Let S be as set of real numbers , and let ${x_n}$ be a sequence which converges to l. Suppose that for every $n inmathbb{N},x_n$ is an upper bound for S . prove l is an upper bound of S

And also Suppose ${x_n}$ is a sequence in S such that $x_n to 1$ and 1 is an upper bound of S. Show that $1 = text{lub}(S)$

*my idea*:

since {x_n} is converges then $|x_n-l|<epsilon,$ for each $n ge N$

Also $x_n $ is upper bound of S then $x<x_n, forall nin N$

but how to we processed from here

## Answer

Suppose otherwise that $1$ is not an upper bound of $S$, then $a > 1$ for some $a in S$. Thus set $epsilon = a – 1 > 0$, then there exists $N_0 in mathbb{N}$ we have if $n ge N_0 implies |x_n -1|< epsilon = a – 1implies x_n – 1 < a – 1implies x_n < a$. Thus $x_n$ is no longer an upper bound of $S$, contradiction. Thus $1$ is an upper bound of $S$. Note that I took $l = 1$ to make it easier for you to follow. You can adjust it to any $l$.

Let’s tackle your second question.

Suppose ${x_n}$ is a sequence in $S$ such that $x_n to 1$, and $1$ is an upper bound of $S$. Show that $1 = text{lub}(S)$.

For this one, we also do a contradiction proof as above. So suppose that $1 neq text{lub}(S)$, then $1 > text{lub}(S) = mimplies 1 – m > 0$. Again, let $epsilon = 1 – m > 0 implies |x_n – 1| < 1 – m implies 1 – x_n < 1 – m implies m < x_n implies m$ is no longer an upper bound for $S$. Contradiction, and thus $1 = text{lub}(S)$.