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'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!

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.

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

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).

• 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? – manan Jul 6 '16 at 3:23
• ** I meant $|s_n - s| > |s_n|$ above. – manan Jul 6 '16 at 3:33
• 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. – Ethan Alwaise Jul 6 '16 at 3:35
• Yeah, just saw that. Independently, does my proof regarding a sequence of nonpositive terms can't converge to a positive limit above make sense? – manan Jul 6 '16 at 3:36
• Instead of using $a_n - a \geq a_n$ for all $n$, use $a_n - a \geq a$ for all $n$. – Ethan Alwaise Jul 6 '16 at 3:39

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?)

• 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? – manan Jul 6 '16 at 3:06
• 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 $s_n\leq t_n$. – Eric Wofsey Jul 6 '16 at 3:09

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.