55 776
Assignments Done
97,2%
Successfully Done
In December 2017
Your physics homework can be a real challenge, and the due date can be really close — feel free to use our assistance and get the desired result.
Be sure that math assignments completed by our experts will be error-free and done according to your instructions specified in the submitted order form.
Our experts will gladly share their knowledge and help you with programming homework. Keep up with the world’s newest programming trends.

Answer on Real Analysis Question for Jim

Question #7517
Prove: A subset S of the real numbers is closed if and only if lim (as n goes to infinity) X_n is an element of S whenever (X_n) is a convergent sequence whose terms are all in S.
Expert's answer
By definition a subset S of R is closed if its complement R\S is open.

1)
Suppose S is closed, so R\S is open.
Let (x_n) be any sequence of elements of
S converging to some x.
We should show that x belogs to S.

Suppose x
belogs to R\S.
Since R\S is open, there exist an open interval U=(x-a, x+a)
that does not intersect S for some a>0.
On the other hand, the sequence
(x_n) converges to x, so there exists N>0 such that x_n belogs to U for all
n>N.
But then x_n with n>N belogs to R\S, which contradicts to the
assumptino that x_N belongs to S.

Hence x belongs to S.


2)
Conversely, suppose S is not closed, and so R\S is not open.
We will find a
sequence of points from S converging to some x belonging to R\S.

Notice
that the assumptino that R\S is not open means that there exists x from R\S such
that for any a>0 the
interval (x-a, x+a) contains a point y_a from
S.
Consider the sequence x_n = y_{1/n}.
Since 1/n converges to 0, it
follows that x_n converges to x.
Thus we obtained as sequence of points from
S converging to some x belonging to R\S.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question