# Let S be as set of real numbers , and let ${x_n}$ be a sequence which converges to l

Let S be as set of real numbers , and let $\{x_n\}$ be a sequence which converges to l. Suppose that for every $n \in\mathbb{N},x_n$ 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 $x<x_n, \forall n\in N$

but how to we processed from here

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 $n \ge N_0 \implies |x_n -1|< \epsilon = a - 1\implies x_n - 1 < a - 1\implies x_n < a$. 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 $l$.

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 $1 > \text{lub}(S) = m\implies 1 - m > 0$. 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 $1 = \text{lub}(S)$.
• @Deepsea..Let $\{x_n\}$ be a sequene in a set S .suppose that $x_n \to l$ and l is an upper bound for S. how to prove that l=lub(S) – Inverse Problem Oct 8 '17 at 4:02