# Sequence converges

1. Homework Statement [/b]
Let Jn : n $$\in$$N be a sequence of intervals Jn=$$\left[$$an,bn$$\right]$$ such that J1$$\supset$$J2$$\supset$$...$$\supset$$Jn$$\supset$$Jn+1$$\supset$$...
suppose also that the sequence xn=an-bn converges to 0 as n tends to infinite.Show that there is exactly one point a such that a$$\in$$Jn for all n $$\in$$N

## The Attempt at a Solution

i don't know how to start it , any clue??

jgens
Gold Member
Well, I'm dreadfully awful at sequence questions, so take my feedback with a grain of salt. Since the sequence $x_n = b_n - a_n$ converges to zero for arbitrarily large $n$, this means that $\mathrm{inf}(b_n) = \mathrm{sup}(a_n) = c$. Can you prove that this number $c$ must always be an element of $[a_n,b_n]$.

Well, I'm dreadfully awful at sequence questions, so take my feedback with a grain of salt. Since the sequence $x_n = b_n - a_n$ converges to zero for arbitrarily large $n$, this means that $\mathrm{inf}(b_n) = \mathrm{sup}(a_n) = c$. Can you prove that this number $c$ must always be an element of $[a_n,b_n]$.

still confusing

jgens
Gold Member
The number $c$ would have the property that $a_n \leq c \leq b_n$ for all natural numbers $n$. What does this tell you about $c$ and its relationship to the interval $[a_n,b_n]$?

Again, I'm awful at these types of proofs, so if another member says something otherwise, I would follow their feedback (I'm just trying making sure that you actually have some feedback).