Solving: Showing Sequence Converges to a Single Point

[ak123456]

1. Homework Statement [/b]
Let Jn : n \inN be a sequence of intervals Jn=\left[an,bn\right] such that J1\supsetJ2\supset...\supsetJn\supsetJn+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\inJn for all n \inN

Homework Equations

The Attempt at a Solution

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

 

[jgens]

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].

 

[ak123456]

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]

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).