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

Solution 1:

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.

Solution 2:

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).

Solution 3:

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.

Solution 4:

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?)