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.
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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.
Alex D
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A convergent (hence bounded) sequence need not converge to either its least upper bound or its greatest lower bound. – egreg Oct 29 '18 at 00:19
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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 $x_0\le b$.
egreg
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