
  • $\begingroup$ How do you know $-\epsilon \lt x_n - l$? $\endgroup$ Oct 22, 2023 at 20:25
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    $\begingroup$ @user129393192 if $|x_n - l| < \epsilon$, then $-\epsilon < x_n - l < \epsilon$. In particular, $x_n - l > -\epsilon$. $\endgroup$
    – kobe
    Oct 22, 2023 at 20:49
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Write

$$|y_n-l|\le |y_n-x_n|+|x_n-l|\le( z_n-x_n)+|x_n-l|$$ Can you take it from here?

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    $\begingroup$ @silva: And as another tiny hint, there's a metric space axiom you use over and over again that you will want to use here. $\endgroup$
    – nomen
    Feb 5, 2015 at 19:43
  • $\begingroup$ @nomen Yeah, I've been taking that inequality over and over. The trick is just using that to get to that final statement I mentioned. $\endgroup$ Feb 5, 2015 at 19:46
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Ok, so this is what I have now:

Let $x_n \le y_n \le z_n \space\space \forall n \in \mathbb N$ and $\lim_{n \to \infty} x_n = \lim_{n \to \infty} z_n = l$ for some $l \in \mathbb R$.

Then we have: $$\begin{align} &\forall \varepsilon_1 > 0,\, \exists N_{\varepsilon_1} \in \mathbb N,\, \forall n \ge N_{\varepsilon_1}\space |x_n - l| < \varepsilon_1 \\ &\forall \varepsilon_2 > 0,\, \exists N_{\varepsilon_2} \in \mathbb N,\, \forall n \ge N_{\varepsilon_2}\space |z_n -l| < \varepsilon_2 \end{align}$$ Let $N = \max\lbrace N_{\varepsilon_1},N_{\varepsilon_2} \rbrace$ and $\varepsilon = \min\lbrace \varepsilon_1, \varepsilon_2 \rbrace$, so that we have $$\forall \varepsilon > 0,\, \exists N \in \mathbb N,\, \forall n \ge N\space |x_n - l| < \varepsilon \space \text{and} \space |z_n -l| < \varepsilon$$

Let $\frac\varepsilon3 > 0$. Then $\exists N \in \mathbb N,\, \forall n \ge N\space |x_n -l| < \frac\varepsilon3 \text{ and } |z_n -l| < \frac\varepsilon3$ so that $$|z_n - x_n| = |z_n - l + l - x_n| \le |z_n - l| + |x_n - l| < \frac\varepsilon3 + \frac\varepsilon3 = \frac{2\varepsilon}3$$ By assumption, $$\begin{align} x_n \le&\, y_n \le z_n \\ 0 \le&\, y_n - x_n \le z_n - x_n \\ &|y_n - x_n| \le |z_n - x_n| \\ |y_n - l| = |y_n - x_n + x_n - l| \le& |y_n - x_n| + |x_n - l| \le |z_n - x_n| + |x_n - l| < \frac{2\varepsilon}3 + \frac\varepsilon3 = \varepsilon \end{align} $$ And so $\lbrace y_n \rbrace$ also converges to $l$, thus, Q.E.D.

I think that that's satisfactory. If there are any ambiguities or mistakes, please let me know. Thanks to everybody for your advice and contributions.

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  • $\begingroup$ Can you just assume that, since both $x_{n}$ and $z_{n}$ converge to the same limit, then for $\frac{\epsilon}{2}$ there is an $N$ such that $|z_{n}-x_{n}|<\frac{\epsilon}{2}$, and from here just going on with $|y_{n}-l|=|y_{n}-x_{n}+x_{n}-l|\leq|y_{n}-x_{n}|+|x_{n}-l|\leq|z_{n}-x_{n}|+|x_{n}-l|\leq\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$? (Here I took $\frac{epsilon}{2}$ as a boundary for $|x_{n}-l|$) $\endgroup$
    – Allonsy
    Dec 13, 2017 at 18:01
  • $\begingroup$ I think you need $\epsilon = \max\{\epsilon_1, \epsilon_2 \}$. Currently written, if it's the min, the inequalities that follow aren't necessarily true. $\endgroup$ Sep 22, 2022 at 14:41
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How about $$ |y_n-l|\leq\max\{|x_n-l|,|z_n-l|\}<\varepsilon $$ where last inequality of the above holds for $n\geq N$ if $N$ is big enough?

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An alternative proof is the following:

Proof: Since $x_n \leq y_n \leq z_n$ then $0\leq y_n-x_n\leq z_n-x_n$, thus $|y_n-x_n|\leq z_n-x_n$.

Combining the above with the fact that $\lim(z_n-x_n)=\lim z_n-\lim x_n=l-l=0 \ $, we get: $$\lim(y_n-x_n)=0$$

Now we can write the terms of $(y_n)$ as the sum of the terms of two converging sequences: $y_n=(y_n-x_n)+x_n$, so we have: $$ \lim y_n=\lim\big((y_n-x_n)+x_n \big)=\lim(y_n-x_n)+\lim x_n=0+l=l $$

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    $\begingroup$ You seem to be using the squeeze theorem in your proof of the squeeze theorem. How do you arrive at that limit y_n-x_n = 0 ?? $\endgroup$ Feb 4, 2018 at 0:12
  • $\begingroup$ @CogitoErgoCogitoSum: I am using the property that: "if $\lim \beta_n=0$ and there is some $\kappa\in\mathbb{N}$, such that for all $n>\kappa$, $\ |\alpha_n|\leq |\beta_n|$, then $\lim\alpha_n=0$". This property is an immediate consequence of the $\epsilon$-$\delta$ definition of the limit of a sequence and it is generally not referred to as the "squeeze theorem". (although, it can obviously be understood as a special case of the squeeze theorem). $\endgroup$
    – KonKan
    Feb 10, 2018 at 2:24
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As $n$ grows, you can get the distance from $x_n$ to $l$ to be less than any $\epsilon > 0$, and the distance from $z_n$ to $l$ to be less than $\epsilon$ as well. Just take the max $N$ of the two indices for $x_n$ and $z_n$ that guarantee this, and you will get that both $x_n$ and $z_n$ are within $\epsilon$ of $l$ for $n \geq N$. What does that say about the distance between $y_n$ and $l$, for $n \geq N$?

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