I started out saying to fix e > 0 and that there exists N such that for all n >= N, |xn - x| < e and |yn - y| < e. But I don't know how to go from there, i.e. how can I utilize the x < y part to continue with my proof? Thank you!
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For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – emonHR Oct 02 '21 at 19:43
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Hint: Since $x<y$, you know there is a $\delta>0$ such that $0<\delta<y-x$. – Clayton Oct 02 '21 at 19:44
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That misses the important information: For almost all $n$, so for all $n>N$ for some $N$. Any easy proof: Consider: If for any $N$ theres is a $n>N$ so that $x_n\geq y_n$ then you can find subsequences $x_{n_k} \geq y_{n_k}$. Also as $x_n$ and $y_n$ converge to $x,y$ these subsequences have the same limits.
Now, as the limit is monotonous this would mean that $x\geq y$.
If you want to continue your path: Find some $N$ so that for $n>N$ you have $|x_n-x|< (y-x)/2$ and $|y_n-y|< (y-x)/2$. Then $x_n < x+(y-x)/2 = y-(y-x)/2 < y_n$.
Lazy
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Oh yes I forgot to include that part though I had it written down for some reason. Yes, that makes much more sense, thank you dearly! – TreausreDragon Oct 02 '21 at 19:51