2
$\begingroup$

I'm stuck with the proof of the following:

Suppose that $(s_n)$ converges to $s$, $(t_n)$ converges to $t$, and $s_n \leq t_n \: \forall \: n$. Prove that $s \leq t$.

I've tried starting with $s_n \leq t_n \: \forall : n$ and the definitions of each limit (i.e. $|s_n - s| \leq \epsilon \: \forall \: n > N_1$), but I'm not really getting very far. Any help is appreciated!

$\endgroup$
2
$\begingroup$

Suppose $s>t$ write $c=s-t$ there exists $N$ such that $n>N$ implies $|s_n-s|<c/4$. There exists $N'$ such that $n>N'$ implies $|t_n-t|<c/4$, take $n>\sup(N,N')$ $s_n-t_n=s_n-s+s-t+t-t_n\geq s-t-|s_n-s|-|t-t_n|\geq c-c/4-c/4\geq c/2$ contradiction.

$\endgroup$
  • $\begingroup$ This makes sense, thanks! How did you know to define $c = s-t$? $\endgroup$ – manan Jul 6 '16 at 3:05
  • $\begingroup$ it is my experience, ... keep working, you'll easily solve this type of questions. $\endgroup$ – Tsemo Aristide Jul 6 '16 at 3:06
2
$\begingroup$

Since $\{s_n\}$ converges to $s$ and $\{t_n\}$ converges to $t$, $\{t_n - s_n\}$ converges to $t - s$. Since $s_n \leq t_n$ for all $n$, each term $t_n - s_n$ is nonnegative. It thus suffices to show that a sequence of nonnegative terms cannot converge to a negative limit (use proof by contradiction).

$\endgroup$
  • $\begingroup$ Thanks for the reply! I've been trying to prove the second half (a sequence of nonnegative terms can't converge to a positive limit), and I've made some progress. Let $S = \{ i \in \mathbb R : i \leq 0 \}$ and let $\lim S = s > 0$. Let $N > 0$ and $\epsilon > 0$. Then for $n > N$, $|s_n - s| < \epsilon$. Since $s_n \leq 0$, $s_n - s < s_n$. So, $|s_n - s| > s_n \: \forall \: n > N$, a contradiction as we assumed $|s_n - s| < \epsilon$. Does this make sense / what could I improve? $\endgroup$ – manan Jul 6 '16 at 3:23
  • $\begingroup$ ** I meant $|s_n - s| > |s_n|$ above. $\endgroup$ – manan Jul 6 '16 at 3:33
  • $\begingroup$ I misspoke, a sequence of nonnegative terms cannot converge to a negative limit. So assume you have a sequence $\{a_n\}$ where each $a_i \geq 0$ and assume it converges to a limit $a < 0$ and try to get a contradiction. $\endgroup$ – Ethan Alwaise Jul 6 '16 at 3:35
  • $\begingroup$ Yeah, just saw that. Independently, does my proof regarding a sequence of nonpositive terms can't converge to a positive limit above make sense? $\endgroup$ – manan Jul 6 '16 at 3:36
  • $\begingroup$ Instead of using $a_n - a \geq a_n$ for all $n$, use $a_n - a \geq a$ for all $n$. $\endgroup$ – Ethan Alwaise Jul 6 '16 at 3:39
1
$\begingroup$

Hint: Suppose $s>t$ for a contradiction. Then intuitively, for $n$ large, $s_n$ is very close to $s$ and $t_n$ is very close to $t$, so $s_n$ would have to be greater than $t_n$. Can you find an $\epsilon$ so that if you knew $s_n$ were within $\epsilon$ of $s$ and $t_n$ were within $\epsilon$ of $t$, then $s_n$ would be greater than $t_n$? (If you have trouble doing this, you might consider a concrete example: suppose $s=1$ and $t=0$. Then what does $\epsilon$ need to be?)

$\endgroup$
  • $\begingroup$ So in the concrete case, you'd need $\epsilon = 1$, right? Also, did you mean "suppose $s > t$ for a contradiction" in your first sentence? $\endgroup$ – manan Jul 6 '16 at 3:06
  • $\begingroup$ Oops, yes, corrected on that second point. In the concrete case, you can't quite take $\epsilon=1$. For instance, you could have $s_n=t_n=1/2$ and then $s_n$ would be within $1$ of $s$ and $t_n$ would be within $1$ of $t$, but $s_n\leq t_n$. $\endgroup$ – Eric Wofsey Jul 6 '16 at 3:09
1
$\begingroup$

Here is a proof without contradiction: for every $\varepsilon>0$, there exists $N$ large enough so that $s-\varepsilon/2\leqslant s_{n}\leqslant s+\varepsilon/2$ and $t-\varepsilon/2\leqslant t_{n}\leqslant t+\varepsilon/2$ for all $n>N$.

Therefore, from the hypothesis that $s_{n}\leqslant t_{n}$, $$ s-\varepsilon/2\leqslant s_{n}\leqslant t_{n}\leqslant t+\varepsilon/2, $$ from which it can be concluded that $$ s\leqslant t+\varepsilon. $$ Since $\varepsilon>0$ is arbitrary, the proof is done.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.