# 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 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)\$.