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The question is: Show that if $x_n \leq y_n \leq z_n$ for all $n \in \mathbb N$, and if $lim x_n = limz_n = l$, then $limy_n =l$ as well.

So far for my solution I have: Using the definition of convergence, we know $|x_n - l| < \epsilon$ for some $\epsilon > 0$, and $|z_n - l| < \epsilon $ for some $\epsilon > 0$.

Using this, we have $|x_n - l| - |z_n - l| < \epsilon - \epsilon$ which goes to $|x_n - l - z_n + l| < 0$, thus $|x_n - z_n| < 0$.

We know absolute value cant be less than 0, but I'm not sure where I could incorporate a $\leq$ sign so that the final inequality becomes $|x_n - z_n| \leq 0$, proving that $x_n = z_n$, therefor $x_n = y_n = z_n$, so $limy_n = l$ as well. Can I do this, or am I going in the wrong direction completely? Thanks!

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    $\begingroup$ u shouldn't be able to prove $x_n = z_n$ because it's not always true. There's a mistake here: $|x_n - l | < \epsilon$, $|z_n - l| < \epsilon$ doesn't mean $ |x_n - l | - |z_n - l| < 0$, you can't "substract" same-sign inequalities, only add them. $\endgroup$ – Dominik Kutek Oct 26 '19 at 18:54
  • $\begingroup$ If $a < \epsilon$ and $b < \epsilon$ does not mean that $a -b < \epsilon - \epsilon$. $\endgroup$ – copper.hat Oct 26 '19 at 18:55
  • $\begingroup$ Furthermore the step where you merge the two absolute values into one is also incorrect; this isn't always true. Example: $|(-3)-0| - |6-0| = -3 \neq |(-3)-0-6+0|$ $\endgroup$ – WaveX Oct 26 '19 at 19:05
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No that's not a proper way, indeed for example

$$(3<4) \quad \land \quad (2<4) \quad \not\Rightarrow \quad (3-2)<0$$

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That is wrong, for several reasons:

  • Asserting that $\lim_{x\to\infty}x_n=\lim_{n\to\infty}z_n=l$ does not meant that $\lvert x_n-l\rvert,\lvert z_n-l\rvert<\varepsilon$ for some $\varepsilon>0$; it means that you have that inequality for every $\varepsilon>0$, if $n$ is large enough.
  • There is no way you can prove that $\lvert x_n-l-z_n+l\rvert<0$; an absolute value is always greater than or equal to $0$.

You can prove it as follows. Take $\varepsilon>0$. If $n$ is large enough, then $\lvert x_n-l\rvert,\lvert z_n-l\rvert<\varepsilon$, which means that $x_n,z_n\in(l-\varepsilon,l+\varepsilon)$. But then, since $x_n\leqslant y_n\leqslant z_n$, $y_n\in(l-\varepsilon,l+\varepsilon)$, which means that $\lvert y_n-l\rvert<\varepsilon$ (if $n$ is large enough).

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Jose has explained the two errors in your attempt. Here is a hint to get you on the right track.

Fix $\epsilon > 0$. Your goal is to show that for all large $n$ you have $|y_n - l| < \epsilon$.

The key observation is $|y_n - l| \le \max\{|x_n - l|, |z_n - l|\}$. (For instance, if $y_n \le l$ then $|y_n - l| \le |x_n - l|$.) Now try to show that the right-hand side is smaller than $\epsilon$ for all large $n$, using the fact that $\lim_{n \to \infty} x_n = \lim_{n \to \infty}z_n = l$.

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Using the definition of convergence, we know $|x_n−l|<ϵ$ for some $ϵ>0$.

That is, sadly, not the definition for convergence and is actually a pretty serious botch.

If $x_n= \frac 1{2^n}$ then what is $\lim x_n$. Well by your definition, $0 < x_n \le 1$ so $-39 < x_n - 39 \le -38$ so $|x_n-39| < 39$. So for $\epsilon = 39$ we have $|x_n -39| < \epsilon$ so $\lim x_n = 39$.

