Show that if $x_n \le y_n \le z_n$ for all $n \in \mathbb N$, and if $lim(x_n) = lim (z_n) = l$, then $lim (y_n)= l$ as well.

Proof. Since $lim(x_n) = lim (z_n) = l$. Then follows that $|x_n -l| < \frac{\epsilon}{2}$ whenever $n \ge N_1$ and similarly $|l-z_n | < \frac{\epsilon}{2}$ whenever $n \ge N_2$. Choose $max \{ N_1,N_2\}$ and by the triangle inequality we get $$|x_n - z_n| \le|x_n -l|+ |l-z_n| \lt \epsilon .$$

since $$|x_n - z_n| \lt \epsilon $$ it follows that $x_n=z_n$ for sufficiently large $n \ge \max\{N_1, N_2\}$. Hence it follows that $x_n=y_n= z_n$ (from the fact that $x_n=z_n$ and $x_n \le y_n \le z_n$). And therefore $lim(x_n) = lim (y_n) = l$.

Question 1: Is my attempt at the proof correct? If not why not? Also if not can you give a correct version or atleast a few hints?

Question 2: I use the property that $|a-b|< \epsilon$ is equivalent to $a=b$ but it doesn't sit well with me below that $x_n=z_n$ since $(x_n), (z_n)$ could have different elements and still converge at the same value.) Why am I getting contradictory results here?

Thanks in advance.

I really appreciate this site helping me learn how to write proofs in real analysis :)


Firstly, $z_n$ does not equal $x_n$ for sufficiently large $n$. For an answer to your second question $\frac{1}{n}$ and $\frac{1}{2n}$ both converge to zero, but they are never equal for a given $n$. Use the definition of convergence instead. What do you need to show that $\lim_{n\to \infty} y_n=l$?

  • $\begingroup$ Yes that is correct $\endgroup$ – Red Jul 15 '18 at 2:55
  • $\begingroup$ I completely agree with you that $z_n$ does not necessarily equal to $x_n$ but as I have shown the inequality $|x_n - z_n| \lt \epsilon$ would imply that $x_n = z_n$ doesn't it?? $\endgroup$ – Red Jul 15 '18 at 2:59
  • $\begingroup$ @Red: why would the inequality imply $z_n=x_n$? $\endgroup$ – Paramanand Singh Jul 15 '18 at 6:06

Most elementary results of convergence for sequences either depend on the triangle inequality (via an $\epsilon/2$ trick) or the inequality $0\leq\frac{|x|}{1+|x|}\leq 1$ for all $x$ (to avoid cases where division by $0$ may occur).

The Squeeze Theorem however is probably best proved by a different approach.

Try to show that if $x\leq y\leq z$ then $|y-l|\leq \max\left\{|x-l|, |z-l|\right\}$ no matter what $l$ may be. This is probably best proved with a picture.

  • $\begingroup$ Is the aproach I have taken completely wrong or can it be tweaked for a correct proof? $\endgroup$ – Red Jul 15 '18 at 3:11
  • $\begingroup$ @Red you have claimed that the sequences $\{x_n\}$ and $\{z_n\}$ eventually agree. But an easy counter example is $x_n=-1/n$, $z_n=1/n$ and $y_n=0$ for all $n$. $\endgroup$ – Robert Wolfe Jul 15 '18 at 3:12
  • $\begingroup$ I completely agree with this, the sequences need not agree eventually.But why is the triangle inequality gving me that result anyway? I feel like I have used it wrong but I cannot figure out where my mistake llies? as I said in th second part of my question $\endgroup$ – Red Jul 15 '18 at 3:16
  • 1
    $\begingroup$ @Red if $a$ and $b$ satisfy $|a-b|<\epsilon$ for every $\epsilon>0$ then $a=b$. However you have not proved that $|x_n-z_n|<\epsilon$ for every $\epsilon>0$. You have proved that for every $\epsilon>0$ there is an $n$ such that $|x_n-z_n|<\epsilon$. $\endgroup$ – Robert Wolfe Jul 15 '18 at 3:27
  • $\begingroup$ I see my mistake. Thanks! $\endgroup$ – Red Jul 15 '18 at 3:28

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