if a bounded sequence xn in [a,b] converges to x0 then x0 belongs to [a,b]

Question

If a bounded sequence xn in [a,b] converges to x0 then prove that x0 belongs to [a,b]. I don't understand here how can x0 belong to [a,b], it may be slightly greater than b too. Please help.

Answer 1

A convergent sequence is surely bounded. Since $a\le x_n\le b$, permanence of sign tells you that $$ a\le\lim_{n\to\infty}x_n\le b $$

If you want a detailed proof, assume by contradiction that $x_0<a$. Then there exists $N$ such that, for $n>N$, $$ |x_n-x_0|<\frac{a-x_0}{2} $$ which implies $$ x_n<x_0+\frac{a-x_0}{2}=\frac{a+x_0}{2}<a $$ Contradiction.

Similarly for proving that $x_0\le b$.

Answer 2

The sequence is convergent and thus bounded, so there exists $M$ and $g$ such that

$$ |a_{n}|<M\qquad\text{and}\qquad\lim_{n\rightarrow\infty}a_{n}=g. $$

This holds because we can find $M$ such that $[a,b]\in[-M,M].$ Then

$$ |g|=|\lim_{n\rightarrow\infty}a_{n}|=\lim_{n\rightarrow\infty}|a_{n}|<\lim_{n\rightarrow\infty}M=M $$ So $$ g\in[-M,M]\qquad\text{and}\qquad{a_{n}}\in[-M,M], $$ as desired.