The actual definition is that for any $\epsilon > 0$ we can find some $N$ so that whenever $n > N$ we will have $|x_n -l| < \epsilon$.

So instead of saying:

"Using the definition of convergence, we know |xn−l|<ϵ for some ϵ>0, and |zn−l|<ϵ for some ϵ>0"

we should say:

Using the definition of convergence, we know that for any $\epsilon > 0$ there are $N_1$ and $N_2$ so that for all $n > N_1$ we'd have $|x_n−l|<ϵ$, and for all $n > N_2$ we'd have $ |z_n−l|<ϵ $. ANd for all $n > \max(N_1, N_2)$ we'd have both $|x_n - l|< \epsilon$ and $|z_n -l|< \epsilon$.

Now you claim that $|x_n-l|< e$ and $|z_n-l| < e$ implies $|x_n-1|-|z_n-l| < e-e$. This is utterly wrong. Negatives flip the inequalities so $-|z_n - l| > -\epsilon$. And if you have $|x_n-1|$ is less then $\epsilon$. And $-|z_n-l| > -\epsilon$ you can't say any thing about how they combine.

For example $4 < 5$ and $1 < 4$ so $4- 1 < 5-4$. Really?

But what you can say is $|x_n -l| + |z_n-l| < \epsilon + \epsilon$.

The you claim that $|x_n-l| - |z_n -l| = |(x_n -l)-(z_n-1)|$. You can not combine absolute values that way. Consider $|5|-|-3| = 5 - 3 =2$ and $|(5)-(-3)| = |5+3| = 8$.

What you need to do is use addition and the triangle inequality: $|a-b| + |b-c| \ge |a-c|$.

so $|(x_n-l)+ (l- z_n)| \le |x_n-l|+|z_n - 1| < \epsilon + \epsilon = 2\epsilon$.

How do we put this all together?

........

We want to find an $\mathscr N$ so that for all $n >\mathscr N$ we have $|y_n - l|\epsilon$.

And we know we can talk about $|z_n - l|$ and $|y_n - l|$.

So $|y_n - l|= |(y_n - x_n) +(x_n -l)| \le |(y_n-x_n)| + |x_n-l| \le |(z_n - x_n)| + |x_n-l|=$

$|(z_n-l) + (l-x_n)| + |x_n-l| \le |z_n-l| + |x_n-l| + |x_n-l|$.

Now we can "trap " the $z_n, x_n$ close to $l$.

Now $\frac \epsilon 3 > 0$ so there are $N_1$, and $N_2$ so that if $n> \max(N_1, N_2)$the we have $|z_n -l| < \frac \epsilon 3$ and $|x_n -l|<\frac \epsilon 3$.

SO $|y_n -l| \le |z_n-l| + |x_n-l| + |x_n-l|<3\frac \epsilon 3 = \epsilon$.

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Let $\epsilon>0$ As $x_n\rightarrow l$ then you can find $N_1 \in \mathbb{N}$ such that whenever $n\geq N_1$ we have $|x_n-l|<\epsilon$ Similarly, $\exists N_2 \in \mathbb{N}$ such that whenever $n\geq N_2$ then $|z_n-l|< \epsilon$.

Therefore, since $x_n\leq y_n \leq z_n$ for each $n\in \mathbb{N}$ it follows that for $N=$ $max(N_1,N_2)$ we have $N\geq N_1$ and $N\geq N_2$. Therefore

$-\epsilon+l$ $\leq$ $x_n$ $\leq y_n$ $\leq$ $z_n$ $\leq$ $\epsilon+ l$ hence $|y_n -l| \leq$ $\epsilon$. As $\epsilon>0$ is arbitrary, $y_n \rightarrow l$.

Further note if $a<K$ and $b<K$ it is not necessarily the case that $a-b<0$. Consider the following counterexample $5<10$, $1<10$ but $4<0$ is clearly false. Further note that showing $x_n = z_n$ is not necessarily possible given the conditions of the hypothesis.

Further, "$|x_n-l|<\epsilon$ for some $\epsilon>0$ " does not follow from the definition. You need to be precise.

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